Homework Help is Here – Start Your Trial Now!
Engineering
Civil Engineering
2. A beam of rectangular cross section has b= 22 in. and h = 15 in. It is to carry a total factored load of 3600 lb/ft uniformly distributed over its 26 ft span, and in addition the beam will be subjected to a uniformly distributed torsion of 1800 ft-lb/ft at factored loads. Closed stirrup-ties will be used to provide for flexural shear and torsion, placed with the stirrup steel centroid 1.75 in. from each concrete face. The corresponding flexural effective depth will be approximately 12.5 in. Design the transverse reinforcement for this beam and calculate the increment of longitudinal steel area needed to provide for torsion, using f = 4000 psi and f₁ = 60,000 psi. Tip: For transverse reinforcement, use No. 4 bars. to save time it will be given as * = 0.0313 Vu = must be estimated from shear analysis, but wil and T₁ = To make the calculation simple, just consider the spacing required to carry the maximum V₂ and T₁ at d from the support (increasing spacing through the beam length can be skipped). The longitudinal reinforcement also needs to be calculated. Tur tu 1/2 = 23.4 ft-hips => Tuot2=73,4 -d·tu = 21.53 Ve= ver. 1/2 = 46.8 kips => Viatd = 46.8-2.14 Xo=18.5 Yo=11.5 Aoh: 212.75 A0 = 180.84 ph=60 At/s = 0.0159 in²/in =43.05 0.017253059 (wrong Tu value) Avis + 2A+1s = 0.063 in²/in S= 6.35 or bin I'Min spacing governed by d12 = 6.25 & chosen spacing is ( this value Using form for A. A. 0.952

2. A beam of rectangular cross section has b= 22 in. and h = 15 in. It is to carry a total factored load of 3600 lb/ft uniformly distributed over its 26 ft span, and in addition the beam will be subjected to a uniformly distributed torsion of 1800 ft-lb/ft at factored loads. Closed stirrup-ties will be used to provide for flexural shear and torsion, placed with the stirrup steel centroid 1.75 in. from each concrete face. The corresponding flexural effective depth will be approximately 12.5 in. Design the transverse reinforcement for this beam and calculate the increment of longitudinal steel area needed to provide for torsion, using f = 4000 psi and f₁ = 60,000 psi. Tip: For transverse reinforcement, use No. 4 bars. to save time it will be given as * = 0.0313 Vu = must be estimated from shear analysis, but wil and T₁ = To make the calculation simple, just consider the spacing required to carry the maximum V₂ and T₁ at d from the support (increasing spacing through the beam length can be skipped). The longitudinal reinforcement also needs to be calculated. Tur tu 1/2 = 23.4 ft-hips => Tuot2=73,4 -d·tu = 21.53 Ve= ver. 1/2 = 46.8 kips => Viatd = 46.8-2.14 Xo=18.5 Yo=11.5 Aoh: 212.75 A0 = 180.84 ph=60 At/s = 0.0159 in²/in =43.05 0.017253059 (wrong Tu value) Avis + 2A+1s = 0.063 in²/in S= 6.35 or bin I'Min spacing governed by d12 = 6.25 & chosen spacing is ( this value Using form for A. A. 0.952

Steel Design (Activate Learning with these NEW titles from Engineering!)
Steel Design (Activate Learning with these NEW titles from Engineering!)
6th Edition
ISBN:9781337094740
Author:Segui, William T.
Publisher:Segui, William T.
Chapter10: Plate Girders
Section: Chapter Questions
Problem 10.7.8P
See similar textbooks
icon
Related questions
Question

I have the correct answer provided, just lookng for a more detailed breadown of how the answer was obtained thanks.

2. A beam of rectangular cross section has b= 22 in. and h = 15 in. It is to carry a total factored load of 3600 lb/ft uniformly distributed over its 26 ft span, and in addition the beam will be subjected to a uniformly distributed torsion of 1800 ft-lb/ft at factored loads. Closed stirrup-ties will be used to provide for flexural shear and torsion, placed with the stirrup steel centroid 1.75 in. from each concrete face. The corresponding flexural effective depth will be approximately 12.5 in. Design the transverse reinforcement for this beam and calculate the increment of longitudinal steel area needed to provide for torsion, using f = 4000 psi and f₁ = 60,000 psi. Tip: For transverse reinforcement, use No. 4 bars. to save time it will be given as * = 0.0313 Vu = must be estimated from shear analysis, but wil and T₁ = To make the calculation simple, just consider the spacing required to carry the maximum V₂ and T₁ at d from the support (increasing spacing through the beam length can be skipped). The longitudinal reinforcement also needs to be calculated.
Transcribed Image Text:2. A beam of rectangular cross section has b= 22 in. and h = 15 in. It is to carry a total factored load of 3600 lb/ft uniformly distributed over its 26 ft span, and in addition the beam will be subjected to a uniformly distributed torsion of 1800 ft-lb/ft at factored loads. Closed stirrup-ties will be used to provide for flexural shear and torsion, placed with the stirrup steel centroid 1.75 in. from each concrete face. The corresponding flexural effective depth will be approximately 12.5 in. Design the transverse reinforcement for this beam and calculate the increment of longitudinal steel area needed to provide for torsion, using f = 4000 psi and f₁ = 60,000 psi. Tip: For transverse reinforcement, use No. 4 bars. to save time it will be given as * = 0.0313 Vu = must be estimated from shear analysis, but wil and T₁ = To make the calculation simple, just consider the spacing required to carry the maximum V₂ and T₁ at d from the support (increasing spacing through the beam length can be skipped). The longitudinal reinforcement also needs to be calculated.
Tur tu 1/2 = 23.4 ft-hips => Tuot2=73,4 -d·tu = 21.53 Ve= ver. 1/2 = 46.8 kips => Viatd = 46.8-2.14 Xo=18.5 Yo=11.5 Aoh: 212.75 A0 = 180.84 ph=60 At/s = 0.0159 in²/in =43.05 0.017253059 (wrong Tu value) Avis + 2A+1s = 0.063 in²/in S= 6.35 or bin I'Min spacing governed by d12 = 6.25 & chosen spacing is ( this value Using form for A. A. 0.952
Transcribed Image Text:Tur tu 1/2 = 23.4 ft-hips => Tuot2=73,4 -d·tu = 21.53 Ve= ver. 1/2 = 46.8 kips => Viatd = 46.8-2.14 Xo=18.5 Yo=11.5 Aoh: 212.75 A0 = 180.84 ph=60 At/s = 0.0159 in²/in =43.05 0.017253059 (wrong Tu value) Avis + 2A+1s = 0.063 in²/in S= 6.35 or bin I'Min spacing governed by d12 = 6.25 & chosen spacing is ( this value Using form for A. A. 0.952
Expert Solution
steps

Step by step

Solved in 2 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Similar questions
Recommended textbooks for you
Steel Design (Activate Learning with these NEW ti…
Steel Design (Activate Learning with these NEW ti…
Civil Engineering
ISBN:
9781337094740
Author:
Segui, William T.
Publisher:
Cengage Learning
SEE MORE TEXTBOOKS