Two Way (Punching) shear capacity question

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CE_Gator

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1. Factor Loads 1.2*372+1.6*117=634 kips (634,000 lbs)

2. Find Area resisting shear = width of column + 2(d/2) = 22+2(20.5/2)=(42.5 in.)^2=1806 in^2

3. Find R (upward force from earth) R=Pu*B1*B2/Af =(634000*42.5*42.5)/114^2=88116 lbs

4. Punching Shear Vu, Vu=Pu-R/Area of Shear = (634000-88116)/1806=302.22 lbs/in^2

Ans. A

 
1. Factor Loads 1.2*372+1.6*117=634 kips (634,000 lbs)

2. Find Area resisting shear = width of column + 2(d/2) = 22+2(20.5/2)=(42.5 in.)^2=1806 in^2

3. Find R (upward force from earth) R=Pu*B1*B2/Af =(634000*42.5*42.5)/114^2=88116 lbs

4. Punching Shear Vu, Vu=Pu-R/Area of Shear = (634000-88116)/1806=302.22 lbs/in^2
The area resisting shear is a vertical area (visualize the 4 walls with height = d along the perimeter of the critical section 42.5 x 42.5. Step 2 uses a horizontal (plan) area

 
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