The resultant internal loading acting on the cross section of C are Normal force of 0 N, Shear force of 233.33 N and Bending moment of 433.325 N.
We know that Moment = Force * Distance.
Given that,
Force along span C = 600 N / m
First chart a free body diagram as shown in Fig 1.
The distributed load along span c is replaced by a point load F at its midpoint.
Force at midpoint of span C = 600 * 2 N
Take moments about point A
Σ [tex]M_{A}[/tex] = 0
-600 * ( 2 ) * [ 1 + ( ( 1 + 1 ) / 2 ) ] + [tex]B_{y}[/tex] ( 4.5 ) - 900 ( 6 ) = 0
- 2400 + 4.5 [tex]B_{y}[/tex] - 5400 = 0
[tex]B_{y}[/tex] = 1733.33 N
Consider a section at Point C and draw a free body diagram as shown in Fig 2.
In fig 2,
[tex]V_{C}[/tex] = Shear force at point C
[tex]N_{C}[/tex] = Normal force at point C
[tex]M_{C}[/tex] = Bending moment at point C
Applying equilibrium condition to the system. Consider equilibrium of Forces along X direction
Σ [tex]F_{x}[/tex] = 0
[tex]N_{C}[/tex] = 0 N
Consider equilibrium of Forces along Y direction
Σ [tex]F_{y}[/tex] = 0
[tex]V_{C}[/tex] - 600 ( 1 ) + [tex]B_{y}[/tex] - 900 = 0
[tex]V_{C}[/tex] = 600 - 1733.33 + 900
[tex]V_{C}[/tex] = 233.33 N
Take moment about point C
Σ [tex]M_{C}[/tex] = 0
- [tex]M_{C}[/tex] - 600 ( 1 ) ( 1 / 2 * 1 ) + [tex]B_{y}[/tex] ( 2.5 ) - 900 ( 4 ) = 0
- [tex]M_{C}[/tex] - 300 + 2.5 ( 1733.33 ) - 3600 = 0
[tex]M_{C}[/tex] = 433.325 N
The resultant internal loading acting on the cross section are Normal force, Shear force and Bending moment
Therefore, The resultant internal loading acting on the cross section of C are as follows
Normal force [tex]N_{C}[/tex] = 0 N
Shear force [tex]V_{C}[/tex] = 233.33 N
Bending moment [tex]M_{C}[/tex] = 433.325 N
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