3. (20 points) A simply supported beam with rectangular section (shown here) has

Question

3. (20 points) A simply supported beam with rectangular section (shown here) has a span of L-30 ft. A uniformly distributed dead load of wa 1.25 kips/ft (including its self- weight) and a live load of w1=2.3 kip/ft are applied to this beam. We want to design U shape stirrups for this beam using No. 4 bars with fy.-40 ksi. For this purpose, use fc- 4 ksi, d = 24, b=10", h = 27". Based on ACI requirements, first compute the ultimate uniform load (Wu) using basic load factors and then determine the followings for shear y o o o reinforcement of the beam: a. Location where no stirrup is required and also location where only stirrups with maximum spacing, Smax, are required. (5 points) b. Two different values for stirrup spacing (s) for half of this beam, one spacing from one end to L/4 and another spacing from L/4 to L/2. (15 points) You do not need to provide a sketch of stirrups along the beam.

Explanation / Answer

The ultimate load acting on beam = (1.2*1.25)+(1.6*2.3)=5.18 kip/ft

given f'c = 4 ksi

d=24", b=10", h=27"

shear strength of concrete alone = 2*sqrt(f'c)*b*d=2*sqrt(4000)*10*24/1000=30.3 kips

design shear strength of concrete =Vc= 0.75*30.2=22.65 kips

a)Per Cl 11.4.6.1, shear reinforcement is not required required in the location where Vu < 0.5Vc

this means ,in the beam,the region where shear force is less than 0.5*22.65 = 11.32 kips, no shear reinforcement is required

Also, maximum shear reinforcement spacing is required in the region where 0.5Vc<Vu<Vc , i.e., the maximum shear reinforcement spacing is required where shear force 11.32 kips<Vu<22.65 kips

The region where no shear reinforcement is required is in the region 2 ft on either side of the beam midspan,because the shear force in that region is less than11.32 kips

the region where shear reinforcement with maximum spacing is required is L/3 to L/2.5 and 0.6L to 2L/3

b) maximum shear force in the region from end of beam to L/4 = 5.18*15=77.7 kips

shear reinforcement should be designed for a force of 77.7-22.65=55.05 kips

given U shaped #4 stirrups with fyt = 40 ksi

Let spacing required be s

55.05 = 0.75*40*0.4*24/s

s = 5.2"

Provide 5" spacing of stirrups between end and L/4

maximum shear force in the region L/4 to L/2 = 38.85 kips

shear reinforcement should be deigned for a force of 38.85-22.65=16.2 kips

Let spacing required be s

16.2=0.75*40*0.4*24/s

s=17.77 in

d/2=24/2=12"

provide stirrups at 12" spacing in the region L/4 to L/2

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.