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		<title>Masonry Columns, Piers, Pilasters</title>
		<link>http://howtoengineer.com/masonry-column-pier-pilaster/</link>
		<comments>http://howtoengineer.com/masonry-column-pier-pilaster/#comments</comments>
		<pubDate>Thu, 14 Mar 2013 13:00:27 +0000</pubDate>
		<dc:creator>Ryan Freund</dc:creator>
				<category><![CDATA[Columns Pilasters Piers]]></category>
		<category><![CDATA[Masonry]]></category>
		<category><![CDATA[Masonry Columns]]></category>
		<category><![CDATA[Masonry Piers]]></category>
		<category><![CDATA[Pilasters]]></category>
		<category><![CDATA[Reinforced Masonry]]></category>

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		<description><![CDATA[<p><p><a href="http://howtoengineer.com">How To Engineer - Engineers In Training</a></p><p>Masonry Columns, Piers and Pilasters How to analyze and design reinforced masonry columns, piers and pilasters and clarifying effective spacing. For analysis of masonry members utilizing flanges and compression reinforcement see Masonry &#8211; Compression Reinforcement and Effective Flanges References Ref 1: MSJC &#8211; 2005, TMS 402 (AKA ACI 530) &#8211; 05 (2008 version is very similar). Found here Ref 2: Masonry Structures Behavior and Design 2nd Edition by Robert Drysdale, Hamid, and Baker.  (the Third edition found here) Ref 3: Reinforced Masonry Engineering Handbook Clay and Concrete Masonry 6th Edition, Max L Porter (James E. Amrhein original author). Found here Definitions Based on MSJC 05 (all should be similar for 08) Column (MSJC-05 Section 1.6) – An isolated vertical member whose horizontal dimension measured at right angles to its thickness does not exceed 3 times its thickness and whose height is greater than 4 times its thickness. Pier (MSJC-05 Section 1.6) - An isolated vertical member whose horizontal dimension measured at right angles to its thickness is not less than 3 times its thickness nor greater than 6 times its thickness and whose height is less than 5 times its length. In reference to MSJC these are typically part of wall frames (see Ref 3 page 250) and is why there dimensions are [...]</p></p><p>The post <a href="http://howtoengineer.com/masonry-column-pier-pilaster/">Masonry Columns, Piers, Pilasters</a> appeared first on <a href="http://howtoengineer.com">How To Engineer</a>.</p>]]></description>
				<content:encoded><![CDATA[<p><a href="http://howtoengineer.com">How To Engineer - Engineers In Training</a></p><h1>Masonry Columns, Piers and Pilasters</h1>
<p>How to analyze and design reinforced masonry columns, piers and pilasters and clarifying effective spacing.</p>
<p>For analysis of masonry members utilizing flanges and compression reinforcement see <a title="Masonry Compression Reinforcement and Effective Flanges" href="http://wp.me/p2Puq4-co" target="_blank">Masonry &#8211; Compression Reinforcement and Effective Flanges</a></p>
<h3>References</h3>
<p>Ref 1: MSJC &#8211; 2005, TMS 402 (AKA ACI 530) &#8211; 05 (2008 version is very similar). Found <a title="The Masonry Society Bookstore - MSJC" href="https://www.2paysecure.com/tms/cart/scripts/prodView.asp?idproduct=130" target="_blank">here</a></p>
<p>Ref 2: Masonry Structures Behavior and Design 2nd Edition by Robert Drysdale, Hamid, and Baker.  (the Third edition found <a title="The Masonry Society Bookstore Design and Behavior" href="https://www.2paysecure.com/tms/cart/scripts/prodView.asp?idproduct=139" target="_blank">here</a>)</p>
<p>Ref 3: Reinforced Masonry Engineering Handbook Clay and Concrete Masonry 6th Edition, Max L Porter (James E. Amrhein original author). Found <a title="The Masonry Society Bookstore - Handbook" href="https://www.2paysecure.com/tms/cart/scripts/prodView.asp?idproduct=158" target="_blank">here</a></p>
<h2>Definitions</h2>
<p>Based on MSJC 05 (all should be similar for 08)</p>
<h4>Column</h4>
<p>(MSJC-05 Section 1.6) – An isolated vertical member whose horizontal dimension measured at right angles to its thickness does not exceed 3 times its thickness and whose height is greater than 4 times its thickness.</p>
<h4>Pier</h4>
<p>(MSJC-05 Section 1.6) - An isolated vertical member whose horizontal dimension measured at right angles to its thickness is not less than 3 times its thickness nor greater than 6 times its thickness and whose height is less than 5 times its length. In reference to MSJC these are typically part of wall frames (see Ref 3 page 250) and is why there dimensions are defined as such. This however may be confusing as many times a pier is referred to a column which is built integrally with a wall (see pg 403 of Ref 2). Having said that I believe if the column is built integrally with the wall it would be considered a &#8220;flush pilaster&#8221; (see Ref 3).</p>
<h4>Pilaster</h4>
<p>-  While Ref 1 does not give an explicit definition of a pilaster, it does infer that a pilaster is built integrally with the wall. Ref 2 states that a pilaster is a column that is built integrally with a wall and interacts with the wall to resist an out-of-plane lateral load, it is called a pilaster. Pilasters maybe &#8216;flush&#8217; (in the plane of the wall) or project out from the wall in one or both directions.</p>
<h4>Pedestal</h4>
<p>– Upright compression member with a ratio of unsupported height to average least lateral dimension not exceeding 3. This term is typically found in concrete design and</p>
<p>refers to short foundation elements. Many may refer to pedestals as piers or foundation piers.</p>
<h4>Foundation Piers</h4>
<p>- (Ref 1) An isolated vertical foundation member whose horizontal dimension measured at right angles to its thickness does not exceed 3 times its thickness and whose height is equal to or less than 4 times its thickness.</p>
<h4>Wall</h4>
<p>- (Ref 1) A vertical element with a horizontal length to thickness ratio greater than 3, used to enclose space.</p>
<h4>Loadbearing Wall</h4>
<p>- (Ref 1) a Wall carrying vertical loads greater than 200 plf in addition to its own weight.</p>
<h4>Discussion</h4>
<p>As you can see these elements may be referred to one another as they all are used to resist axial and flexure. I would need to check the 2011 version of the MSJC but it should also be noted that Piers are found in the Strength Design of Masonry only and Pilaster are only found in the ASD method. However lets look at how the analysis and design differs for each. I think some main points to consider are &#8211; Is the element isolated or part of wall construction? If the element is isolated it would generally be considered a column or pier. If it is part of a wall system then it would generally be considered a pilaster or flush pilaster. The main difference then between these is that for column/pier elements there is &#8211; added vulnerability due to isolation, typically compression reinforcement and potential for bi-axial bending. Another key to remember is that if the reinforcement is going to be considered to be effective in compression, the enforcement must be tied (confined) for any of the above mentioned elements.</p>
<p>Now lets look at how each classification effects the design.</p>
<h2>Design</h2>
<h3>Dimensional Constraints / Effective Widths</h3>
<p>This can be hard to keep track of as well. How much of the wall should be taken as effective to resist axial and flexural loads? What are the limits on effective flange widths? It is interesting to note that Ref 3 pg 177 limits the effective flange for pilasters to 3x the wall thickness each side of the pilaster however allows for 6x wall thickness for effective flange width on flanged masonry shear walls.</p>
<p>Effective Width</p>
<ol>
<li><span style="font-size: 13px; line-height: 19px;">Walls &#8211; (ref 1 Sect. 2.3.3.3) </span>
<ol>
<li><span style="font-size: 13px; line-height: 19px;">Out-of-plane Forces &#8211; Running Bond Effective width shall be the lesser of:</span>
<ol>
<li><span style="font-size: 13px; line-height: 19px;">Center &#8211; Center spacing of bars</span></li>
<li><span style="font-size: 13px; line-height: 19px;">6 x wall thickness</span></li>
<li><span style="font-size: 13px; line-height: 19px;">72 inches</span></li>
<li>H/6 in the case of a flanged shear wall. Where H is the total wall height (between lateral supports transferring load to the load to the shear wall).</li>
</ol>
</li>
<li><span style="font-size: 13px; line-height: 19px;">Axial Concentrated Load &#8211; Effective width</span>
<ol>
<li>4 x wall thickness + bearing width or length of wall</li>
</ol>
</li>
<li>Design for Axial and Flexure
<ol>
<li>The reinforcement is rarely tied and typically only resists tension forces.</li>
<li>For ASD if the load is concentrated then the 4x the wall thickness limit would apply otherwise the limits presented in point 1 would apply and the reinforcement would need to be tied if it going to be relied upon to carry compression forces.</li>
</ol>
</li>
<li>In-Plane Forces &#8211; Flanged Shear Walls
<ol>
<li>ASD &#8211; The flange is limited to 6x wall thickness each side of the web (Ref 3 pg 199)</li>
<li>Wall intersections shall meet Ref 1 &#8211; Section 1.9.4.1</li>
</ol>
</li>
</ol>
</li>
<li><span style="font-size: 13px; line-height: 19px;">Flush Pilasters  </span>
<ol>
<li>Effective Wdith
<ol>
<li>Tpcially taken as 4 x wall thickness (Ref 3 page 178)</li>
</ol>
</li>
</ol>
</li>
<li>Pilasters
<ol>
<li>Effective width
<ol>
<li>Ref 1 and 2 &#8211; 6 x Wall thickness each side of web</li>
<li>Ref 3 &#8211; 3 x Wall thickness each side of web</li>
</ol>
</li>
<li>Consideration for flanges to be effective (see Ref 1 &#8211; 1.9.4.2).
