1) Deflection at C due to self weight of wire AC = dlc
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.
Using (2) we get: dlB= 0.15 mm
2)
Maximum strees is given by formula
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=362mm4
Using (1) we get: σ = 4.95×104 Pa
Maximum deflection is given by formula
In our case, E=29,000Gpa
Using (3) we get: δ = 6×10-5 mm
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