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Answer to Question #91196 in Mechanical Engineering for Nathan

Question #91196
1)A steel wire ABC 16m long having cross sectional area of 4mm2 weighs 20N . If the modulus of elasticity for the wire material is 200Gpa find the deflections at C and B

2)A max stress in a 6.36mm steel wide flange beam 100mm long moment of inertia 362mm4,modulus of elasticity 29,000Gpa with a load of 15kN.Calculate the max stress and deflection
Expert's answer

1) Deflection at C due to self weight of wire AC = dlc

@$dl_c = \frac { Wl} {2AE} (1)@$

Using (1) we get: dlc= 0.2 mm

Deflection at B:

Now deflection at B is due to two reasons: i) due to self

weight of AB and ii) due to weight of BC.

@$dl_B = \frac { 0.5W×0.5l} {2AE} + \frac { 0.5W×0.5l} {AE} (2)@$

Using (2) we get: dlB= 0.15 mm

2)

Maximum strees is given by formula

@$σ= \frac { y q L^2} {8I} (1)@$

where y is distance to extreme point from neutral, q is the uniform load per length unit of beam (N/m, N/mm, lb/in), L is the length of beam (m, mm, in), I is the moment of inertia of the cross section

@$q=\frac { W} {L} (2)@$

In our case, y=6.36mm, L=100 mm, W=15kN, I=362mm4

Using (1) we get: σ = 4.95×104 Pa

Maximum deflection is given by formula

@$δ= \frac { 5 q L^4} {384 E I} (3)@$

In our case, E=29,000Gpa

Using (3) we get: δ = 6×10-5 mm

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