<ol>
<li>Flanges  (wall intersections) must be capable to transfer the required shear stress</li>
<li>Atleast 50% of the masonry units at the interface shall interlock; or Steel connectors grouted into the wall (1/4&#8243;x1.5&#8243; x28&#8243; inculding 2&#8243; long 90 deg bend at each end to form a U or Z shape; or Intersecting bond beams shall be provided in intersecting walls at a maximum spacing of 48&#8243; oc (As = 0.1 in^2 and shall be developed on each side of intersection)</li>
</ol>
</li>
</ol>
</li>
<li>Piers
<ol>
<li>Limits on Width
<ol>
<li>The width should be greater than 3 x thickness but less than 6 times its thickness
<ol>
<li>I believe if it were less than 3 x thickness it would then be a column and if it were greater than 6 it would then be a wall.</li>
</ol>
</li>
</ol>
</li>
<li>Design limitations &#8211; Strength Design
<ol>
<li>Max factored axial load <img src="//s0.wp.com/latex.php?latex=0.3A_nf%27_m+&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="0.3A_nf&#039;_m " title="0.3A_nf&#039;_m " class="latex" /> (per Ref 1 3.3.4.3.1)</li>
<li>Max effective height should not exceed 25 x nominal thickness unless the pier is designed according to provisions of Ref 1 Section 3.3.5 (walls, must consider P-Delta effects)</li>
<li>Nominal thickness should not exceed 16&#8243;</li>
<li>Provide (1) bar in the end cells and minimum area of longitudinal reinforcement should be 0.0007bd</li>
<li>Uniformly distribute reinforcement</li>
</ol>
</li>
</ol>
</li>
<li>Column
<ol>
<li>Limit on Width (Ref 1 &#8211; 2.1.6)
<ol>
<li>Horizontal dimension measured at right angles to its thickness does not exceed 3 times its thickness. Minimum side dimension should be 8&#8243;</li>
</ol>
</li>
<li>Limits on Height
<ol>
<li>Height is greater than 4 times its thickness</li>
<li>Ratio of Effective Height / least nominal dimension &lt; 25 (Ref 1 &#8211; 2.1.6.2)</li>
</ol>
</li>
<li>ASD Requirements (Ref 1 Section 2.1.6
<ol>
<li>Min Side Dimension &#8211; 8&#8243;</li>
<li>Height / least nominal dimension &lt;= 25</li>
<li>Min Moment should be 0.1 x each side dimension</li>
<li>Min longitudinal reinforcement <img src="//s0.wp.com/latex.php?latex=%3D+0.0025A_n&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="= 0.0025A_n" title="= 0.0025A_n" class="latex" /> and shall at least 4 bars</li>
<li>Max long reinf <img src="//s0.wp.com/latex.php?latex=%3D+0.04A_n&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="= 0.04A_n" title="= 0.04A_n" class="latex" /></li>
<li>Compression reinforcement shall met lateral tie requirements (as always)</li>
</ol>
</li>
<li>SD (strength design) Requirements (Ref 1 Section 3.3.4.4)
<ol>
<li>Same as ASD except that the provisions limit the distance between lateral support should be limited to 30 x the nominal width where as ASD uses &#8216;effective height&#8217; (meaning considering end restraints) should be &lt; 24 x thickness. Also max reinforcement shall be per Ref 1 Section 3.3.5 but not to exceed  <img src="//s0.wp.com/latex.php?latex=%3D+0.04A_n&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="= 0.04A_n" title="= 0.04A_n" class="latex" /></li>
</ol>
</li>
</ol>
</li>
<li>Lateral Ties
<ol>
<li><span style="font-size: small;"><span style="line-height: 19px;">Requirements (Ref 1 Section 2.1.6.5) If </span>reinforcement<span style="line-height: 19px;"> is to be considered effective in resisting </span></span>compressive<span style="font-size: small;"><span style="line-height: 19px;"> forces lateral ties must be provided.</span></span>
<ol>
<li>1/4&#8243; diameter (min)</li>
<li>Vertical spacing &lt; smaller of -
<ol>
<li>16&#8243;</li>
<li>48 lateral tie bar diameters</li>
<li>least cross-sectional dimension</li>
</ol>
</li>
<li>Every corner and alternate longitudinal bar shall have lateral support provided by the corner of a lateral tie (angle should be less than 135 deg). Except for a circle arrangement which is permitted. No bar bar shall be farther than 6&#8243; (each side) from a laterally supported bar. (i.e. the middle bar should be less than 6&#8243; from 2 bars that are laterally supported by a lateral tie &#8216;corner&#8217;)</li>
<li>Place lateral ties in mortar joint or grout</li>
<li>Lap length shall be 48 tie diameters (min)</li>
<li>Locate the first and last lateral tie at 1/2 spacing.  Also above and below single beams framing into the vertical member.</li>
<li>Where beams or brackets frame into a column from four directions ties may be terminated 3&#8243; (max) from the lowest reinforcement in the shallowest of the beam members framing into the column.</li>
</ol>
</li>
</ol>
</li>
</ol>
<p>&nbsp;</p>
<p>The post <a href="http://howtoengineer.com/masonry-column-pier-pilaster/">Masonry Columns, Piers, Pilasters</a> appeared first on <a href="http://howtoengineer.com">How To Engineer</a>.</p>]]></content:encoded>
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		<title>Complex Geometries &#8211; Stable Feature Behind MSE Wall</title>
		<link>http://howtoengineer.com/complex-geometries-stable-feature-behind-wall/</link>
		<comments>http://howtoengineer.com/complex-geometries-stable-feature-behind-wall/#comments</comments>
		<pubDate>Mon, 11 Feb 2013 15:00:48 +0000</pubDate>
		<dc:creator>Ryan Freund</dc:creator>
				<category><![CDATA[Earth Retention]]></category>
		<category><![CDATA[Retaining Walls]]></category>
		<category><![CDATA[Complex Geometry]]></category>
		<category><![CDATA[MSE]]></category>
		<category><![CDATA[Narrow Wall]]></category>
		<category><![CDATA[SRW]]></category>
		<category><![CDATA[Trial Wedge Method]]></category>

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		<description><![CDATA[<p><p><a href="http://howtoengineer.com">How To Engineer - Engineers In Training</a></p><p>Stable Feature Behind MSE Wall / Narrow Wall This will be one of several  posts regarding MSE walls with complex geometry. The main references for this post and the posts to follow will be the research conducted by The Federal Highway Administration (FHWA) and mainly the publications listed below. The methods presented are simplified / preliminary methods. Ultimately a limit equilibrium analysis should be run. However this is beyond the scope&#8230; for now. References FHWA-NHI-10-024 Design and Construction of MSE walls and Reinforced Soil Slopes FHWA-NHI-00-043 MSE Walls and Reinforced Soils Slopes Design and Construction Guidelines (Now superseded by publication above) FHWA-CFL/TD-06-001 Shored MSE Wall Systems (SMSE) Design Guidelines All of which can be found here on FHWA website. &#160; Overview I am posting my notes that correlate to the spreadsheet for now and a brief overview. We will elaborate more on the procedure in the near future. The stable feature prevents external lateral earth pressures from exerting force on the reinforced soil mass. Therefore traditional sliding and overturning are not generally a design concern. For the trial wedge we define the geomtry and solve for the forces in the horizontal and vertical directions. We have two unknowns and two equations of equilibrium which we solve using matrix algebra (in excel) to determine the active force on the wall. Required checks are as [...]</p></p><p>The post <a href="http://howtoengineer.com/complex-geometries-stable-feature-behind-wall/">Complex Geometries &#8211; Stable Feature Behind MSE Wall</a> appeared first on <a href="http://howtoengineer.com">How To Engineer</a>.</p>]]></description>
				<content:encoded><![CDATA[<p><a href="http://howtoengineer.com">How To Engineer - Engineers In Training</a></p><h1>Stable Feature Behind MSE Wall / Narrow Wall</h1>
<p>This will be one of several  posts regarding MSE walls with complex geometry. The main references for this post and the posts to follow will be the research conducted by The Federal Highway Administration (FHWA) and mainly the publications listed below. The methods presented are simplified / preliminary methods. Ultimately a limit equilibrium analysis should be run. However this is beyond the scope&#8230; for now.</p>
<h2>References</h2>
<p><span style="font-size: 13px; line-height: 19px;">FHWA-NHI-10-024 Design and Construction of MSE walls and Reinforced Soil Slopes</span></p>
<p>FHWA-NHI-00-043 MSE Walls and Reinforced Soils Slopes Design and Construction Guidelines (Now superseded by publication above)</p>
<p>FHWA-CFL/TD-06-001 Shored MSE Wall Systems (SMSE) Design Guidelines</p>
<p>All of which can be found <a title="FHWA MSE Publications " href="http://www.fhwa.dot.gov/engineering/geotech/library_listing.cfm" target="_blank">here</a> on FHWA website.</p>
<p>&nbsp;</p>
<p><span style="font-size: 1.5em; line-height: 19px;">Overview</span></p>
<p>I am posting my notes that correlate to the spreadsheet for now and a brief overview. We will elaborate more on the procedure in the near future.</p>
<p>The stable feature prevents external lateral earth pressures from exerting force on the reinforced soil mass. Therefore traditional sliding and overturning are not generally a design concern. For the trial wedge we define the geomtry and solve for the forces in the horizontal and vertical directions. We have two unknowns and two equations of equilibrium which we solve using matrix algebra (in excel) to determine the active force on the wall.</p>
<p>Required checks are as follows:</p>
<ol>
<li><span style="font-size: 13px; line-height: 19px;">Pullout of geogrid</span></li>
<li><span style="font-size: 13px; line-height: 19px;">Reinforced soil and stable feature interface failure plane stability</span></li>
<li><span style="font-size: 13px; line-height: 19px;">Bearing</span></li>
<li><span style="font-size: 13px; line-height: 19px;">Global / Slope stability</span></li>
</ol>
<h2>Calculation</h2>
<div id="attachment_738" class="wp-caption alignnone" style="width: 310px"><a href="http://howtoengineer.com/wp-content/uploads/2013/02/Force-Diagram-Stable-Feature-Behind-MSE-Wall.jpg"><img class="size-medium wp-image-738" alt="Force Diagram - Stable Feature Behind MSE Wall" src="http://howtoengineer.com/wp-content/uploads/2013/02/Force-Diagram-Stable-Feature-Behind-MSE-Wall-300x181.jpg" width="300" height="181" /></a><p class="wp-caption-text">Force Diagram &#8211; Stable Feature Behind MSE Wall</p></div>
<p>Calculation assumes that the stable features extends up or very near the top grade surface. The wall face, top grade and stable feature slopes may all be adjusted. Currently the failure angle must be &#8216;manually&#8217; adjusted to find the maximum active pressure acting on the back of the wall facing units.</p>
<p>&nbsp;</p>
<h2>Notes</h2>
<p><a href="http://howtoengineer.com/wp-content/uploads/2013/02/Trial-Wedge-with-Stable-Feature-Behind-Wall-Notes-HTE.pdf">Trial Wedge with Stable Feature Behind Wall Notes HTE</a></p>
<h2>Spreadsheet</h2>
<p><a href="http://howtoengineer.com/wp-content/uploads/2013/02/Trial-Wedge-for-Stable-Feature.xlsx">Trial Wedge for Stable Feature</a></p>
<p>As always please use with caution and check for errors!</p>
<p>&nbsp;</p>
<p>&nbsp;</p>
<p>&nbsp;</p>
<p>The post <a href="http://howtoengineer.com/complex-geometries-stable-feature-behind-wall/">Complex Geometries &#8211; Stable Feature Behind MSE Wall</a> appeared first on <a href="http://howtoengineer.com">How To Engineer</a>.</p>]]></content:encoded>
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		<title>Torsion in Rectangular Prism</title>
		<link>http://howtoengineer.com/torsion-in-rectangular-prism/</link>
		<comments>http://howtoengineer.com/torsion-in-rectangular-prism/#comments</comments>
		<pubDate>Tue, 29 Jan 2013 16:00:11 +0000</pubDate>
		<dc:creator>Ryan Freund</dc:creator>
				<category><![CDATA[Mechanics of Materials]]></category>
		<category><![CDATA[Torsion]]></category>
		<category><![CDATA[Torsion in Rectangular Shape]]></category>

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		<description><![CDATA[<p><p><a href="http://howtoengineer.com">How To Engineer - Engineers In Training</a></p><p>Torsion in Rectangular Prism I hope to expand on this post in the future and relate the equations that are derived for find stresses and strains of a rectangular prism (bar) to that of an I shaped section. Then relating that to AISC&#8217;s design guide 9 (see here for AISC design guides). Explanation of Stress &#8211; Strain Behavior We recall that for circular shafts that section planes remain plane (and circular) and that all radii lines remain straight; no bulging or cupping. Therefore shear stress occurs along a set of concentric circles. This however is not possible with a rectangular shaft as if the shear stress was tangent to a set of concentric circles the shear stress would not be perpendicular to the boundary. The shear stress must be perpendicular to the boundary. Why? Because if it is not than components perpendicular to the boundary will exist which creates a &#8216;complementary&#8217; shear stress on the outside free edge of the prism (see Figure below). From this we conclude that the shear stress must be perpendicular to the periphery (boundary) of the section. We can also conclude that the shear stress must also be zero at the corners as neither one of the two perpendicular components can exist (see Figure below). For a circular section it [...]</p></p><p>The post <a href="http://howtoengineer.com/torsion-in-rectangular-prism/">Torsion in Rectangular Prism</a> appeared first on <a href="http://howtoengineer.com">How To Engineer</a>.</p>]]></description>
