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Villanova University Dept. of Civil & Environmental Engineering CEE 4606 - Capstone II Structural Engineering 1 CEE 4606 - Capstone II Structural Engineering.

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Presentation on theme: "Villanova University Dept. of Civil & Environmental Engineering CEE 4606 - Capstone II Structural Engineering 1 CEE 4606 - Capstone II Structural Engineering."— Presentation transcript:

1 Villanova University Dept. of Civil & Environmental Engineering CEE 4606 - Capstone II Structural Engineering 1 CEE 4606 - Capstone II Structural Engineering Lecture 10 – Masonry Shear Walls

2 Villanova University Dept. of Civil & Environmental Engineering CEE 4606 - Capstone II Structural Engineering 2 Time for a Pop Quiz… You should be familiar with… 1.Differences between unreinforced and reinforced masonry 2.MSJC code 3.Components of masonry 4.Properties/manufacturing of CMU 5.Components of mortar

3 Villanova University Dept. of Civil & Environmental Engineering CEE 4606 - Capstone II Structural Engineering 3 Time for a Pop Quiz… You should be familiar with… 6.Types of mortar 7.Proportion and property specifications for mortar 8.Differences between mortar and grout 9.Types of reinforcement 10.Definition of f’ m

4 Villanova University Dept. of Civil & Environmental Engineering CEE 4606 - Capstone II Structural Engineering 4 Time for a Pop Quiz… You should be familiar with… 11.Unit strength method vs. prism test method 12.Bond patterns for masonry 13.Gross, bedded, and net areas

5 Villanova University Dept. of Civil & Environmental Engineering CEE 4606 - Capstone II Structural Engineering 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 Villanova University Dept. of Civil & Environmental Engineering CEE 4606 - Capstone II Structural Engineering 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 Villanova University Dept. of Civil & Environmental Engineering CEE 4606 - Capstone II Structural Engineering 7 Reinforced Shear Walls Shear provisions are a function of the M/Vd ratio (essentially the h/d ratio) h d H VMVM h H VMVM d Flexure likely to control Shear likely to conrol

8 Villanova University Dept. of Civil & Environmental Engineering CEE 4606 - Capstone II Structural Engineering 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 Villanova University Dept. of Civil & Environmental Engineering CEE 4606 - Capstone II Structural Engineering 9 Reinforced Shear Wall Example 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. 8’-0” 6’-8” H (wind) 80” Avg. d=72”

10 Villanova University Dept. of Civil & Environmental Engineering CEE 4606 - Capstone II Structural Engineering 10 Reinforced Shear Wall Example (continued) Assume wall has no horizontal shear reinforcement Assume wall is underreinforced (technically must check) M = A s F s jd (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 F v = (3000) 1/2 = 55 psi > 35 psi F v = (4/3)(35) = 46.7 psi V max = bdF v = (7.625)(72)(46.7) H = 25.6 kips Shear governs ---> H=25.6 kips 8’-0” 6’-8” H 80” Avg. d=72”

11 Villanova University Dept. of Civil & Environmental Engineering CEE 4606 - Capstone II Structural Engineering 11 Reinforced Shear Wall Example (continued) Assume wall has sufficient horizontal steel to take full shear corresponding to flexural capacity M/Vd = 96/72 =1.333 > 1.0 F v = 1.5(3000) 1/2 = 82 psi > 75 psi V max = F v bd = [(4/3)(75)](7.625)(72) = 54.9 kips H = 34.1 kips < 54.9 OK A v /s = V max /F s d = 34.1/[[(1.333)(24)](72)] = 0.0148 sq. in. per inch With 2 #4 bars, A v =0.40 in. 2 s =.40/.0148 = 27.0” -----> Use 2#4 @ 24” spacing 8’-0” 6’-8” H 80” Avg. d=72”

12 Villanova University Dept. of Civil & Environmental Engineering CEE 4606 - Capstone II Structural Engineering 12 Reinforced Shear Wall Example (continued) 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 8’-0” 6’-8” H 80” Avg. d=72”


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