Use Area Method. The beam subjected to loading as shown is pinned at D, has roller supports at A and E, and has an internal hinge at W. Draw the complete shear and bending moment diagram of the beam directly below its FBD. Plot the shear and bending moment values, indicate the degree of each curve, and determine the absolute maximum shear and bending moment. Summarize your answers using a table similar to the one shown below. Note that W is not a change-of-load point. 7 kN 7 kN-m 2 m 3 m 2 kN/m 3 m 42 kN-m 2 kN 6 m 2 kN/m

Structural Analysis
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Chapter2: Loads On Structures
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Use Area Method. The beam subjected to loading as shown is pinned at D, has roller
supports at A and E, and has an internal hinge at W. Draw the complete shear and
bending moment diagram of the beam directly below its FBD. Plot the shear and
bending moment values, indicate the degree of each curve, and determine the absolute
maximum shear and bending moment. Summarize your answers using a table similar to
the one shown below. Note that W is not a change-of-load point.
M₁
7 kN-m
2 m
7 kN
V₁-10x²
10
==
+
-x³
3 m
W
D
2 kN/m
+
V₂
= 400x 922.5
M₂ = 200x²922.5x +922.5
V3-100x+1650.723305
3 m
-
M3-50x² + 1650.723305x
-5667.728176
V₁ = -500x + 4850.723305
Figure 1. Given Problem
M₁ = -250x² + 4850.723305x
-17467.728176
42 kN-m
2 kN
6 m
x = 0
x = 1.5
x=8
x = 8
2 kN/m
x 1.5 V₂ = -322.5 N
x = 13
V₁ = ON
M₁ = ON m
Figure 2. Sample Table
3 m
x = 5.5 V₂ = 1277.5 N
M₂ 1898.75 N·m
=
x = 5.5 V₂ = 1100.7233 N
M3 1898.75 N·m
=
V₁ = -22.5 N
M₁ = -11.25 N·m
M₂=-11.25 N·m
V3 850.7233 N
=
=
M3 4338.0583 N•m
V4850.7233 N
M₁5338.0583 N·m
V4-1649.2767 N
M₁ = 3341.8748 N.m
Transcribed Image Text:Use Area Method. The beam subjected to loading as shown is pinned at D, has roller supports at A and E, and has an internal hinge at W. Draw the complete shear and bending moment diagram of the beam directly below its FBD. Plot the shear and bending moment values, indicate the degree of each curve, and determine the absolute maximum shear and bending moment. Summarize your answers using a table similar to the one shown below. Note that W is not a change-of-load point. M₁ 7 kN-m 2 m 7 kN V₁-10x² 10 == + -x³ 3 m W D 2 kN/m + V₂ = 400x 922.5 M₂ = 200x²922.5x +922.5 V3-100x+1650.723305 3 m - M3-50x² + 1650.723305x -5667.728176 V₁ = -500x + 4850.723305 Figure 1. Given Problem M₁ = -250x² + 4850.723305x -17467.728176 42 kN-m 2 kN 6 m x = 0 x = 1.5 x=8 x = 8 2 kN/m x 1.5 V₂ = -322.5 N x = 13 V₁ = ON M₁ = ON m Figure 2. Sample Table 3 m x = 5.5 V₂ = 1277.5 N M₂ 1898.75 N·m = x = 5.5 V₂ = 1100.7233 N M3 1898.75 N·m = V₁ = -22.5 N M₁ = -11.25 N·m M₂=-11.25 N·m V3 850.7233 N = = M3 4338.0583 N•m V4850.7233 N M₁5338.0583 N·m V4-1649.2767 N M₁ = 3341.8748 N.m
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