				<content:encoded><![CDATA[<p><a href="http://howtoengineer.com">How To Engineer - Engineers In Training</a></p><h1>Torsion in Rectangular Prism</h1>
<p>I hope to expand on this post in the future and relate the equations that are derived for find stresses and strains of a rectangular prism (bar) to that of an I shaped section. Then relating that to AISC&#8217;s design guide 9 (see <a title="AISC Design Guides" href="http://www.aisc.org/SearchTaxonomy/DesignGuides.aspx?id=4442">here</a> for AISC design guides).</p>
<h2>Explanation of Stress &#8211; Strain Behavior</h2>
<p>We recall that for circular shafts that section planes remain plane (and circular) and that all radii lines remain straight; no bulging or cupping. Therefore shear stress occurs along a set of concentric circles. This however is not possible with a rectangular shaft as if the shear stress was tangent to a set of concentric circles the shear stress would not be perpendicular to the boundary. The shear stress must be perpendicular to the boundary. Why? Because if it is not than components perpendicular to the boundary will exist which creates a &#8216;complementary&#8217; shear stress on the outside free edge of the prism (see Figure below). From this we conclude that the shear stress must be perpendicular to the periphery (boundary) of the section.</p>
<div id="attachment_729" class="wp-caption alignnone" style="width: 310px"><a href="http://howtoengineer.com/wp-content/uploads/2013/01/Torsion-Shear-Stress-in-Non-Circular-Prism.jpg"><img class="size-medium wp-image-729" alt="Torsion - Shear Stress in Non-Circular Prism" src="http://howtoengineer.com/wp-content/uploads/2013/01/Torsion-Shear-Stress-in-Non-Circular-Prism-300x168.jpg" width="300" height="168" /></a><p class="wp-caption-text">Torsion &#8211; Shear Stress in Non-Circular Prism</p></div>
<p>We can also conclude that the shear stress must also be zero at the corners as neither one of the two perpendicular components can exist (see Figure below).</p>
<div id="attachment_728" class="wp-caption alignnone" style="width: 310px"><a href="http://howtoengineer.com/wp-content/uploads/2013/01/Torsion-Shear-Stress-in-Non-Circular-Prism-Corner-Stress.jpg"><img class="size-medium wp-image-728" alt="Torsion - Shear Stress in Non-Circular Prism Corner Stress" src="http://howtoengineer.com/wp-content/uploads/2013/01/Torsion-Shear-Stress-in-Non-Circular-Prism-Corner-Stress-300x270.jpg" width="300" height="270" /></a><p class="wp-caption-text">Torsion &#8211; Shear Stress in Non-Circular Prism Corner Stress</p></div>
<p>For a circular section it was also shown that the plane cross sections remained undistorted in their own plane. Meaning there was no distortion of the concentric circles and radii, the radii remained straight.  We cannot directly prove this to be true for a non-circular prism but we can look at what it means if we assume there is some distortion.</p>
<p>For a non-circular section, we imagine a perpendicular grid pattern across a section. These grids then become distortion in the plane of the section. These shear stress then cause shear stresses to appear on sections parallel to the longitudinal axis. These stresses do not add up to a twisting moment and are &#8216;useless&#8217;. This does not constitue a proof but it should be noted that nature opposes an applied action (Torque) through the minimum elastic energy i.e. the simplest way possible or via the simplest stresses possible. It is shown via energy methods that these stress do NOT occur and or reasoning is correct.</p>
<div id="attachment_730" class="wp-caption alignnone" style="width: 310px"><a href="http://howtoengineer.com/wp-content/uploads/2013/01/Torsion-In-Plane-Distortion.png"><img class="size-medium wp-image-730" alt="Torsion - In Plane Distortion" src="http://howtoengineer.com/wp-content/uploads/2013/01/Torsion-In-Plane-Distortion-300x241.png" width="300" height="241" /></a><p class="wp-caption-text">Torsion &#8211; In Plane Distortion</p></div>
<h2> Calculating Stress</h2>
<p>Too be continued&#8230;.. Refer to <a title="Roark's Formula's For Stress and Strain" href="http://www.amazon.com/Roarks-Formulas-Stress-Strain-Edition/dp/0071742476/ref=sr_1_sc_1?ie=UTF8&amp;qid=1359343727&amp;sr=8-1-spell&amp;keywords=roraks+formulas+for+stress+and">Roark</a> for now <img src='http://howtoengineer.com/wp-includes/images/smilies/icon_smile.gif' alt=':)' class='wp-smiley' /> </p>
<p>The post <a href="http://howtoengineer.com/torsion-in-rectangular-prism/">Torsion in Rectangular Prism</a> appeared first on <a href="http://howtoengineer.com">How To Engineer</a>.</p>]]></content:encoded>
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		<title>Mechanics of Materials &#8211; Basics</title>
		<link>http://howtoengineer.com/mechanics-of-materials-basics/</link>
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		<pubDate>Mon, 28 Jan 2013 16:10:34 +0000</pubDate>
		<dc:creator>Ryan Freund</dc:creator>
				<category><![CDATA[Mechanics of Materials]]></category>

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		<description><![CDATA[<p><p><a href="http://howtoengineer.com">How To Engineer - Engineers In Training</a></p><p>Mechanics of Materials &#8211; Basics I would like to describe the process or logic or algorithm if you will in which must be followed in order to develop theories for mechanics of materials.  The process / logic / algorithm can go in either direction but you CANNOT skip steps. This idea is fact but was nicely presented in Dr. Madhukar Vable in his textbook Intermediate Mechanics of Materials. Structural Analysis Algorithm (Logic) Displacements Kinematics Strains Material Models Stresses Static Equivalency Internal Forces and Moments Equilibrium / Energy Methods External Forces and Moments So, you may start with external forces then using equilibrium or energy methods you may determine the internal forces and moments. From here you may determine stresses by static equivalency. Then strains may be determined by the material model and finally displacements by use of kinematics. You may start at any step in the above algorithm and can proceed in any direction but you cannot skip a step. Meaning, if you know the internal forces and moments you can not determine the displacements with out considering the stresses and strains. &#160;</p></p><p>The post <a href="http://howtoengineer.com/mechanics-of-materials-basics/">Mechanics of Materials &#8211; Basics</a> appeared first on <a href="http://howtoengineer.com">How To Engineer</a>.</p>]]></description>
				<content:encoded><![CDATA[<p><a href="http://howtoengineer.com">How To Engineer - Engineers In Training</a></p><h1>Mechanics of Materials &#8211; Basics</h1>
<p>I would like to describe the process or logic or algorithm if you will in which must be followed in order to develop theories for mechanics of materials.  The process / logic / algorithm can go in either direction but you CANNOT skip steps. This idea is fact but was nicely presented in Dr. Madhukar Vable in his textbook Intermediate Mechanics of Materials.</p>
<h2>Structural Analysis Algorithm (Logic)</h2>
<ol>
<li><span style="line-height: 13px;">Displacements</span>
<ul>
<li>Kinematics</li>
</ul>
</li>
<li>Strains
<ul>
<li>Material Models</li>
</ul>
</li>
<li>Stresses
<ul>
<li>Static Equivalency</li>
</ul>
</li>
<li>Internal Forces and Moments
<ul>
<li>Equilibrium / Energy Methods</li>
</ul>
</li>
<li>External Forces and Moments</li>
</ol>
<p>So, you may start with external forces then using equilibrium or energy methods you may determine the internal forces and moments. From here you may determine stresses by static equivalency. Then strains may be determined by the material model and finally displacements by use of kinematics. You may start at any step in the above algorithm and can proceed in any direction but you cannot skip a step. Meaning, if you know the internal forces and moments you can not determine the displacements with out considering the stresses and strains.</p>
<p>&nbsp;</p>
<p>The post <a href="http://howtoengineer.com/mechanics-of-materials-basics/">Mechanics of Materials &#8211; Basics</a> appeared first on <a href="http://howtoengineer.com">How To Engineer</a>.</p>]]></content:encoded>
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		<title>Stability &#8211; AISC&#8217;s Approximate Second-Order Analysis B1 B2 Method</title>
		<link>http://howtoengineer.com/stability-aiscs-direct-analysis-method-b1-b2-hand-calc-method/</link>
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		<pubDate>Tue, 22 Jan 2013 16:00:42 +0000</pubDate>
		<dc:creator>Ryan Freund</dc:creator>
				<category><![CDATA[Stability - Steel Structures]]></category>
		<category><![CDATA[Steel]]></category>
		<category><![CDATA[AISC]]></category>
		<category><![CDATA[Approximate Second-Order Analysis]]></category>
		<category><![CDATA[B1 B2]]></category>
		<category><![CDATA[DAM]]></category>
		<category><![CDATA[Direct Analysis Method]]></category>
		<category><![CDATA[Hand Calc]]></category>
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		<description><![CDATA[<p><p><a href="http://howtoengineer.com">How To Engineer - Engineers In Training</a></p><p>AISC&#8217;s Approximate Second-Order Analysis B1 B2 Method Lets look at a very simple building with a simple moment frame to resist lateral loads. We will complete the analysis using AISC&#8217;s approximate second-order analysis more commonly known as the B1 &#8211; B2 method. We will demonstrate this method with the use of an example on a very simple building. Before we begin lets discus the B1, B2 method. The B1-B2 Method is an approximate second-order analysis using the multipliers B1 and B2 (no way!). The procedure can be found in Appendix 8 of the 2010 AISC 350 Steel Construction Manual. B1 Discussion:      See AISC Eqn A-8-3 Where: coefficient assuming no lateral translation of the frame. See Eqn A-8-4 . However I would conservatively use 1.0 and be done with it! 1 (LRFD), 1.6 (ASD) = may be determined from a first-order estimate is permitted (for use in eqn A-8-3 only). = axial load (using ASD/LRFD Load Combo&#8217;s) assuming the structure is &#8220;restrained against lateral translation&#8221; (first order). = axial load due to lateral translation of the structure only (using ASD/LRFD Load Combo&#8217;s) (first order). = coefficient assuming no lateral translation of the frame. = Flexural rigidity required by the analysis i.e because we are using [...]</p></p><p>The post <a href="http://howtoengineer.com/stability-aiscs-direct-analysis-method-b1-b2-hand-calc-method/">Stability &#8211; AISC&#8217;s Approximate Second-Order Analysis B1 B2 Method</a> appeared first on <a href="http://howtoengineer.com">How To Engineer</a>.</p>]]></description>
				<content:encoded><![CDATA[<p><a href="http://howtoengineer.com">How To Engineer - Engineers In Training</a></p><h1>AISC&#8217;s Approximate Second-Order Analysis B1 B2 Method</h1>
<p>Lets look at a very simple building with a simple moment frame to resist lateral loads. We will complete the analysis using AISC&#8217;s approximate second-order analysis more commonly known as the B1 &#8211; B2 method. We will demonstrate this method with the use of an example on a very simple building. Before we begin lets discus the B1, B2 method. The B1-B2 Method is an approximate second-order analysis using the multipliers B1 and B2 (no way!). The procedure can be found in Appendix 8 of the 2010 AISC 350 Steel Construction Manual.</p>
<h3>B1 Discussion:</h3>
<p><img src="//s0.wp.com/latex.php?latex=B1%3D%5Cfrac%7BC_m%7D%7B1-%5Calpha+P_r%2FP_%7Be1%7D%7D&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="B1=&#92;frac{C_m}{1-&#92;alpha P_r/P_{e1}}" title="B1=&#92;frac{C_m}{1-&#92;alpha P_r/P_{e1}}" class="latex" />      See AISC Eqn A-8-3</p>
<p>Where:</p>
<ul>
<li><img src="//s0.wp.com/latex.php?latex=C_m+%3D+&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="C_m = " title="C_m = " class="latex" /> coefficient assuming no lateral translation of the frame. See Eqn A-8-4 . However I would conservatively use 1.0 and be done with it!</li>
<li><img src="//s0.wp.com/latex.php?latex=%5Calpha%3D+&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="&#92;alpha= " title="&#92;alpha= " class="latex" /> 1 (LRFD), 1.6 (ASD)</li>
<li><img src="//s0.wp.com/latex.php?latex=P_r+%3D+P_%7Bnt%7D%2BP_%7Blt%7D&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="P_r = P_{nt}+P_{lt}" title="P_r = P_{nt}+P_{lt}" class="latex" /> = may be determined from a first-order estimate is permitted (for use in eqn A-8-3 only).
<ul>
<li><img src="//s0.wp.com/latex.php?latex=P_%7Bnt%7D&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="P_{nt}" title="P_{nt}" class="latex" /> = axial load (using ASD/LRFD Load Combo&#8217;s) assuming the structure is &#8220;restrained against lateral translation&#8221; (first order).</li>
<li><img src="//s0.wp.com/latex.php?latex=P_%7Blt%7D&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="P_{lt}" title="P_{lt}" class="latex" /> = axial load due to lateral translation of the structure only (using ASD/LRFD Load Combo&#8217;s) (first order).</li>
</ul>
</li>
<li><img src="//s0.wp.com/latex.php?latex=P_%7Be1%7D+%3D+%5Cfrac+%7B%5Cpi+EI%2A+%7D+%7B+%28K_1+L%29%5E2+%7D+&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="P_{e1} = &#92;frac {&#92;pi EI* } { (K_1 L)^2 } " title="P_{e1} = &#92;frac {&#92;pi EI* } { (K_1 L)^2 } " class="latex" /> = coefficient assuming no lateral translation of the frame.
