Problem 3 (30 pts) The beam shown below is subjected to unfactored service live

Question

Problem 3 (30 pts) The beam shown below is subjected to unfactored service live loads, PLI and P12 (1) Compute the stirrup spacing required by ACI for beam section in the region B-C. Neglect the self weight of the beam. [10 pts (2) Does the #3 stumps with the spacing determined in the previous step meet the ACI minimum transverse stirrup requirement? Check it only in the region B-C. [10 pts] (3) Compute the shear strength provided by the stirrups if the spacing is 4 in. [10 pts] h srengsh provided by he st Use = 4 ksi and fy = 60 ksi. Be sure to factor the loads to compute shear force. be 20 in 5 in d- 215 in. #3 closed stirrups b-10 in PLI- 25 kips PL: 37.5 kips 5 ft 5 ft.

Explanation / Answer

Let us determine the vertical reaction force at support C due to service live loads

RC = (25/2)+(37.5*17/10)=76.25 kips

Service level Shear force in region BC = RC-37.5=76.25-37.5=38.75 kips

Factored shear force in region BC = 1.6*38.75=62 kips

1) beam web width = 10 in

beam effective depth = 21.5 in

grade of concrete = 4000 psi

nominal shear strength of concrete=2*sqrt(4000)*21.5*10=27.2 kips

Factored shear strength of concrete = 0.75*27.2=20.4 kips

The shear force for which stirrups have to be designed = 62-20.4=41.6 kips

Area of stirrup = 2*0.11=0.22 in2

grade of steel = 60 ksi

Let spasing required be s

41.6=0.75*60*0.22*21.5/s

s=5.11"

provide stirrups at 5" spacing

2) minimum shear reinforcement per ACI = 0.75*sqrt(4000)*10*5/60000=0.039 in2 or 50*10*5/60000=0.04 in2

Shear reinforcement provided = 0.22 in2 is greater than minimum shear reinforcement per ACI

3)If spacing is 4 in,shear strength provided by stirrups

=0.75*60*0.22*21.5/4=53.2 kips

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

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