Table 1: Use these estimated material values unless asked to calculate them. Con
ID: 1710869 • Letter: T
Question
Table 1: Use these estimated material values unless asked to calculate them. Configuration before deflection is shown at right. The material's stress-strain diagram is shown below. a) What is the material's Modulus of Elasticity? b) How much shorter does the bar get under the 2 kN compressive load? c) Does the material yield? Why or why not? d) If the material's Poisson Ratio is 0.35, what is the precise new diameter under the 2 kN compressive load? e) What is the maximum shear stress induced in the shaft? Where does this condition occur? What is the simultaneous axial stress at this state? f) What is the strain energy in this bar?Explanation / Answer
a)material's modulus of elasticity=slope of stress strain curve within elastic limit=40000/0.004=1*10^7 Pa
b)Applied load = 2 kN
Initial diameter = 0.08 m
Area of cross section= (pi)*d^2/4=0.005 m^2
Axial compressive stress = 2000/0.005=400kPa
The applied pressure is much higher than the strength of the material.Therefore the material reaches its maximum deformation per the stress straim diagram and then fails
Maximum strain=0.9
Initial length = 0.5 m
Amount by which the bar shorten = 0.9*0.5=0.45m
Final length of bar=0.05m
c)yes the material yields and then crushes because the apploed load and hence pressure is way too high than the material's strength
d)Poisson's ratio = 0.35
Starin in transverse direction=0.9*0.35=0.315
Final diameter =0.08 + (0.08*0.315)=0.1052m
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.