<ul>
<li><img src="//s0.wp.com/latex.php?latex=EI%2A+&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="EI* " title="EI* " class="latex" /> = Flexural rigidity required by the analysis i.e because we are using DAM =</li>
<li><img src="//s0.wp.com/latex.php?latex=EI%2A+%3D+0.8+%5Ctau_b+EI+&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="EI* = 0.8 &#92;tau_b EI " title="EI* = 0.8 &#92;tau_b EI " class="latex" /> where <img src="//s0.wp.com/latex.php?latex=%5Ctau&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="&#92;tau" title="&#92;tau" class="latex" /> is defined in AISC Section C2.3 (adjustments to stiffness). I prefer to avoid adjusting <img src="//s0.wp.com/latex.php?latex=%5Ctau&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="&#92;tau" title="&#92;tau" class="latex" /> as it seems to become an iterative process. Therefore I add an additional notional load =
<ul>
<li><img src="//s0.wp.com/latex.php?latex=N_%7B+%5Ctau%7D+%3D+0.001+%5Calpha+Y_i+%5Ctext%7Bwhere%7D+Y_i+&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="N_{ &#92;tau} = 0.001 &#92;alpha Y_i &#92;text{where} Y_i " title="N_{ &#92;tau} = 0.001 &#92;alpha Y_i &#92;text{where} Y_i " class="latex" /> is defined in Section C2.2b (essentially the gravity load at level &#8216;i&#8217;) and <img src="//s0.wp.com/latex.php?latex=%5Calpha&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="&#92;alpha" title="&#92;alpha" class="latex" /> is as defined above.</li>
</ul>
</li>
</ul>
</li>
<li><img src="//s0.wp.com/latex.php?latex=K_1&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="K_1" title="K_1" class="latex" /> is based on the assumption of no lateral translation so we conservatively use 1.0.</li>
</ul>
<p>B1 accounts for for <img src="//s0.wp.com/latex.php?latex=P-%5Cdelta&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="P-&#92;delta" title="P-&#92;delta" class="latex" /> effects in compression non-sway compression members. These are the moments caused by local displacements due to axial load. The AISC commentary suggests that if B1 is &gt; than 1.2 than a rigorous second order analysis should be undertaken. This is due to the fact that B1 captures the local effects of second order forces/deformation but does not capture what effect these &#8216;local&#8217; deformations may have on the overall structure. This is semi obvious in the fact that the we are using results from a first order analysis and also none of the variables relate to the rest of the structure.</p>
<h3>B2 Discussion:</h3>
<p><img src="//s0.wp.com/latex.php?latex=B2%3D%5Cfrac%7B1%7D%7B1-%5Cfrac%7B%5Calpha+P_%7Bstory%7D%7D%7BP_%7Bestory%7D%7D%7D&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="B2=&#92;frac{1}{1-&#92;frac{&#92;alpha P_{story}}{P_{estory}}}" title="B2=&#92;frac{1}{1-&#92;frac{&#92;alpha P_{story}}{P_{estory}}}" class="latex" />       See AISC Eqn A-8-6</p>
<p>Where:</p>
<ul>
<li><img src="//s0.wp.com/latex.php?latex=%5Calpha+&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="&#92;alpha " title="&#92;alpha " class="latex" /> is as defined above.</li>
<li><img src="//s0.wp.com/latex.php?latex=P_%7Bstory%7D+%3D+&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="P_{story} = " title="P_{story} = " class="latex" /> total vertical load supported by the story (using ASD/LRFD load Combos) including loads in columns not part of the lateral force resisting system. This is essential the total gravity load on the story under evaluation.</li>
<li><img src="//s0.wp.com/latex.php?latex=P_%7Bestory%7D+%3D+&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="P_{estory} = " title="P_{estory} = " class="latex" /> the &#8220;elastic critical buckling strength for the story in the direction of translation being considered, determined by buckling analysis&#8221; or
<ul>
<li><img src="//s0.wp.com/latex.php?latex=P_%7Bestory%7D+%3D+R_M+%5Cfrac%7BH+L%7D+%7B%5CDelta_H%7D+&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="P_{estory} = R_M &#92;frac{H L} {&#92;Delta_H} " title="P_{estory} = R_M &#92;frac{H L} {&#92;Delta_H} " class="latex" />
<ul>
<li><img src="//s0.wp.com/latex.php?latex=R_M+%3D+1-0.15%28P_%7Bmf%7D+%2F+P_%7Bstory%7D%29&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="R_M = 1-0.15(P_{mf} / P_{story})" title="R_M = 1-0.15(P_{mf} / P_{story})" class="latex" />
<ul>
<li><img src="//s0.wp.com/latex.php?latex=P_%7Bmf%7D+&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="P_{mf} " title="P_{mf} " class="latex" /> = total vertical load in columns (in the story under evaluation) that are part of the moment frames (=0 for braced frame systems).</li>
</ul>
</li>
<li>L = height of story</li>
<li><img src="//s0.wp.com/latex.php?latex=%5CDelta_H+%3D+&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="&#92;Delta_H = " title="&#92;Delta_H = " class="latex" /> Inter-story drift. Use first order analysis and stiffness as required by analysis i.e For DAM use reduced stiffness as discussed above (see the B1 discussion). Where drift varies across the story, the maximum drift may be used conservatively or a weighted average based on vertical load. It is important to realize here that this drift or deflection should include the deflection of the columns in the frame and ALSO the diaphragm deflection. The commentary words this as any &#8220;horizontal framing system that increases over-turning effect&#8221;. This makes sense as columns that are not part of the frame (usually called leaning columns) will displace greater than the columns that are part of the moment frame. This displacement coupled with gravity load will increase the demand on the frame.</li>
<li>H = Story shear (lateral force), produced by the lateral force used to compute the inter-story drift. Once again, coordinate the use of total story shear or individual force on the frame. As mentioned for <img src="//s0.wp.com/latex.php?latex=P_%7Bstory%7D+&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="P_{story} " title="P_{story} " class="latex" />.</li>
<li>AISC provides a user note that says H and <img src="//s0.wp.com/latex.php?latex=%5CDelta_H&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="&#92;Delta_H" title="&#92;Delta_H" class="latex" /> &#8220;may be based on any lateral loading that provides a representative value of the story lateral stiffness.&#8221; As you can see the equation is really using the lateral stiffness of the structure (kip/in).</li>
</ul>
</li>
</ul>
</li>
</ul>
<p>B2 accounts for <img src="//s0.wp.com/latex.php?latex=P-%5CDelta&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="P-&#92;Delta" title="P-&#92;Delta" class="latex" /> effects on forces and moments in all members. These effects are due to lateral displacement of the structure. We also notice that B2 uses several variables which related back to the overall structure, mainly the the story shear, gravity load and deflection. Furthermore we see that the deflection is based on not only on the deflection of the frame but also of the diaphragm which indirectly accounts for &#8220;leaning columns&#8221;.</p>
<h3>Short Summary</h3>
<p>We can see that B2 applies to all members part of the Lateral Force Resisting System (LFRS), meaning any member with <img src="//s0.wp.com/latex.php?latex=P_%7Blt%7D+and+M_%7Blt%7D&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="P_{lt} and M_{lt}" title="P_{lt} and M_{lt}" class="latex" /> (members not part of the LFRS will not have these forces) and B1 applies only to compression members of the LFRS. We see that by using the DA method we eliminate having to use sidesway alignment chart (fig C-A-7-2) to try and determine K (effective length). However the B1-B2 method can be trick when when B1 gets large and multiple members frame into a column. Why? Well because the moments should be balanced and thus the column (compression member) will be multiplied by B1 and the beams would not be. So this moment then needs to be distributed to these connecting elements. I would suggest reading the Summary at the end of the commentary to the B1-B2 method  (Comm 8 pg 16.1-526) they discuss the how to apply the method in more &#8216;global&#8217; terms and will give you a better feel for applying the method. It is too long to repeat here.</p>
<p>&nbsp;</p>
<h1>Example</h1>
<p>Alright so lets define some parameters and loads.</p>
<p>Lets use a 1-story, 3-bay x 4-bay rectangular building. Bay size is 25&#8242;x25&#8242;. The columns are pinned at the base and have a moment connection from beam to column. For this analysis we will use a &#8220;Wind Only&#8221; moment frame or flexible moment connection. There is definitely some debate on using this type of system. Essentially beams are designed as simply supported for gravity loads and fixed for lateral loads. For a more complete discussion on &#8220;wind only&#8221; or flexible moment connection- moment frames see - <a title="Wind Only Moment Frames (Flexible Moment Connection)" href="http://wp.me/p2Puq4-a2">Wind Only Moment Frames Discussion</a>.</p>
<p>Size: 1-Story- 3&#215;4 &#8211; 25&#8242;x25&#8242; Bays (75&#8242; x 125&#8242; Building). Height is 15&#8242; columns with 5&#8242; parapet.</p>
<p>Gravity Loads:</p>
<ul>
<li>Dead load: Say 30 psf just to give it some weight.</li>
<li>Live load: Say 100 psf again, too add some weight.</li>
</ul>
<p>Lateral load:</p>
<ul>
<li>Check seismic, but lets use wind for now.</li>
<li>Wind: say 20 psf. Most people will forget that there is a 1.5 multiplier on the parapet when designing for the LFRS so lets use 30 psf on  the parapet.</li>
</ul>
<h2>Analysis</h2>
<p>We have 2 separate analysis to perform.</p>
<ul>
<li>Gravity load on analysis</li>
<li>Lateral load only analysis</li>
</ul>
<h3>Gravity Load Analysis</h3>
<p>Lets place the moment frame on grid lines 2 and 4 in the north-south direction.<br />
For the gravity load analysis we would use all load combinations and assume the frame is restrained against lateral movement. Therefore we would not have moments due to lateral forces at the moment connections. For simplicity we will use our dead and live load. This would typically be snow load as this is a one story roof and remember to account for drift load as well.</p>
<h4>Typical Beam;</h4>
<p>Dead load; <img src="//s0.wp.com/latex.php?latex=w_%7BDL%7D%3D30%5C%3Bpsf+%5C%3Bx+%5C%3B25+%5C%3Bft+%3D+750%5C%3B+plf&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="w_{DL}=30&#92;;psf &#92;;x &#92;;25 &#92;;ft = 750&#92;; plf" title="w_{DL}=30&#92;;psf &#92;;x &#92;;25 &#92;;ft = 750&#92;; plf" class="latex" /><br />
Live load; <img src="//s0.wp.com/latex.php?latex=w_%7BLL%7D%3D100%5C%3B+psf%5C%3B+x%5C%3B+25%5C%3B+ft+%3D+2500%5C%3B+plf&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="w_{LL}=100&#92;; psf&#92;; x&#92;; 25&#92;; ft = 2500&#92;; plf" title="w_{LL}=100&#92;; psf&#92;; x&#92;; 25&#92;; ft = 2500&#92;; plf" class="latex" /><br />
Total load; <img src="//s0.wp.com/latex.php?latex=w+%3D+3.25%5C%3B+klf&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="w = 3.25&#92;; klf" title="w = 3.25&#92;; klf" class="latex" /><br />
Shear; <img src="//s0.wp.com/latex.php?latex=V_r+%3D+%5B3.25%5C%3B+klf%5C%3B+x+25%5C%3B+ft%5D+%2F+2+%3D+40.6+kip&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="V_r = [3.25&#92;; klf&#92;; x 25&#92;; ft] / 2 = 40.6 kip" title="V_r = [3.25&#92;; klf&#92;; x 25&#92;; ft] / 2 = 40.6 kip" class="latex" /><br />
Moment; <img src="//s0.wp.com/latex.php?latex=M_r+%3D+%5B3.25%5C%3B+klf%5C%3B+x+%2825%5C%3B+ft%29%5E2%5D+%2F+8+%3D+254+kip-ft&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="M_r = [3.25&#92;; klf&#92;; x (25&#92;; ft)^2] / 8 = 254 kip-ft" title="M_r = [3.25&#92;; klf&#92;; x (25&#92;; ft)^2] / 8 = 254 kip-ft" class="latex" /><br />
Unbraced length say 5ft. (practically fully braced for positive moment)</p>
<p>Required moment of inertia &#8211; dead load; <img src="//s0.wp.com/latex.php?latex=I_r%3D%5Cfrac+%7B5wL%5E4%7D%7B%28384E%5CDelta%29%7D+%3D+%5Cfrac+%7B5%5C%3Bx%5C%3B3.25klf%5C%3B25ft%5E4%5C%3B1728%7D%7B%28384%5C%3Bx%5C%3B29000ksi%5C%3Bx%5C%3B25%5C%3Bx%5C%3B12%2F240%29%7D%3D+788in%5E4&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="I_r=&#92;frac {5wL^4}{(384E&#92;Delta)} = &#92;frac {5&#92;;x&#92;;3.25klf&#92;;25ft^4&#92;;1728}{(384&#92;;x&#92;;29000ksi&#92;;x&#92;;25&#92;;x&#92;;12/240)}= 788in^4" title="I_r=&#92;frac {5wL^4}{(384E&#92;Delta)} = &#92;frac {5&#92;;x&#92;;3.25klf&#92;;25ft^4&#92;;1728}{(384&#92;;x&#92;;29000ksi&#92;;x&#92;;25&#92;;x&#92;;12/240)}= 788in^4" class="latex" /></p>
<p>Required moment of inertia &#8211; live load; <img src="//s0.wp.com/latex.php?latex=I_r%3D%5Cfrac+%7B5%5C%3Bx%5C%3B2.5klf%5C%3B25ft%5E4%5C%3B1728%7D%7B%28384%5C%3Bx%5C%3B29000ksi%5C%3Bx%5C%3B25%5C%3Bx%5C%3B12%2F360%29%7D%3D+909in%5E4&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="I_r=&#92;frac {5&#92;;x&#92;;2.5klf&#92;;25ft^4&#92;;1728}{(384&#92;;x&#92;;29000ksi&#92;;x&#92;;25&#92;;x&#92;;12/360)}= 909in^4" title="I_r=&#92;frac {5&#92;;x&#92;;2.5klf&#92;;25ft^4&#92;;1728}{(384&#92;;x&#92;;29000ksi&#92;;x&#92;;25&#92;;x&#92;;12/360)}= 909in^4" class="latex" /> (Controls)</p>
<h4>Typical Column &#8211; Exterior;</h4>
