1. The following simply supported beam was shown to be adequate in flexure: fé ' = 4.5 ksi, normal weight concrete and fy = 60 ksi, As = 9.41 in², P = 0 for part (1) and (2) (1) Find V and V using ACI318-19 Table 22.5.5.1 and Table 21.2.1, then compare with V @ d from support. Are stirrups needed? (2) Find V at 15 ft from end. What % of the value found in part (1) is this value? Are stirrups needed? (3) Find V if Pu is not zero, but is (a) 150 k compression; (b) 150 k tension. including beam wit D=1.6ITK/ft L-2.650k/ft Pu As=9.41in² 38'span Pu

Structural Analysis
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Author:KASSIMALI, Aslam.
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Chapter2: Loads On Structures
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1. The following simply supported beam was shown to be adequate in flexure:
fé
' = 4.5 ksi, normal weight concrete and fy = 60 ksi, As
= 9.41 in², P = 0 for part (1) and (2)
(1) Find V and V using ACI318-19 Table 22.5.5.1 and Table 21.2.1, then compare with V @ d from
support. Are stirrups needed?
(2) Find V at 15 ft from end. What % of the value found in part (1) is this value? Are stirrups needed?
(3) Find V if Pu is not zero, but is (a) 150 k compression; (b) 150 k tension.
including beam wit
D=1.6ITK/ft L-2.650k/ft
Pu
As=9.41in²
38'span
Pu
Transcribed Image Text:1. The following simply supported beam was shown to be adequate in flexure: fé ' = 4.5 ksi, normal weight concrete and fy = 60 ksi, As = 9.41 in², P = 0 for part (1) and (2) (1) Find V and V using ACI318-19 Table 22.5.5.1 and Table 21.2.1, then compare with V @ d from support. Are stirrups needed? (2) Find V at 15 ft from end. What % of the value found in part (1) is this value? Are stirrups needed? (3) Find V if Pu is not zero, but is (a) 150 k compression; (b) 150 k tension. including beam wit D=1.6ITK/ft L-2.650k/ft Pu As=9.41in² 38'span Pu
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