1. A rectangular beam is 15 in. wide and 30 in. deep. For f = 4000 psi and f₁ = 60,000 psi, determine the required spacing of No. 4 (No. 13) closed stirrups for a factored shear of 80 kips and factored torsional moment of 50 ft-kips. The stirrup centroid is located 1.75 in. from each concrete face. The effective depth is 26.5 in. Given parameters: P=0.85, E=29000000 psi, E₁ = 57000 √√f psi = 3604997 psi V = 80 kip T=50 ft-kip 0,= 0.90 %=0.75 Sy Ey = = 0.002 E 2=1.0 Tip: To estimate the required spacing, the required steel area due to torsion and due to shear must be combined. Therefore, 2A+ + A₁ = As No.4. Using the equations in the textbook, A+ shall be estimated in function of s. I.e., A₁ = xxx.yy × s. In principle, A must be estimated from shear analysis, but to save time it will be given as Av S = 0.0355m² in Xo: 6·2·1.75=11.5 40:6-2·1.75=26.5 Aoh: XOYO = 304.75 Ph: 2x0tYO) : 76 = Ao = Aoh-0.85-759.04 Using torsion form, At/s = 0.02574 in²/in Avls: 0.0355in/i0 2 Alls + Av/s = 0.08697 in²/0 2.0.2-0.4), Since stirrup area => 2.0.2 = (0.4), consequent S= 4.60in So, s = Yin ". This is ( max spacing for torsion, thus
1. A rectangular beam is 15 in. wide and 30 in. deep. For f = 4000 psi and f₁ = 60,000 psi, determine the required spacing of No. 4 (No. 13) closed stirrups for a factored shear of 80 kips and factored torsional moment of 50 ft-kips. The stirrup centroid is located 1.75 in. from each concrete face. The effective depth is 26.5 in. Given parameters: P=0.85, E=29000000 psi, E₁ = 57000 √√f psi = 3604997 psi V = 80 kip T=50 ft-kip 0,= 0.90 %=0.75 Sy Ey = = 0.002 E 2=1.0 Tip: To estimate the required spacing, the required steel area due to torsion and due to shear must be combined. Therefore, 2A+ + A₁ = As No.4. Using the equations in the textbook, A+ shall be estimated in function of s. I.e., A₁ = xxx.yy × s. In principle, A must be estimated from shear analysis, but to save time it will be given as Av S = 0.0355m² in Xo: 6·2·1.75=11.5 40:6-2·1.75=26.5 Aoh: XOYO = 304.75 Ph: 2x0tYO) : 76 = Ao = Aoh-0.85-759.04 Using torsion form, At/s = 0.02574 in²/in Avls: 0.0355in/i0 2 Alls + Av/s = 0.08697 in²/0 2.0.2-0.4), Since stirrup area => 2.0.2 = (0.4), consequent S= 4.60in So, s = Yin ". This is ( max spacing for torsion, thus
Chapter9: Application Of Influence Lines
Section: Chapter Questions
Problem 11P
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Transcribed Image Text:1. A rectangular beam is 15 in. wide and 30 in. deep. For f = 4000 psi and f₁ = 60,000 psi,
determine the required spacing of No. 4 (No. 13) closed stirrups for a factored shear of 80 kips
and factored torsional moment of 50 ft-kips. The stirrup centroid is located 1.75 in. from each
concrete face. The effective depth is 26.5 in.
Given parameters:
P=0.85, E=29000000 psi, E₁ = 57000 √√f psi = 3604997 psi
V = 80 kip
T=50 ft-kip
0,= 0.90
%=0.75
Sy
Ey
=
= 0.002
E
2=1.0
Tip: To estimate the required spacing, the required steel area due to torsion and due to shear must
be combined. Therefore, 2A+ + A₁ = As No.4. Using the equations in the textbook, A+ shall be
estimated in function of s. I.e., A₁ = xxx.yy × s. In principle, A must be estimated from shear
analysis, but to save time it will be given as
Av
S
= 0.0355m²
in
Transcribed Image Text:Xo: 6·2·1.75=11.5
40:6-2·1.75=26.5
Aoh: XOYO = 304.75
Ph: 2x0tYO) : 76
=
Ao = Aoh-0.85-759.04
Using torsion form,
At/s = 0.02574 in²/in
Avls: 0.0355in/i0
2 Alls + Av/s = 0.08697 in²/0
2.0.2-0.4),
Since stirrup area => 2.0.2 = (0.4), consequent
S= 4.60in
So, s = Yin
". This is ( max spacing for torsion, thus
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