<p>Dead load = <img src="//s0.wp.com/latex.php?latex=P_%7BDL%7D%3D%5C%3B30psf%5C%3Bx%5C%3B25ft%5C%3Bx%5C%3B12.5ft%5C%3B%3D9.375kip&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="P_{DL}=&#92;;30psf&#92;;x&#92;;25ft&#92;;x&#92;;12.5ft&#92;;=9.375kip" title="P_{DL}=&#92;;30psf&#92;;x&#92;;25ft&#92;;x&#92;;12.5ft&#92;;=9.375kip" class="latex" /><br />
Live load = <img src="//s0.wp.com/latex.php?latex=P_%7BLL%7D%3D%5C%3B100psf%5C%3Bx%5C%3B25ft%5C%3Bx%5C%3B12.5ft%5C%3B%3D31.25kip&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="P_{LL}=&#92;;100psf&#92;;x&#92;;25ft&#92;;x&#92;;12.5ft&#92;;=31.25kip" title="P_{LL}=&#92;;100psf&#92;;x&#92;;25ft&#92;;x&#92;;12.5ft&#92;;=31.25kip" class="latex" /><br />
Required axial load = <img src="//s0.wp.com/latex.php?latex=P_r%3D%5C%3B40.6kip&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="P_r=&#92;;40.6kip" title="P_r=&#92;;40.6kip" class="latex" /></p>
<h4>Typical Column &#8211; Interior;</h4>
<p>Dead load = <img src="//s0.wp.com/latex.php?latex=P_%7BDL%7D%3D%5C%3B30psf%5C%3Bx%5C%3B25ft%5C%3Bx%5C%3B25ft%5C%3B%3D18.75kip&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="P_{DL}=&#92;;30psf&#92;;x&#92;;25ft&#92;;x&#92;;25ft&#92;;=18.75kip" title="P_{DL}=&#92;;30psf&#92;;x&#92;;25ft&#92;;x&#92;;25ft&#92;;=18.75kip" class="latex" /><br />
Live load = <img src="//s0.wp.com/latex.php?latex=P_%7BLL%7D%3D%5C%3B100psf%5C%3Bx%5C%3B25ft%5C%3Bx%5C%3B25ft%5C%3B%3D62.5kip&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="P_{LL}=&#92;;100psf&#92;;x&#92;;25ft&#92;;x&#92;;25ft&#92;;=62.5kip" title="P_{LL}=&#92;;100psf&#92;;x&#92;;25ft&#92;;x&#92;;25ft&#92;;=62.5kip" class="latex" /><br />
Required axial load = <img src="//s0.wp.com/latex.php?latex=P_r%3D%5C%3B81.25kip&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="P_r=&#92;;81.25kip" title="P_r=&#92;;81.25kip" class="latex" /></p>
<p>To Be Cont&#8230;.</p>
<p>&nbsp;</p>
<p>The post <a href="http://howtoengineer.com/stability-aiscs-direct-analysis-method-b1-b2-hand-calc-method/">Stability &#8211; AISC&#8217;s Approximate Second-Order Analysis B1 B2 Method</a> appeared first on <a href="http://howtoengineer.com">How To Engineer</a>.</p>]]></content:encoded>
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		<title>Portal Method &#8211; Moments and Deflection</title>
		<link>http://howtoengineer.com/portal-method-moments-and-deflection/</link>
		<comments>http://howtoengineer.com/portal-method-moments-and-deflection/#comments</comments>
		<pubDate>Mon, 21 Jan 2013 14:00:40 +0000</pubDate>
		<dc:creator>Ryan Freund</dc:creator>
				<category><![CDATA[Analysis]]></category>
		<category><![CDATA[Approximate Methods]]></category>

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		<description><![CDATA[<p><p><a href="http://howtoengineer.com">How To Engineer - Engineers In Training</a></p><p>Portal Method &#8211; Moments and Deflection The portal method is an analysis method for used for finding approximate internal forces in indeterminate structures subject to a lateral load. Portal Frame Assumptions: Zero moment at the center of each girder. Therefore a hinge is placed here. Zero moment at the center of each column, if the column has a fixed base. Therefore a hinge is placed here (or at the bottom of the frame if the frame is hinged at the base). The interior columns hinges have twice the shear force as the exterior columns. This is due to the fact that the frame is considered to be a superposition of 2 portals. I also like to envision this due to the fact that the frame acts like a short cantilevered beam where shear is the predominate deformation force. If you recall the shear stress distribution of a rectangular beam, it is greater in the middle of the beam than at the extremes. I admit this may not be techinically correct but it helps me remember the method. Portal Method Notes: Portal Method &#8211; Approx Deflection Using Energy Methods I will elaborate on this method in the future. For now take a look at the notes (yes [...]</p></p><p>The post <a href="http://howtoengineer.com/portal-method-moments-and-deflection/">Portal Method &#8211; Moments and Deflection</a> appeared first on <a href="http://howtoengineer.com">How To Engineer</a>.</p>]]></description>
				<content:encoded><![CDATA[<p><a href="http://howtoengineer.com">How To Engineer - Engineers In Training</a></p><h1>Portal Method &#8211; Moments and Deflection</h1>
<p>The portal method is an analysis method for used for finding approximate internal forces in indeterminate structures subject to a lateral load.</p>
<p>Portal Frame Assumptions:</p>
<ul>
<li><span style="line-height: 13px;">Zero moment at the center of each girder. Therefore a hinge is placed here.</span></li>
<li>Zero moment at the center of each column, if the column has a fixed base. Therefore a hinge is placed here (or at the bottom of the frame if the frame is hinged at the base).</li>
<li>The interior columns hinges have twice the shear force as the exterior columns.
<ul>
<li>This is due to the fact that the frame is considered to be a superposition of 2 portals.</li>
<li>I also like to envision this due to the fact that the frame acts like a short cantilevered beam where shear is the predominate deformation force. If you recall the shear stress distribution of a rectangular beam, it is greater in the middle of the beam than at the extremes. I admit this may not be techinically correct but it helps me remember the method.</li>
</ul>
</li>
</ul>
<p>Portal Method Notes:</p>
<p><a href="http://howtoengineer.com/wp-content/uploads/2013/01/Portal-Method-Approx-Deflection-Using-Energy-Methods.pdf">Portal Method &#8211; Approx Deflection Using Energy Methods</a></p>
<p>I will elaborate on this method in the future. For now take a look at the notes (yes I know the handwriting is awful).</p>
<p>Here is a spreadsheet for single story, pin-base moment frames. Currently it only accounts for bending deflection. Not all that useful but kinda fun none the less.</p>
<p><a href="http://howtoengineer.com/wp-content/uploads/2013/01/Portal-Method-Lateral-Load-Pin-Base-Single-Story-Deflection.xlsx">Portal Method &#8211; Lateral Load &#8211; Pin Base &#8211; Single Story Deflection</a></p>
<p>&nbsp;</p>
<p>The post <a href="http://howtoengineer.com/portal-method-moments-and-deflection/">Portal Method &#8211; Moments and Deflection</a> appeared first on <a href="http://howtoengineer.com">How To Engineer</a>.</p>]]></content:encoded>
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		<title>Stability &#8211; AISC&#8217;s Direct Analysis Method</title>
		<link>http://howtoengineer.com/stability-aiscs-direct-analysis-method/</link>
		<comments>http://howtoengineer.com/stability-aiscs-direct-analysis-method/#comments</comments>
		<pubDate>Tue, 25 Dec 2012 14:34:16 +0000</pubDate>
		<dc:creator>Ryan Freund</dc:creator>
				<category><![CDATA[Stability - Steel Structures]]></category>
		<category><![CDATA[Stability Analysis]]></category>
		<category><![CDATA[AISC]]></category>
		<category><![CDATA[DAM]]></category>
		<category><![CDATA[Direct Analysis Method]]></category>
		<category><![CDATA[Stability]]></category>
		<category><![CDATA[Steel Design]]></category>

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		<description><![CDATA[<p><p><a href="http://howtoengineer.com">How To Engineer - Engineers In Training</a></p><p>Stability &#8211; AISC&#8217;s Direct Analysis Method Intro In the 14th Edition of the American Steel Construction Manual the Direct Analysis Method (DAM) is moved into the main specification from the appendix. I know many are not used to this new approach and some will say that if it&#8217;s not broke don&#8217;t fix it but I like the DA Method. Like anything else you need to put some time into learning it but it&#8217;s really not so bad in the end. It is a very interesting approach in that most design codes / manuals do not get into the analysis meaning that they don&#8217;t tell you how to get the required/design forces but rather they give an allowable/capacity of the member to which you are designing. However in DAM, AISC is assisting you in your analysis to make the design simpler. Essentially we need to address two different types of deflection/deformation associated with axial load that amplify moments in the structure. The first is P-δ (I may refer to this as P-d) which are moments associated with the axial load and deflection due to column curvature (Think of these as local displacements). The second is P-Δ (I may refer to this as P-D) [...]</p></p><p>The post <a href="http://howtoengineer.com/stability-aiscs-direct-analysis-method/">Stability &#8211; AISC&#8217;s Direct Analysis Method</a> appeared first on <a href="http://howtoengineer.com">How To Engineer</a>.</p>]]></description>
				<content:encoded><![CDATA[<p><a href="http://howtoengineer.com">How To Engineer - Engineers In Training</a></p><h1>Stability &#8211; AISC&#8217;s Direct Analysis Method</h1>
<h3>Intro</h3>
<p>In the 14th Edition of the American Steel Construction Manual the Direct Analysis Method (DAM) is moved into the main specification from the appendix. I know many are not used to this new approach and some will say that if it&#8217;s not broke don&#8217;t fix it but I like the DA Method. Like anything else you need to put some time into learning it but it&#8217;s really not so bad in the end. It is a very interesting approach in that most design codes / manuals do not get into the analysis meaning that they don&#8217;t tell you how to get the required/design forces but rather they give an allowable/capacity of the member to which you are designing. However in DAM, AISC is assisting you in your analysis to make the design simpler. Essentially we need to address two different types of deflection/deformation associated with axial load that amplify moments in the structure. The first is P-δ (I may refer to this as P-d) which are moments associated with the axial load and deflection due to column curvature (Think of these as local displacements). The second is P-Δ (I may refer to this as P-D) moments which are caused by axial load and the translation of the end of the column (i.e. interstory drift) think of these as global displacements. Neither of these moments will show up in a first order elastic analysis. Well they may&#8230;sorta. I don&#8217;t want to get off track so I will explain what I mean later (this refers to common FEM models and placing multiple nodes along a member). We will now cover the DAM for a computer based approach and a simplified hand method.</p>
<div id="attachment_597" class="wp-caption alignnone" style="width: 307px"><a href="http://howtoengineer.com/wp-content/uploads/2012/12/Pd-and-PD-Sketch.jpg"><img class="size-medium wp-image-597" alt="Pd and PD Sketch" src="http://howtoengineer.com/wp-content/uploads/2012/12/Pd-and-PD-Sketch-297x300.jpg" width="297" height="300" /></a><p class="wp-caption-text">Pd and PD Sketch</p></div>
<h3>Overview</h3>
<p>The direct analysis method is basically accounting for (3) issues:</p>
<ol>
<li>Effects of initial geometric imperfections</li>
<li>Second-Order effects &#8211; Axial-Displacement Moments P-D and P-d (as shown above).</li>
<li>Effects of material non-linearity - In-elasticity due to residual stresses.</li>
</ol>
<p>AISC actually states that there are (5) requirements. Below these requirements are listed and how they are addressed (AISC C-C1.1).</p>
<h4>Considerations:</h4>
<ol>
<li>Consider all deformations
<ul>
<li>Note that this says &#8216;consider&#8217; not necessarily include, i.e. column shears deformation, in-plane &#8216;rigid&#8217; diaphragm displacement.</li>
<li>The model or analysis shall &#8216;consider&#8217; all deformations.</li>
</ul>
</li>
<li>Consider P-d and P-D
<ul>
<li>Perform a rigerous second order analysis</li>
<li>Use B1, B2 Method</li>
</ul>
</li>
<li>Consider geometric imperfections
<ul>
<li>This typically this stems from column out of plumbness</li>
<li>This may be directly modeled in the analysis</li>
<li>A notional load may be applied to the analysis</li>
<li>Use KL = L</li>
</ul>
</li>
<li>Consider stiffness due to inelasticity. This is typically due to residual stresses in framing members. Therefore some elements may soften &#8216;inelastically&#8217;  prior to reaching their design strength.
<ul>
<li>Apply a stiffness reduction factor</li>
<li>Use KL = L</li>
</ul>
</li>
<li>Consider uncertainty in strength and stiffness
<ul>
<li>Apply a stiffness reduction factor.</li>
<li>Use KL = L</li>
</ul>
</li>
</ol>
<h2> Applying the Direct Analysis Method</h2>
<p>First we will look at applying this method in a strict sense and assuming the use of a computer model. Then we will get to a more conservative hand calc method.</p>
<ol>
<li><span style="line-height: 13px;">Model your structure and apply all loads. Set up your load combinations according to LRFD or ASD (Most likely see IBC load combo&#8217;s).</span></li>
<li>Run a first-order analysis and determine deflections.
<ul>
<li>Amplify the ASD loads x 1.6</li>
<li>Modify the stiffness of all members. For a first trial run use a 0.8 factor. This would be applied to axial (0.8*EA) and flexural (0.8EI) stiffness.
<ul>
<li>AISC states that the stiffness reduction need only be applied to members that contribute to the stability of the structure however they can be applied to all members to prevent artificial distortion.</li>
</ul>
</li>
</ul>
</li>
<li>Run a second-order analysis.
