show work and value of fy 60000 psi Problem FE 3 A simply su

show work and value of fy: 60,000 psi

Problem FE 3. A simply supported beam has a clear span22 ft and is subjected to an external uniform nominal dead load of 0.9kips/ft (does mot include the weight of the beam) and a uniformly distributed nominal live load of 1.2 kips ft. Assume = 4,000 psi, and fy 12 in 20 in. 17 in. 60,000 psi. 3.5 The length stirrups are needed in each direction is most likely a) 8.1 ft b) 11.0 ft c) 5.5 ft d) 22.0 ft 3.6 Assuming #3 stirrups the maximum spacing of the stirrups at the critical section is a) 13.8 in b) 8.5 in c) 12 in d) Not enough information 3.7 Assuming #4 stirrups the maximum spacing of the stirrups at the critical section is a) 12.0 in b) 25.0 in c) 8.5 in d) Not enough information

Solution

3.5)Dead load on beam = 0.9kip/ft

self weight of beam = 150*(12/12)*(20/12)/1000=0.25 kip/ft

Total dead load = 0.9+0.25=1.15 kip/ft

Live load = 1.2 kip/ft

Factored dead load = 1.2*1.15=1.38 kip/ft

Factored live load = 1.6*1.2=1.92 kip/ft

Total design load on beam = 1.38+1.92=3.3 kip/ft

shear strength of section without rebar = 2*sqrt(f\'c)*b*d=2*sqrt(4000)*12*17/1000= 25.8 kips

Design shear strength = 0.75*25.8=19.35 kips

stirrups will be required at the section where shear force at the section is greater than 0.5 times the shear strength of concrete alone

Therefore, stirrups will be required at sections where design shear force>0.5*19.35=9.68 kips

Let the section be at a distnace x from left support where shear force =9.68 kips

3.3*(22/2) - 3.3x = 9.68

x=8 ft

Correct option is (a)

3.6) Maximum shear force in beam = 3.3*(22/2)=36.3 kips

shear strength of concrete=19.35 kips

shear force for which stirrups have to be designed = 36.3-19.35=16.95 kips

Area of stirrups = 0.11*2=0.22 in²

16.95=0.75*60*0.22*17/s

s=9.93 in

Correct option is option (b)

3.7)Maximum shear force in beam = 3.3*(22/2)=36.3 kips

shear strength of concrete=19.35 kips

shear force for which stirrups have to be designed = 36.3-19.35=16.95 kips

Area of stirrups = 0.2*2=0.4 in²

16.95=0.75*60*0.4*17/s

s=18 in

Correct option is option (a)