CEE 4606 - Capstone II Structural Engineering

1. CEE 4606 - Capstone IIStructural Engineering Lecture 10 – Masonry Shear Walls

2. Time for a Pop Quiz… You should be familiar with… • Differences between unreinforced and reinforced masonry • MSJC code • Components of masonry • Properties/manufacturing of CMU • Components of mortar

3. Time for a Pop Quiz… You should be familiar with… • Types of mortar • Proportion and property specifications for mortar • Differences between mortar and grout • Types of reinforcement • Definition of f’m

4. Time for a Pop Quiz… You should be familiar with… • Unit strength method vs. prism test method • Bond patterns for masonry • Gross, bedded, and net areas

5. Shear Walls • Essentially act as vertical beams that resist gravity and in-plane lateral loads • Generally, shear walls must be checked for flexure and shear • Deflections (stiffness) may also be critical • difficult, consider both shear and flexural displacements • see text

6. Unreinforced Shear Walls • Generally not very efficient • Limited tension capability • Work best with high axial loads • Basic Design Criteria: • Net Tensile Stress (Flexural – Axial) • Net Compressive Stress (Flexural + Axial) • Shear Stress

7. Reinforced Shear Walls • Shear provisions are a function of the M/Vd ratio (essentially the h/d ratio) d H Flexure likely to control d H h h Shear likely to conrol V M V M

8. Reinforced Shear Walls – Allowable Shear Stress Provisions (MSJC 2.3.5.2.2) • For walls with in-plane flexural reinforcement and no shear reinforcement • For walls with in-plane flexural reinforcement and shear reinforcement to resist the full shear

9. Reinforced Shear Wall Example 80” • Fully grouted 8” (nominal) CMU wall • Type S mortar, f’m=3000 psi • The wall has no axial stress (ignore self-weight) • Vertical reinforcement is 2#8 at each end • Determine the maximum horizontal load H that can be applied to the wall. Avg. d=72” 6’-8” H (wind) 8’-0”

10. Reinforced Shear Wall Example (continued) 80” • Assume wall has no horizontal shear reinforcement • Assume wall is underreinforced (technically must check) • M = AsFsjd (Assume j=0.9) • M = [(2)(.79)][(1.333)(24)](.9)(72) • M = 3275 in.-kips • H = M/h = 3275/96 = 34.1 kips • M/Vd = 96/72 =1.333 > 1.0 • Fv = (3000)1/2 = 55 psi > 35 psi • Fv = (4/3)(35) = 46.7 psi • Vmax = bdFv = (7.625)(72)(46.7) • H = 25.6 kips • Shear governs ---> H=25.6 kips Avg. d=72” 6’-8” H 8’-0”

11. Reinforced Shear Wall Example (continued) 80” • Assume wall has sufficient horizontal steel to take full shear corresponding to flexural capacity • M/Vd = 96/72 =1.333 > 1.0 • Fv = 1.5(3000)1/2 = 82 psi > 75 psi • Vmax = Fvbd = [(4/3)(75)](7.625)(72) • = 54.9 kips • H = 34.1 kips < 54.9 OK • Av/s = Vmax/Fsd • = 34.1/[[(1.333)(24)](72)] • = 0.0148 sq. in. per inch • With 2 #4 bars, Av=0.40 in.2 • s = .40/.0148 = 27.0” • -----> Use 2#4 @ 24” spacing Avg. d=72” 6’-8” H 8’-0”

12. Reinforced Shear Wall Example (continued) 80” • Summary • Only vertical reinforcement (2#8) • H = 25.6 kips (shear governs) • Vertical reinforcement (2#8) and horizontal reinforcement (2#4 @ 24”): • H = 34.1 kips Avg. d=72” 6’-8” H 8’-0”