<ul>
<li>Amplify the ASD loads x 1.6</li>
<li>Modify the stiffness of all members. For a first trial run use a 0.8 factor. This would be applied to axial (0.8*EA) and flexural (0.8EI) stiffness.
<ul>
<li>AISC states that the stiffness reduction need only be applied to members that contribute to the stability of the structure however they can be applied to all members to prevent artificial distortion.</li>
</ul>
</li>
<li>So this is just a mouse click away right? Well not quite. You should really know what your analysis software is doing. It is difficult if not impossible in some situations for software programs to perform a rigorous second-order analysis. For the program to perform this analysis it usually needs to run an iterative process on many nodes which may not be realistic. Therefore the program may use a geometric stiffness method which only accounts for P-D moments. Therefore P-d moments are still unaccounted for. However these moments may be &#8220;semi&#8221; captured if the column element is broken into several nodes. This way the deflection between nodes is captured in the analysis. AISC recognizes this practical problem and states that the P-d effects on the structure may be neglected if the second order drift to first order drift ratio (also known as B2) is equal or less than 1.7, also no more than 1/3 of the total gravity load on the structure is supported by columns that are part of the moment-resisting frames in the direction of translation being considered.</li>
<li> In the commentary they equate the 1.7 to a 1.5 limit with no stiffness reduction.</li>
<li>Furthermore P-d effects must be considered to individual members subject to compression and flexure. In this case B-1 could be used.</li>
</ul>
</li>
<li>Find the drift ratio (B2) of second order to first order drift. This will be used to determine what sort of notional loads will need to be applied.</li>
<li>Notional loads &#8211; Initial Imperfections
<ul>
<li>Initial imperfections may be directly applied in the model. Typically an out of plumbness of 1/500 is used the maximum specified in the Code of Standard Practice.</li>
<li>If not modeled directly notional loads may be applied. These are lateral loads Ni =0.002*α*Yi. These loads are distributed over the level in the same maner as the gravity load.
<ul>
<li>Ni = notional load at level i</li>
<li>α = 1.0 (LRFD); α = 1.6 (ASD)</li>
<li>Yi = gravity load applied at level i under each respective load combination</li>
</ul>
</li>
<li>If B2 (drift ratio) is &lt;= 1.7 then the notional loads may be applied as a minimum. Meaning that they are applied to gravity only load combinations but are not applied if the &#8216;actual&#8217; lateral loads i.e. wind/EQ forces are greater than the notional load.</li>
</ul>
</li>
<li>Adjust stiffness
<ul>
<li>For all stiffness that contribute to the stability of  the structure a 0.8 factor shall be applied i.e. EI, AE, etc.</li>
<li>Additionally for flexure the stiffness should be multiplied by 0.8<img src="//s0.wp.com/latex.php?latex=%5Ctau_b&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="&#92;tau_b" title="&#92;tau_b" class="latex" />:
<ul>
<li>For <img src="//s0.wp.com/latex.php?latex=%5Calpha+P_r%2FP_y+%5Cleq+0.5+%5Ctext%7B+than+%7D+%5Ctau_b+%3D1.0+&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="&#92;alpha P_r/P_y &#92;leq 0.5 &#92;text{ than } &#92;tau_b =1.0 " title="&#92;alpha P_r/P_y &#92;leq 0.5 &#92;text{ than } &#92;tau_b =1.0 " class="latex" /></li>
<li>For <img src="//s0.wp.com/latex.php?latex=%5Calpha+P_r%2FP_y+%3E+0.5+%5Ctext%7B+than+%7D+%5Ctau_b+%3D4%28%5Calpha+P_r%2FP_y%29%5B1-%28%5Calpha+P_r%2FP_y%29%5D+&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="&#92;alpha P_r/P_y &gt; 0.5 &#92;text{ than } &#92;tau_b =4(&#92;alpha P_r/P_y)[1-(&#92;alpha P_r/P_y)] " title="&#92;alpha P_r/P_y &gt; 0.5 &#92;text{ than } &#92;tau_b =4(&#92;alpha P_r/P_y)[1-(&#92;alpha P_r/P_y)] " class="latex" />
<ul>
<li>α = 1.0 (LRFD); α = 1.6 (ASD)</li>
<li>Pr = required axial compressive strength of the member</li>
<li>Py = axial yield strength = Fy*Ag (yield stress x gross area)</li>
</ul>
</li>
</ul>
</li>
<li>In lieu of using taub a notional load of 0.001*α*Yi may be applied to the structure in similar fashion as the notional loads for initial imperfections. However these notional loads are additive for all load combinations.</li>
</ul>
</li>
<li>Rerun second order analysis and check drift ratio, B2. Update any parameters based on the new drift ratio.</li>
<li>Design members using K=1. No Alignment Chart Required, Yeah!!!</li>
</ol>
<h2> Applying the Direct Analysis Method</h2>
<p>Now for a simplified hand calc to demonstrate the use of B1 and B2.<br />
This post got to be a tad long so I&#8217;m going to break this into a separate post <a title="B1 B2 DAM Hand Calc" href="http://wp.me/p2Puq4-ag">here</a>.</p>
<h3>References:</h3>
<p>AISC 14th Edition CSC&#8217;s &#8220;Simple Guide to Direct Analysis&#8221; and webinar. Note that CSC&#8217;s Fastrak software does perform a rigorous second order analysis. RISA&#8217;s Practical Analysis with the AISC 13th Edition by Josh Plummer AISC Engineering Journal 3th Q 2008 &#8220;A comparison of Frame Stability Analysis Methods&#8221;</p>
<p>The post <a href="http://howtoengineer.com/stability-aiscs-direct-analysis-method/">Stability &#8211; AISC&#8217;s Direct Analysis Method</a> appeared first on <a href="http://howtoengineer.com">How To Engineer</a>.</p>]]></content:encoded>
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		<title>Surcharge Analysis &#8211; Elastic Methods &#8211; Strip Load</title>
		<link>http://howtoengineer.com/surcharge-analysis-elastic-methods-strip-load/</link>
		<comments>http://howtoengineer.com/surcharge-analysis-elastic-methods-strip-load/#comments</comments>
		<pubDate>Sun, 23 Dec 2012 14:39:41 +0000</pubDate>
		<dc:creator>Ryan Freund</dc:creator>
				<category><![CDATA[Earth Retention]]></category>
		<category><![CDATA[Lateral Earth Pressure]]></category>
		<category><![CDATA[Retaining Walls]]></category>
		<category><![CDATA[Boussinesq]]></category>
		<category><![CDATA[Surchage]]></category>

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		<description><![CDATA[<p><p><a href="http://howtoengineer.com">How To Engineer - Engineers In Training</a></p><p>Strip Load Surcharge Analysis using Elastic Methods Here we will dig deeper into analyzing strip loads with elastic methods. First to avoid confusion between seeing similar equations which use different reference angles, we will set our own nomenclature. Nomenclature K = Constant typically between 1 and 2, depending on the stiffness of the wall. It may be appropriate to use K=1 for flexible walls such as cantilevered sheet pile, segmental walls. It may be appropriate to use K=2 for rigid walls. Refer to CivilTech Software manual (http://www.civiltechsoftware.com/downloads/sh_manu.pdf). &#160; Below are (2) common equations which result in the same solution but are shown differently. I will refer to these equations as the integrated method or equations. Most codes and design guides refer to these equations but it seems too differ on who is credited. Some refer to them as Boussinesq as modified by Tang or Spangler, others Mindlin or Terzaghi (although Terzaghi is mostly responsible for modifying point and line load elastic equations). Note that β is located at half of θ and not half the width of the strip load. The other In Joseph Bowles&#8217; &#8220;Foundation Analysis and Design&#8221; he discuses different methods of analyzing offset surcharges using elastic methods. He discuss the [...]</p></p><p>The post <a href="http://howtoengineer.com/surcharge-analysis-elastic-methods-strip-load/">Surcharge Analysis &#8211; Elastic Methods &#8211; Strip Load</a> appeared first on <a href="http://howtoengineer.com">How To Engineer</a>.</p>]]></description>
				<content:encoded><![CDATA[<p><a href="http://howtoengineer.com">How To Engineer - Engineers In Training</a></p><h1>Strip Load Surcharge Analysis using Elastic Methods</h1>
<p>Here we will dig deeper into analyzing strip loads with elastic methods. First to avoid confusion between seeing similar equations which use different reference angles, we will set our own nomenclature.</p>
<div id="attachment_532" class="wp-caption alignnone" style="width: 255px"><a href="http://howtoengineer.com/wp-content/uploads/2012/12/Strip-Load-RSF-Nomenclature.jpg"><img class="size-medium wp-image-532" title="Strip Load - Elastic Methods RSF Nomenclature" alt="Strip Load - Elastic Methods RSF Nomenclature" src="http://howtoengineer.com/wp-content/uploads/2012/12/Strip-Load-RSF-Nomenclature-245x300.jpg" width="245" height="300" /></a><p class="wp-caption-text">Strip Load &#8211; Elastic Methods RSF Nomenclature</p></div>
<h3>Nomenclature</h3>
<p><img src="//s0.wp.com/latex.php?latex=%5Ctext%7Bq+%3D+load%2C+psf%7D&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="&#92;text{q = load, psf}" title="&#92;text{q = load, psf}" class="latex" /><br />
<img src="//s0.wp.com/latex.php?latex=%5Calpha+%3D+%5Ctext%7Bangle+between+wall+face+and+near+side+of+strip+load%7D&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="&#92;alpha = &#92;text{angle between wall face and near side of strip load}" title="&#92;alpha = &#92;text{angle between wall face and near side of strip load}" class="latex" /><br />
<img src="//s0.wp.com/latex.php?latex=%5Cdelta+%3D+%5Ctext%7Bangle+between+wall+face+and+far+side+of+strip+load%7D&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="&#92;delta = &#92;text{angle between wall face and far side of strip load}" title="&#92;delta = &#92;text{angle between wall face and far side of strip load}" class="latex" /><br />
<img src="//s0.wp.com/latex.php?latex=%5Ctheta+%3D+%5Ctext%7Bangle+at+point+of+evaluation+between+near+side+and+far+side+of+strip+load%7D&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="&#92;theta = &#92;text{angle at point of evaluation between near side and far side of strip load}" title="&#92;theta = &#92;text{angle at point of evaluation between near side and far side of strip load}" class="latex" /><br />
<img src="//s0.wp.com/latex.php?latex=%5Cbeta+%3D+%5Calpha+%2B+1%2F2+%28%5Ctheta%29&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="&#92;beta = &#92;alpha + 1/2 (&#92;theta)" title="&#92;beta = &#92;alpha + 1/2 (&#92;theta)" class="latex" /><br />
<img src="//s0.wp.com/latex.php?latex=%5Ctext%7BX1+%3D+distance+from+wall+to+near+side+of+surcharge%7D&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="&#92;text{X1 = distance from wall to near side of surcharge}" title="&#92;text{X1 = distance from wall to near side of surcharge}" class="latex" /><br />
<img src="//s0.wp.com/latex.php?latex=%5Ctext%7BX2+%3D+distance+from+wall+to+far+side+of+surcharge%7D&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="&#92;text{X2 = distance from wall to far side of surcharge}" title="&#92;text{X2 = distance from wall to far side of surcharge}" class="latex" /><br />
K = Constant typically between 1 and 2, depending on the stiffness of the wall. It may be appropriate to use K=1 for flexible walls such as cantilevered sheet pile, segmental walls. It may be appropriate to use K=2 for rigid walls. Refer to CivilTech Software manual (http://www.civiltechsoftware.com/downloads/sh_manu.pdf).</p>
<p>&nbsp;</p>
<p>Below are (2) common equations which result in the same solution but are shown differently. I will refer to these equations as the integrated method or equations. Most codes and design guides refer to these equations but it seems too differ on who is credited. Some refer to them as Boussinesq as modified by Tang or Spangler, others Mindlin or Terzaghi (although Terzaghi is mostly responsible for modifying point and line load elastic equations).</p>
<p><img src="//s0.wp.com/latex.php?latex=%5Csigma_h+%3D+%5Cfrac%7BKQ%7D%7B%5Cpi%7D+%28%5Ctheta-sin%28%5Ctheta%29cos%282%5Cbeta%29%29&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="&#92;sigma_h = &#92;frac{KQ}{&#92;pi} (&#92;theta-sin(&#92;theta)cos(2&#92;beta))" title="&#92;sigma_h = &#92;frac{KQ}{&#92;pi} (&#92;theta-sin(&#92;theta)cos(2&#92;beta))" class="latex" /> Note that β is located at half of θ and not half the width of the strip load.</p>
