For the following beams, draw the simplified free-body diagram and below it the shear and bending moment diagrams. Show all necessary solutions. Let the analysis be from left to right. Determine the points of zero shear (if any), their distance from point A, and the bending moments at these points. Indicate the value of all the points, degree of all curves, and the maximum shear and maximum bending moment of the beam. Use the method of sections. Beam ABCD has a hanger connected at B and is supported by a roller at A and a pin at C. Provide solution and diagram for each section.
For the following beams, draw the simplified free-body diagram and below it the shear and bending moment diagrams. Show all necessary solutions. Let the analysis be from left to right. Determine the points of zero shear (if any), their distance from point A, and the bending moments at these points. Indicate the value of all the points, degree of all curves, and the maximum shear and maximum bending moment of the beam. Use the method of sections. Beam ABCD has a hanger connected at B and is supported by a roller at A and a pin at C. Provide solution and diagram for each section.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:For the following beams, draw the simplified free-body diagram and below it the shear and bending
moment diagrams. Show all necessary solutions. Let the analysis be from left to right. Determine
the points of zero shear (if any), their distance from point A, and the bending moments at these
points. Indicate the value of all the points, degree of all curves, and the maximum shear and
maximum bending moment of the beam.
Use the method of sections. Beam ABCD has a hanger connected at B and is supported by a
roller at A and a pin at C. Provide solution and diagram for each section.
Parameters
2 3 (4)
ABCDE
2
0
2
1
-
1 2
3 4 5
L1
=
L1 =
L2 = 20 +(2+3)
= 200+ (A + B) x 2 N·m
L2= 20+ (B+C) x 2 N/m
L3= 200 + (C + D) x 2 N/m
L4 = 75+ (C + D) x 2 N/m
M = 80+ (D+E) N.m
200+ (1+2) x 2 = 206 N/m
x 2 = 30 N/m
L3= 200+ (3+4) x 2 = 214 N/m
L4 = 75+ (3+4) x 2 = 89 N/m
M = 80 +(4+5) = 89 N·m
100 N
L1
M
B
L2
3 m
3 m
1.5 m
Provide a summary of the equations in the following format.
Simplified Equation
Exploratory Distance
<shear equation>
x=<min value>
<bending moment equation>
x= <max value>
2.5 m
Shear (N)
L4
D
L3
Bending Moment (N-m)
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May I ask where did 45 N in step 1 come from?
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