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The shaft is supported by a smooth thrust bearing at a and a smooth journal bearing at
b. determine the resultant internal loadings acting on the cross section at
c. 600n/m b c lm -lsml sm 900

Sagot :

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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