<p>The other</p>
<p><img src="//s0.wp.com/latex.php?latex=%5Csigma_h+%3D+%5Cfrac%7BKQ%7D%7B%5Cpi%7D+%28%5Ctheta-sin%28%5Ctheta%29cos%28%5Ctheta%2B2%5Calpha%29%29&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="&#92;sigma_h = &#92;frac{KQ}{&#92;pi} (&#92;theta-sin(&#92;theta)cos(&#92;theta+2&#92;alpha))" title="&#92;sigma_h = &#92;frac{KQ}{&#92;pi} (&#92;theta-sin(&#92;theta)cos(&#92;theta+2&#92;alpha))" class="latex" /></p>
<p>In Joseph Bowles&#8217; &#8220;Foundation Analysis and Design&#8221; he discuses different methods of analyzing offset surcharges using elastic methods. He discuss the work of Spangler and the factor of two suggested by Mindlin. In summary he suggests that Spangler&#8217;s experiments were not accurate and possibly flawed due to the setup geometry. Also the equations derived by Spangler use a Poisson&#8217;s ratio of 0.5 which may not be correct for all soils. However Spangler&#8217;s equation for the strip load results in the same equation as shown above. Bowles also states that Mindlins reasoning for the factor 2 (Mindlin says that a rigid wall produced a mirror effect) is wrong.<br />
Bowles suggests to only use the original Boussinesq equation.</p>
<p><img src="//s0.wp.com/latex.php?latex=%5Csigma_r+%3D+P%2F%282%5Cpi%29+%283r%5E2z%2FR%5E5+-+%281-2%5Cmu%29%2F%28R%28R%2Bz%29%29&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="&#92;sigma_r = P/(2&#92;pi) (3r^2z/R^5 - (1-2&#92;mu)/(R(R+z))" title="&#92;sigma_r = P/(2&#92;pi) (3r^2z/R^5 - (1-2&#92;mu)/(R(R+z))" class="latex" /><br />
<img src="//s0.wp.com/latex.php?latex=%5Csigma_x+%3D+%5Csigma_r+%28x%2Fr%29&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="&#92;sigma_x = &#92;sigma_r (x/r)" title="&#92;sigma_x = &#92;sigma_r (x/r)" class="latex" /> [EQN - 1]</p>
<p>With these equations one can &#8220;discretize&#8221; the strip load. Meaning you divide the strip load into a series of concentrated loads. Then apply the Boussinesq equation to each point load. Then the sum the results to find the resulting pressure at a certain elevation on the wall. You then find these resulting pressures at a certain number of elevations on the wall and sum these to find the resulting total force. It should be noted that Poisson&#8217;s ratio should be modified so that it represents a plan strain condition.</p>
<p><img src="//s0.wp.com/latex.php?latex=%5Cmu%27%3D%5Cmu%2F%281-%5Cmu%29&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="&#92;mu&#039;=&#92;mu/(1-&#92;mu)" title="&#92;mu&#039;=&#92;mu/(1-&#92;mu)" class="latex" /> where <img src="//s0.wp.com/latex.php?latex=%5Cmu+&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="&#92;mu " title="&#92;mu " class="latex" /> represents a triaxial condition<br />
Refer to Excavating Systems, Planning, Design and Safety, 2009 for the following suggested values of <img src="//s0.wp.com/latex.php?latex=%5Cmu&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="&#92;mu" title="&#92;mu" class="latex" /><br />
Moist clay soils: 0.4-0.5<br />
Saturated clay soils: 0.45-0.5<br />
Cohesionless, medium dense: 0.3-0.4<br />
Cohesionless, loose to medium: 0.2-0.35</p>
<p>Here are some notes on the discretized method (semi-large file &#8211; about 10MB, I will try to shrink later):<br />
<a href="http://howtoengineer.com/wp-content/uploads/2012/12/Boussinesq-Strip-Load-Discritization-Method-2.pdf">Boussinesq Strip Load Discritization Method 2</a></p>
<p>Now my problem is this &#8211; when I compare the &#8220;discretized&#8221; approach to the integrated I do not get similar values, not even close. I have to think that I am in error somewhere, but I cannot find where. I have compared spreadsheets to other programs to hand calcs. I will continue to search, but for now, I have attached a couple hand calcs showing the discrepancy. I will also attach a spreadsheet comparing techniques.<br />
Attached is a hand calc comparison between the &#8220;integrated&#8221; and discretized approach.<br />
<a href="http://howtoengineer.com/wp-content/uploads/2012/12/Boussinesq-Integrated-vs-Discretized.pdf">Boussinesq Integrated vs Discretized</a><br />
<a title="Elastic Method - Compare Integrated to Discretized" href="http://howtoengineer.com/wp-content/uploads/2012/12/Elastic-Method-Compare-Integrated-to-Discretized.pdf">Elastic Method &#8211; Compare Integrated to Discretized</a></p>
<p><a title="Elastic Methods Nomenclature Notes" href="http://howtoengineer.com/wp-content/uploads/2012/12/Elastic-Methods-Nomenclature-Notes.pdf">Elastic Methods Nomenclature Notes</a></p>
<p><a title="Elastic Methods Nomenclature Notes" href="http://howtoengineer.com/wp-content/uploads/2012/12/Elastic-Methods-Spangler-Derives-Integrated-Method.pdf">Elastic Methods Spangler Derives Integrated Method</a><br />
<a href="http://howtoengineer.com/wp-content/uploads/2012/12/Offset-Surcharge-Strip-Load-Comparison.xlsx">Offset Surcharge Strip Load Comparison Spreadsheet</a></p>
<p>Another problem when using elastic methods is that all equations assume that the load is applied at the same elevation as the top of the wall. How can we handle a slope atop the wall with a load applied at the top of the slope? Well the railroad design manuals (See UPRR Shoring Manual) distribute the load down through the soil at a certain angle (they use a 2V:1H) and find a new uniform load applied at this elevation.</p>
<div id="attachment_579" class="wp-caption alignnone" style="width: 310px"><a href="http://howtoengineer.com/wp-content/uploads/2012/12/Elastic-Methods-Slope-Dia-1.png"><img class="size-medium wp-image-579" alt="Elastic Methods - Slope Diagram" src="http://howtoengineer.com/wp-content/uploads/2012/12/Elastic-Methods-Slope-Dia-1-300x237.png" width="300" height="237" /></a><p class="wp-caption-text">Elastic Methods &#8211; Slope Diagram</p></div>
<p>The post <a href="http://howtoengineer.com/surcharge-analysis-elastic-methods-strip-load/">Surcharge Analysis &#8211; Elastic Methods &#8211; Strip Load</a> appeared first on <a href="http://howtoengineer.com">How To Engineer</a>.</p>]]></content:encoded>
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		<title>Helpful Links for Determining Minimum Design Loads</title>
		<link>http://howtoengineer.com/helpful-links-for-determining-minimum-design-loads/</link>
		<comments>http://howtoengineer.com/helpful-links-for-determining-minimum-design-loads/#comments</comments>
		<pubDate>Wed, 28 Nov 2012 15:18:06 +0000</pubDate>
		<dc:creator>Ryan Freund</dc:creator>
				<category><![CDATA[Loading / Codes]]></category>

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		<description><![CDATA[<p><p><a href="http://howtoengineer.com">How To Engineer - Engineers In Training</a></p><p>Helpful Links for Determining Minimum Design Loads Hopefully these links can save you some time and help get you more accurate design loads. A quick heads-up &#8211; you will usually need to search the town/county/state to see if the Authority Having Jurisdiction (AHJ) has a specific requirement. Wind Load A favorite for Wind Loads in accordance w/ ASCE 7 http://www.atcouncil.org/windspeed/index.php Seismic A favorite for determining your &#8216;base acceleration&#8217; coefficients: http://earthquake.usgs.gov/hazards/designmaps/ Snow This site is no longer free but when I used it, it was useful: http://www.groundsnowbyzip.com/ This is a little dated and really not that useful but I&#8217;ll mention it anyway: http://www.fs.fed.us/t-d/snow_load/states.htm Others This is a &#8216;pay-for&#8217; site but some may use it: http://www.groundsnowbyzip.com/</p></p><p>The post <a href="http://howtoengineer.com/helpful-links-for-determining-minimum-design-loads/">Helpful Links for Determining Minimum Design Loads</a> appeared first on <a href="http://howtoengineer.com">How To Engineer</a>.</p>]]></description>
				<content:encoded><![CDATA[<p><a href="http://howtoengineer.com">How To Engineer - Engineers In Training</a></p><h1>Helpful Links for Determining Minimum Design Loads</h1>
<p>Hopefully these links can save you some time and help get you more accurate design loads. A quick heads-up &#8211; you will usually need to search the town/county/state to see if the Authority Having Jurisdiction (AHJ) has a specific requirement.</p>
<h2>Wind Load</h2>
<p>A favorite for Wind Loads in accordance w/ ASCE 7</p>
<p>http://www.atcouncil.org/windspeed/index.php</p>
<h2>Seismic</h2>
<p>A favorite for determining your &#8216;base acceleration&#8217; coefficients:</p>
<p>http://earthquake.usgs.gov/hazards/designmaps/</p>
<h2>Snow</h2>
<p>This site is no longer free but when I used it, it was useful:</p>
<p>http://www.groundsnowbyzip.com/</p>
<p>This is a little dated and really not that useful but I&#8217;ll mention it anyway:</p>
<p>http://www.fs.fed.us/t-d/snow_load/states.htm</p>
<h2>Others</h2>
<p>This is a &#8216;pay-for&#8217; site but some may use it:</p>
<p>http://www.groundsnowbyzip.com/</p>
<p>The post <a href="http://howtoengineer.com/helpful-links-for-determining-minimum-design-loads/">Helpful Links for Determining Minimum Design Loads</a> appeared first on <a href="http://howtoengineer.com">How To Engineer</a>.</p>]]></content:encoded>
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		<title>Surcharge Analysis &#8211; Elastic Methods</title>
		<link>http://howtoengineer.com/offset-surcharge-elastic-methods/</link>
		<comments>http://howtoengineer.com/offset-surcharge-elastic-methods/#comments</comments>
		<pubDate>Mon, 26 Nov 2012 02:42:42 +0000</pubDate>
		<dc:creator>Ryan Freund</dc:creator>
				<category><![CDATA[Earth Retention]]></category>
		<category><![CDATA[Lateral Earth Pressure]]></category>
		<category><![CDATA[Boussinesq]]></category>
		<category><![CDATA[Elastic Methods]]></category>
		<category><![CDATA[Offset Surcharge]]></category>
		<category><![CDATA[Strip Load]]></category>

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		<description><![CDATA[<p><p><a href="http://howtoengineer.com">How To Engineer - Engineers In Training</a></p><p>Analysis of Offset Surcharges on Retaining Walls Using Elastic Methods Offset surcharges are always up for some debate. Which method to use? What is the line of influence? Is the wall flexible or rigid? Well in this segment we will cover the analysis of surcharge loads on retaining walls using elastic methods. You may commonly here this as a Boussinesq analysis or a Boussinesq as modified by Terzaghi or Tang. Also other equations have been produced by Spangler and Mindlin. Trying to wrap you head around all the different equations and who did what can be confusing. We will try to straighten some of this out and present some of the different approaches and how most of the equations are similar but the equations are presented slightly differently. Constraints and Assumptions There are generally 4 types of surcharges considered &#8211; Point load, Line load, Strip load and Area load. Point loads and area loads have a finite length. Line and strip loads are assumed to be of infinite length parallel to the wall. The back slope is generally considered to be flat. Elastic methods do not consider any soil parameters such as effective shear strength or wall friction. The only way to try and account for different [...]</p></p><p>The post <a href="http://howtoengineer.com/offset-surcharge-elastic-methods/">Surcharge Analysis &#8211; Elastic Methods</a> appeared first on <a href="http://howtoengineer.com">How To Engineer</a>.</p>]]></description>
				<content:encoded><![CDATA[<p><a href="http://howtoengineer.com">How To Engineer - Engineers In Training</a></p><h1>Analysis of Offset Surcharges on Retaining Walls Using Elastic Methods</h1>
<p>Offset surcharges are always up for some debate. Which method to use? What is the line of influence? Is the wall flexible or rigid? Well in this segment we will cover the analysis of surcharge loads on retaining walls using elastic methods. You may commonly here this as a Boussinesq analysis or a Boussinesq as modified by Terzaghi or Tang. Also other equations have been produced by Spangler and Mindlin. Trying to wrap you head around all the different equations and who did what can be confusing. We will try to straighten some of this out and present some of the different approaches and how most of the equations are similar but the equations are presented slightly differently.</p>
