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A V₁ = M₁= 20 4.88 (b) A point #₂ = V₂ = M₂= Note: Express your answers following the standard sign convention for internal shear force and internal bending moment, as described in the textbook. (a) A point #₁ = 10 ft to the right of point A. L₁ 18 ft to the right of point A. -12.32 L number (rtol=0.01, atol=1e-05) kip ? kip - ft kip kip-ft q ? B

A V₁ = M₁= 20 4.88 (b) A point #₂ = V₂ = M₂= Note: Express your answers following the standard sign convention for internal shear force and internal bending moment, as described in the textbook. (a) A point #₁ = 10 ft to the right of point A. L₁ 18 ft to the right of point A. -12.32 L number (rtol=0.01, atol=1e-05) kip ? kip - ft kip kip-ft q ? B

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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A V₁ = M₁ = 4.88 V₂ = M₂ = Note: Express your answers following the standard sign convention for internal shear force and internal bending moment, as described in the textbook. (a) A point #₁ = 10 ft to the right of point A. 20 LE F (b) A point #₂ = 18 ft to the right of point A. -12.32 L number (rtol=0.01, atol=1e-05) kip kip-ft kip kip - ft q ? B
Transcribed Image Text:A V₁ = M₁ = 4.88 V₂ = M₂ = Note: Express your answers following the standard sign convention for internal shear force and internal bending moment, as described in the textbook. (a) A point #₁ = 10 ft to the right of point A. 20 LE F (b) A point #₂ = 18 ft to the right of point A. -12.32 L number (rtol=0.01, atol=1e-05) kip kip-ft kip kip - ft q ? B
A beam supports both a uniform distributed load and a concentrated force, as shown in the figure below. The uniform distributed load has a magnitude q 400 lb/ft and extends across the entire beam. The T concentrated force has a magnitude F 14 kip and acts LF = 17 ft to the right of point A. The beam has an overall length L = 24 ft, and is held in equilibrium by a pin at point A and a roller at point B. Neglect the height and weight of the beam for this analysis. Determine the internal shear force (V) and internal bending moment (M) at the following locations: F A V₁ = 4.88 M. L₁ 20 L Note: Express your answers following the standard sign convention for internal shear force and internal bending moment, as described in the textbook. (a) A point x₁ = 10 ft to the right of point A. kip ? q kin ft B
Transcribed Image Text:A beam supports both a uniform distributed load and a concentrated force, as shown in the figure below. The uniform distributed load has a magnitude q 400 lb/ft and extends across the entire beam. The T concentrated force has a magnitude F 14 kip and acts LF = 17 ft to the right of point A. The beam has an overall length L = 24 ft, and is held in equilibrium by a pin at point A and a roller at point B. Neglect the height and weight of the beam for this analysis. Determine the internal shear force (V) and internal bending moment (M) at the following locations: F A V₁ = 4.88 M. L₁ 20 L Note: Express your answers following the standard sign convention for internal shear force and internal bending moment, as described in the textbook. (a) A point x₁ = 10 ft to the right of point A. kip ? q kin ft B
Expert Solution
Step 1

Given data 

q = 400lb/ft

F = 14 kips

Lf = 17ft

L = 24 ft

Find the shear force and bending moment at 10 ft and 18 ft.

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