A 2.5' Ay Vc when unit load is egment AC: A Ay = 1- 2.5' C X 10 M 7.5' с Msection = 0 X 10 -Vc + - (1 - X 10 VCAC = MB = 0 1 (10-x) - Ay (10) = 0 X By Ay = 1- 10 Equation of Vc when unit load is located at Segment CB: 1 Į M C -) -1=0 C Vc + M section=0 X 10 VCB + VCB = 1- 7.5' -1=0 X 10 X B By= x 10
A 2.5' Ay Vc when unit load is egment AC: A Ay = 1- 2.5' C X 10 M 7.5' с Msection = 0 X 10 -Vc + - (1 - X 10 VCAC = MB = 0 1 (10-x) - Ay (10) = 0 X By Ay = 1- 10 Equation of Vc when unit load is located at Segment CB: 1 Į M C -) -1=0 C Vc + M section=0 X 10 VCB + VCB = 1- 7.5' -1=0 X 10 X B By= x 10
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Construct the influence line for the shear at point C. The x in the table is from point A.
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