<h2>Constraints and Assumptions</h2>
<p>There are generally 4 types of surcharges considered &#8211; Point load, Line load, Strip load and Area load. Point loads and area loads have a finite length. Line and strip loads are assumed to be of infinite length parallel to the wall. The back slope is generally considered to be flat. Elastic methods do not consider any soil parameters such as effective shear strength or wall friction. The only way to try and account for different soil types is by adjusting Poisson&#8217;s ratio or adjusting the factor of safety. However most of the theories assume a Poisson&#8217;s ratio (u) of 0.5 which simplifies the analysis, but more on this latter.</p>
<p>The basis of the elastic methods discussed below is Boussinesq&#8217;s equation for a surcharge on a semi infinite mass</p>
<p>Equation 1:</p>
<div id="attachment_441" class="wp-caption alignnone" style="width: 310px"><a href="http://howtoengineer.com/wp-content/uploads/2012/11/Boussinesq-Equation.jpg"><img class="size-medium wp-image-441" title="Boussinesq Equation" alt="Boussinesq Equation" src="http://howtoengineer.com/wp-content/uploads/2012/11/Boussinesq-Equation-300x123.jpg" width="300" height="123" /></a><p class="wp-caption-text">Boussinesq Equation</p></div>
<div id="attachment_442" class="wp-caption alignnone" style="width: 310px"><a href="http://howtoengineer.com/wp-content/uploads/2012/11/Boussinesq-Equation-Diagram.jpg"><img class="size-medium wp-image-442" title="Boussinesq Equation Diagram" alt="Boussinesq Equation Diagram" src="http://howtoengineer.com/wp-content/uploads/2012/11/Boussinesq-Equation-Diagram-300x223.jpg" width="300" height="223" /></a><p class="wp-caption-text">Boussinesq Equation Diagram</p></div>
<p>Where:</p>
<p>P = point load</p>
<p>v = Poisson&#8217;s ratio</p>
<h2>History</h2>
<p>Most of this is from Bowles 4th Edition Foundation Analysis and Design.</p>
<p>1936 Spangler &#8211; Performed experiments to measure the lateral pressure on a wall due to point loads form a truck behind a rigid retaining wall. Spangler used Boussinesq&#8217;s equation with u=0.5 and found that the actual lateral pressure was approx 2x the pressure found by equation 1 with u=0.5. Bowles points out some flaws of this experiment &#8211; loose soils with a wall of &#8216;finite&#8217; length and &#8216;old&#8217; technology used in the load cells.</p>
<p>1936 Mindlin &#8211; Discusses the results of Spangler. He explains that the 2x factor could be due to the rigid wall producing a mirror load effect. This reasoning is disputed by Bowles.</p>
<p>1954 Terzaghi &#8211; suggests 0.4*H (wall height) is a critical distance behind the wall where two different equations should be used for point loads and line loads. Terzaghi proposed (2) different modified boussinesq equations for point and line loads both using v=0.5.</p>
<p>1962 Teng &#8211; Similar modification claiming that the pressure on rigid walls for strip loads should be 2x the results obtained by the &#8216;integrated&#8217; Boussinesq equation. Some also credit Teng the strip load equation shown below as a plastic&#8217;solution as opposed to the Boussinesq solution which is an elastic solution (See Civiltech Software Design Manual and Foundation Design by Wayne C. Teng 1962).</p>
<p>1972 &#8211; Rehnman and Broms showed that when the soil behind the wall was dense the lateral pressure from point loads was much less than with loose soils. Also gravelly backfill&#8217;s produced larger lateral pressures than finer-grained soils. This observation would mean soil state and Poisson&#8217;s ratio are significant.</p>
<h2> When and How to Apply Elastic Methods</h2>
<p>First the how &#8211; The equations presented below give pressures at a certain height on the wall. Therefore it is suggested that a program or spreadsheet is made that uses these equations to find pressures at set intervals or segments of the wall (sigmah). Then the force is found by multiplying the height of this segment by the pressure (Fi=force at elevation &#8216;i&#8217; = sigmah * hseg). Then you solve for the height above the bottom of the wall to this segment (yi = height to force, Fi). Then you solve for the resultant force and it elevation. Fh=sum(Fi&#8217;s) Then find the elevation Ybar = Sum(yi&#8217;s * Fi&#8217;s)/Sum(Fi&#8217;s). This is similar to finding the centroid of an area by parts. This pressure/force is then superimposed on the soil pressure.</p>
<h3>Zone of Influence</h3>
<p>Second the when &#8211; This is a much more difficult question to answer. There are many different ways that an offset surcharge maybe handled see <a title="Approaches to Analyzing Surcharges" href="http://wp.me/p2Puq4-4h">here</a> for a more indepth look. So if we are specifically looking at when to apply elastic methods the answer would be &#8211; anytime. We can superimpose the results of the elastic method results on to the soil pressure as would be found using Rankine or Coulomb equations (see <a title="Retaining Wall - Lateral Earth Pressure" href="http://wp.me/p2Puq4-Y">here</a> for refresher). However elastic methods will yield a horizontal force for any surcharge applied at any distance behind the retaining wall. This may be overly conservative however it is up to the engineer to determine what distance beyond the wall that the surcharge would have no affect on the retaining wall. Usually this information is given in the governing code. Some examples &#8211; many use the Rankine or Coulomb failure plane as the determining distance however a trial wedge may give different results when finding the failure plane angle. Edward White has suggested that pressures may distribute down at 1V:1H (See Foundation Engineering Handbook by Hans F. Winterkorn and Fang). Sometimes a 2V: 1H is used as in a Meyerhoff or Boussinesq bearing pressure distribution. While a 2V:1H has been found accurate at short depths (See Bowles 3rd Edition p172) and is usually conservative when finding vertical stress at a certain depth but can be unconservation when finding horizontal stresses. Most rail road design manuals suggest at 1.5H:1V influence line. The NCMA recognizes a 2H:1V as a conservative estimate to find the horizontal distance to which you may disregard surcharge loads.</p>
<h2>Elastic Method Equations</h2>
<p>Below are the &#8216;typical&#8217; equations used in most design manuals for using elastic methods. There are links to other posts which will give further discussion, examples and spreadsheet calculations. The equations assume rigid walls (pressures maybe less for flexible walls, Civiltech Software recommends 0.5 for flexible, 0.75 for semi-flexible and 1.0 for rigid &#8211; see further discussion), a Poisson&#8217;s ratio of 0.5, and the pressure maybe combined by method of superposition.<br />
Nomenclature:<br />
σh= horizontal pressure<br />
H = Height of wall (excavation)<br />
x1 = m*H = distance to surcharge (point load, line load, strip load)<br />
zi = n*H = distance from top of wall to elevation under consideration<br />
Q = point load or line load surcharge<br />
q = uniform surcharge</p>
<h2>Point Loads</h2>
<div id="attachment_491" class="wp-caption alignnone" style="width: 310px"><a href="http://howtoengineer.com/wp-content/uploads/2012/11/Point-Load-Section.jpg"><img class="size-medium wp-image-491" title="Point Load Section" alt="Point Load Section" src="http://howtoengineer.com/wp-content/uploads/2012/11/Point-Load-Section-300x204.jpg" width="300" height="204" /></a><p class="wp-caption-text">Point Load Section</p></div>
<div id="attachment_493" class="wp-caption alignnone" style="width: 282px"><a href="http://howtoengineer.com/wp-content/uploads/2012/11/Point-Load-Plan-View1.jpg"><img class="size-full wp-image-493" title="Point Load Plan View" alt="Point Load Plan View" src="http://howtoengineer.com/wp-content/uploads/2012/11/Point-Load-Plan-View1.jpg" width="272" height="256" /></a><p class="wp-caption-text">Point Load Plan View</p></div>
<p>Boussinesq equation as modified by Terzaghi.</p>
<p><img src="//s0.wp.com/latex.php?latex=For+%5Cquad+m+%3E+0.4%3A+%5Csigma_h+%3D+%5Cfrac%7B1.77Q%7D%7BH%5E2%7D+%5Cfrac%7Bm%5E2+n%5E2%7D%7B%28m%5E2%2Bn%5E2%29%5E3%7D&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="For &#92;quad m &gt; 0.4: &#92;sigma_h = &#92;frac{1.77Q}{H^2} &#92;frac{m^2 n^2}{(m^2+n^2)^3}" title="For &#92;quad m &gt; 0.4: &#92;sigma_h = &#92;frac{1.77Q}{H^2} &#92;frac{m^2 n^2}{(m^2+n^2)^3}" class="latex" /></p>
<p><img src="//s0.wp.com/latex.php?latex=For+%5Cquad+m+%5Cleq+0.4%3A+%5Csigma_h+%3D+%5Cfrac%7B0.28Q%7D%7BH%5E2%7D+%5Cfrac%7Bn%5E2%7D%7B%280.16%2Bn%5E2%29%5E3%7D&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="For &#92;quad m &#92;leq 0.4: &#92;sigma_h = &#92;frac{0.28Q}{H^2} &#92;frac{n^2}{(0.16+n^2)^3}" title="For &#92;quad m &#92;leq 0.4: &#92;sigma_h = &#92;frac{0.28Q}{H^2} &#92;frac{n^2}{(0.16+n^2)^3}" class="latex" /></p>
<p>To evaluate a point at an angle to the point load along the wall:</p>
<p><img src="//s0.wp.com/latex.php?latex=%5Csigma%27_h+%3D+%5Csigma_h+cos%5E2%281.1%5Ctheta%29&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="&#92;sigma&#039;_h = &#92;sigma_h cos^2(1.1&#92;theta)" title="&#92;sigma&#039;_h = &#92;sigma_h cos^2(1.1&#92;theta)" class="latex" /></p>
<p>See <a title="Surcharge - Elastic Methods - Point and Line Loads" href="http://wp.me/p2Puq4-7P">here</a> for a thorough discussion</p>
<h2>Line Loads</h2>
<div id="attachment_494" class="wp-caption alignnone" style="width: 310px"><a href="http://howtoengineer.com/wp-content/uploads/2012/11/Line-Load-Section.jpg"><img class="size-medium wp-image-494" title="Line Load Section" alt="Line Load Section" src="http://howtoengineer.com/wp-content/uploads/2012/11/Line-Load-Section-300x279.jpg" width="300" height="279" /></a><p class="wp-caption-text">Line Load Section</p></div>
<p><img src="//s0.wp.com/latex.php?latex=For+%5Cquad+m+%3E+0.4%3A+%5Csigma_h+%3D+%5Cfrac%7B4Q%7D%7B%5Cpi+H%7D%5Cfrac%7Bm%5E2+n%7D%7B%28m%5E2%2Bn%5E2%29%5E2%7D&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="For &#92;quad m &gt; 0.4: &#92;sigma_h = &#92;frac{4Q}{&#92;pi H}&#92;frac{m^2 n}{(m^2+n^2)^2}" title="For &#92;quad m &gt; 0.4: &#92;sigma_h = &#92;frac{4Q}{&#92;pi H}&#92;frac{m^2 n}{(m^2+n^2)^2}" class="latex" /></p>
<p><img src="//s0.wp.com/latex.php?latex=For+%5Cquad+m+%5Cleq+0.4%3A+%5Csigma_h+%3D+%5Cfrac%7B0.203Q%7D%7BH%7D+%5Cfrac%7Bn%7D%7B%280.16%2Bn%5E2%29%5E2%7D&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="For &#92;quad m &#92;leq 0.4: &#92;sigma_h = &#92;frac{0.203Q}{H} &#92;frac{n}{(0.16+n^2)^2}" title="For &#92;quad m &#92;leq 0.4: &#92;sigma_h = &#92;frac{0.203Q}{H} &#92;frac{n}{(0.16+n^2)^2}" class="latex" /></p>
<p>See <a title="Surcharge - Elastic Methods - Point and Line Loads" href="http://wp.me/p2Puq4-7P">here</a> for a thorough discussion</p>
<h2>Strip Loads</h2>
<div id="attachment_495" class="wp-caption alignnone" style="width: 301px"><a href="http://howtoengineer.com/wp-content/uploads/2012/11/Strip-Load-Section.jpg"><img class="size-full wp-image-495" title="Strip Load Section" alt="Strip Load Section" src="http://howtoengineer.com/wp-content/uploads/2012/11/Strip-Load-Section.jpg" width="291" height="292" /></a><p class="wp-caption-text">Strip Load Section</p></div>
<p><img src="//s0.wp.com/latex.php?latex=%5Csigma_h+%3D+%5Cfrac%7B2Q%7D%7B%5Cpi%7D+%28%5Cbeta-sin%28%5Cbeta%29cos%282%5Calpha%29%29&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="&#92;sigma_h = &#92;frac{2Q}{&#92;pi} (&#92;beta-sin(&#92;beta)cos(2&#92;alpha))" title="&#92;sigma_h = &#92;frac{2Q}{&#92;pi} (&#92;beta-sin(&#92;beta)cos(2&#92;alpha))" class="latex" /></p>
<p>See <a title="Surcharge Analysis - Elastic Methods - Strip Loads" href="http://wp.me/p2Puq4-8a">here</a> for further discussion.</p>
<h2>Area Loads</h2>
<p>Will update shortly</p>
<p>The post <a href="http://howtoengineer.com/offset-surcharge-elastic-methods/">Surcharge Analysis &#8211; Elastic Methods</a> appeared first on <a href="http://howtoengineer.com">How To Engineer</a>.</p>]]></content:encoded>
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