Asked by sand
A chandelier of mass 250 kg is hung vertically from a ceiling using a 1.25 m long brass rod of square cross-section of sides 0.4 cm each. How much will the rod stretch when the chandelier is hung? By what factor is the minimum breaking load greater than the weight of the chandelier?
Answers
Answered by
drwls
The cross sectional area is
A = 0.16 cm^2 = 1.6*10^-5 m^2
The stress in the road is
sigma = M*g/A = 1.53*10^8 N/m^2
The amount of stretch is
delta L = L*sigma/E
where E is Young's modular for brass.
The breaking load is
P = A* sigmau
where sigmay is the yield stress for brass.
You will have to look up the modulus and yield stress values. They want you to compute the factor P/(M*g)
A = 0.16 cm^2 = 1.6*10^-5 m^2
The stress in the road is
sigma = M*g/A = 1.53*10^8 N/m^2
The amount of stretch is
delta L = L*sigma/E
where E is Young's modular for brass.
The breaking load is
P = A* sigmau
where sigmay is the yield stress for brass.
You will have to look up the modulus and yield stress values. They want you to compute the factor P/(M*g)
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