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6-13. Draw the influence lines for (a) the moment at C, (b) the shear just to the right of the support at B, and (c) the vertical reaction at B. Solve this problem using the basic method of Sec. 6.1. Assume A is a pin and B is a roller. A 1m-1m+ →1m C 3 m B

6-13. Draw the influence lines for (a) the moment at C, (b) the shear just to the right of the support at B, and (c) the vertical reaction at B. Solve this problem using the basic method of Sec. 6.1. Assume A is a pin and B is a roller. A 1m-1m+ →1m C 3 m B

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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For the solution for this problem, I was wondering how they got 0.75s and the 0.25 in the graph? I thought in the muller method you replace the point of interest with 1 unit load up? Can you please show all calculations so I can see where the 0.75s and 0.25 are coming from
6-13. Draw the influence lines for (a) the moment at C, (b) the shear just to the right of the support at B, and (c) the vertical reaction at B. Solve this problem using the basic method of Sec. 6.1. Assume A is a pin and B is a roller. A 1m 1m- 1m+ 23#UTRU#12 C 3 m Probs. 6-13/14 B
Transcribed Image Text:6-13. Draw the influence lines for (a) the moment at C, (b) the shear just to the right of the support at B, and (c) the vertical reaction at B. Solve this problem using the basic method of Sec. 6.1. Assume A is a pin and B is a roller. A 1m 1m- 1m+ 23#UTRU#12 C 3 m Probs. 6-13/14 B
Step2 b) (1) Moment at 'e' ILD. Provide a at C. After that draw internal hinge at 'c', then apply unit prositive moment deflected shape of the beam. This deflected structure will be ILD for Moment at c. 1x3 0.75 0.75= ↑ 0.75 * Im + Im + 3m 0.75 (+) +1m + Im ✓ curve (B) +3m fig:- ILD for Me. ILD for Shear Force just right of support B: - Provide a vertical slider just right of support B. Then apply unit positive S.F at the vertical slider. Draw the deflected shape of the structure. For determinate structure, if deflected shape has curve portion, then it should not be included in ILD. & Straight line. B 1.0 1m- 7}=0.25 Im (+) 0.25 fig:- :- ILD for (SF) just right of support B.
Transcribed Image Text:Step2 b) (1) Moment at 'e' ILD. Provide a at C. After that draw internal hinge at 'c', then apply unit prositive moment deflected shape of the beam. This deflected structure will be ILD for Moment at c. 1x3 0.75 0.75= ↑ 0.75 * Im + Im + 3m 0.75 (+) +1m + Im ✓ curve (B) +3m fig:- ILD for Me. ILD for Shear Force just right of support B: - Provide a vertical slider just right of support B. Then apply unit positive S.F at the vertical slider. Draw the deflected shape of the structure. For determinate structure, if deflected shape has curve portion, then it should not be included in ILD. & Straight line. B 1.0 1m- 7}=0.25 Im (+) 0.25 fig:- :- ILD for (SF) just right of support B.
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