Masonry Walls – Partial Grout Out of Plane STR

It seems that most text decide to conveniently use a fully grouted wall. Well this is great but fully grouting walls can lead to unnecessary expenses and also add to seismic weight.  Therefore we will look at how to design and analyze a partially grouted wall. We did this once before for ASD so now we will look at Strength design for out-of-plane forces.

References

Ref 1 ACI 530/ASCE 5/TMS 402 direct number references are for the 2008 version however the method does not change up through the 2013 version. Found here

Ref 2 Masrony Structures Behavior and Design by Robert G. Drysdale, Hamid, and Baker 2nd Edition. (the Third edition found here)

Ref 3 Reinforced Masonry Engineering Handbook 6th Edition by Max Porter. Found here

Hand calc and TEDDS Calcs added.

Reinforced Masonry Out of Plane STR – Strain Compatibility Notes

Reinforced Masonry Wall Out-of-Plane Axial Bending and Shear – STR Tedds

Strength Design of masonry walls subject to axial and out of plane bending forces.

The Basics:

The design methodology is similar to reinforced concrete design. We assume the following:

1. Strain compatibility (linear) with a maximum allowable compressive strain.
2. Assume a compressive stress distribution (Compression Block) of the section.
3. The masonry has zero tensile strength and the reinforcement carries all tensile stress. The steel reinforcment has zero compressive strength unless it is tied.

References:

Ref 1 ACI 530/ASCE 5/TMS 402 direct number references are for the 2005 version however the method does not change up through the 2013 version. Found here

Ref 2 Masrony Structures Behavior and Design by Robert G. Drysdale, Hamid, and Baker 2nd Edition. (the Third edition found here)

Ref 3 Reinforced Masonry Engineering Handbook 6th Edition by Max Porter. Found here

 Design/Analysis:

Masonry Out of Plane ASD Diagram

0.) Given or assumed: Section Geometry, spacing /size of reinforcement and material strengths.

1.)    Assume depth to neutral axis (iterate this value), c.

• Because the wall is subject to a moment and axial force we can replace the moment and axial force with a single axial force that is eccentric from the centroid of the section. The eccentric distance is equal to the  Moment (M)/ Axial (P) force. We know that the internal forces must equal the external forces for the section to be in equilibrium. therefore because the external M and P are known we find er=M/P. We then set out to find the position of the neutral axis that will yield a section capacity strength with the same Moment / Axial eccentricity (ratio) as the applied forces.

2.)    Assume masonry strain is equal to ultimate, em=0.0025 (concrete masonry) em=0.0035 (clay masonry).

• The depth of the compression block is 0.8*c (depth of the neutral axis).

3.)    Find stress in steel based on compatibility which must be less than allowable. fs<Fs

• Draw the strain diagram and find the strain in the reinforcement from similar triangles.

Depth of compression block;           a=0.8*c
Effective flange thickness (strength , deflection);      t’_f=min(t_f,a)    t’_fD=min(t_f,c)
Effective height of web (strength, deflection);            y_cw=max(0in,a-t’_f)    y_cwD=max(0in,c-t’_fD)
Strain in steel based on compatibility;                         e_s=e_m/c*(d-c)
Stress in steel;                                             f’_s=E_s*e_s
Usable steel stress;                                         f_s=min(F_y,f’_s)
Compressive force on flange;                                       C_f=0.8*f’_m*t’_f*b
Compressive force on web;      C_w=0.8*f’_m *y_cw *b_w

5.)    Based on equilibrium sum moments and solve for Mns.

Nominal compressive force based on assumed c location and Maximum masonry strain;       P_nsT =C_f+C_w-T;

Nominal moment strength;                                                                                             M_nsT=M_cf+M_cw-M_s1;

Eccentricity of section;                                                                                             e_s=M_nsT/P_nsT;

6.)    Find cracked and effective moment of inertia Icr and Ie

7.)    Assume a deflection (iterate this value)

8.)    Find actual deflection.

• Deflection should be found for ultimate forces and service level forces. The service level deflection should be less than 0.007H (however 0.01 has been found acceptable by SEAOSC/ACI).

9.)    Find moment caused by deflection (iterate till convergent <5%)

10.) Find actual eccentricity of section ea=Mns/Pns.

11.) Find new eccentricity of loads e=(Mu+Mudeflection)/P

12.) Check if eccentricity of section = eccentricity of design loads (e=ea) if not iterate location of N.A. and deflection.

13.) Check Pns>P, Mns>M

14.) Check fa<Fa < where Fa<0.2*f’m or 0.05*f’m (H/d>30)

15.) Note that a slenderness reduction factor is not applied

Questions/Comments:

1.       The axial stress (Pu/Ag ) must be less than 0.2f’m. If the wall’s slenderness ratio exceeds 30 than the axial stress must be less than 0.05f’m.

2.       All walls are designed by an iterative process by checking the moment caused by the deflection due to out-of-plane loads and eccentricity of the DL/LL no matter what the slenderness ratio.

3.       The slenderness reduction factor does not need to be applied (MSJC eqn (3 – 16) and ( 3 – 17)  Discussion – It doesn’t seem like it should be applied as we are accounting for P-d and P-D effects. Also here are my reasons (I only have an older code in front of me ’02) Walls are addressed in section 3.2.5 where they address the limit to axial loads as 0.2f’m and 0.05f’m then stability is addressed with the P-D iterative method and they do not refer to eqn 3-16 and 3-17. However in section 3.2.6 they address walls for in-plane load where they do refer to eqns 3-16 and 3-17. Also in the reinforced masonry engineering handbook’s design example of a wall subject to out of plane forces they do not use the slenderness ratio. They use the iterative P-D Method.

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