# Answer to Question #91196 in Mechanical Engineering for Nathan

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

1) Deflection at *C *due to self weight of wire *AC *= *dl*_{c}

Using (1) we get: *dl*_{c}= 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*.

Using (2) we get: *dl*_{B}= 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

In our case, y=6.36mm, L=100 mm, W=15kN, I=362mm^{4}

Using (1) we get: *σ* = 4.95×10^{4} 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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