5.1 A post-tensioned bonded prestressed beam has the cross section shown in Figure P5.1. It has a span of 75 ft (22.9 m)

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answerhappygod
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5.1 A post-tensioned bonded prestressed beam has the cross section shown in Figure P5.1. It has a span of 75 ft (22.9 m)

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5 1 A Post Tensioned Bonded Prestressed Beam Has The Cross Section Shown In Figure P5 1 It Has A Span Of 75 Ft 22 9 M 1
5 1 A Post Tensioned Bonded Prestressed Beam Has The Cross Section Shown In Figure P5 1 It Has A Span Of 75 Ft 22 9 M 1 (32.42 KiB) Viewed 54 times
5 1 A Post Tensioned Bonded Prestressed Beam Has The Cross Section Shown In Figure P5 1 It Has A Span Of 75 Ft 22 9 M 2
5 1 A Post Tensioned Bonded Prestressed Beam Has The Cross Section Shown In Figure P5 1 It Has A Span Of 75 Ft 22 9 M 2 (27.03 KiB) Viewed 54 times
I NEED COMPLETE SOLUTION FOR THIS QUESTION IN DETAIL ON
A4 PAGES
YOU WILL HAVE TO SOLVE IT BY BOTH METHODS ASKED IN THE
QUESTION THAT IS (A) DETAILED DESIGN METHOD (B) ALTERNATIVE
METHOD
5.1 A post-tensioned bonded prestressed beam has the cross section shown in Figure P5.1. It has a span of 75 ft (22.9 m) and is subjected to a service superimposed dead load Wsp=450 plf (6.6 kN/m) and a superimposed service live load W₁ = 2,300 plf (33.6 kN/m). Design the web reinforcement neces- sary to prevent shear cracking (a) by the detailed design method and (b) by the alternative method at a section 15 ft (4.6 m) from the face of the support. The profile of the prestressing tendon is par- abolic. Use #3 stirrups in your design, and detail the section. The following data are given: A = 876 in.² (5,652 cm²) I = 433,350 in.* (18.03 x 10 cm¹) = 495 in.² (3,194 cm²) 48" (122 cm) Act Go t 18" (46 cm) #7" 3" 39" (99 cm)

c, = 25 in. (63.5 cm) S = 17,300 in.³ (2.83 x 10³ cm³) 38 in. (96.5 cm) сь = S 11,400 in.³ (1.86 x 105 cm³) = Wa = 910 plf (13.3 kN/m) ec = 32 in. (81.3 cm) e = 2 in. (5 cm) f = 5,000 psi (44.5 MPa), normal-weight concrete f = 3,500 psi (24.1 MPa) fy for stirrups = 60,000 psi (41.8 MPa) fp = 270,000 psi (1,862 MPa) low-relaxation strands fps = 243,000 psi (1,675 MPa) fpe = 157,500 psi (1,086 MPa) Aps = twenty-four-in. dia (12.7 mm dia) 7-wire tendons 5.2 Find the shear strengths V, V, and Vw for the beam in Problem 5.1 at 1/10 span intervals along the entire span, and plot the variations in their values along the span in a manner similar to the plot in Figure 5.13.
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