Answer:
The resulting strain is 
.
Explanation:
A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm
Force, F = 44,500 N
Th elastic modulus of Cu to be 110 GPa
The resulting strain is given by the formula as follows :
E is elastic modulus of Cu is are of cross section
So, the resulting strain is .
Answer: its an Ignition coil
Answer:
10.984mm
Explanation:
by elastic modulus
stress=modulus of elasticity*strain
stress=loading/area area" cross-section"
11mm=0.011m
area=π(d/2)^2=π(0.011/2)^2=9.503*10^-5 square meter
stress=55000/(9.503*10^-5)=578.745 MPa
convert MPa and GPa to pascal.
strain=stress/modulus=(578.745*10^6)/(125*10^9)=0.00463............axial strain
v=Poisson ratio
lateral strain=(-v)*axial strain= -0.31*0.00463
lateral strain= -1.4353*10^-3=change in diameter/ original diameter
change in diameter=(-1.4353*10^-3)*0.011= -1.57883*10^-5 m
negative indicates decrease in diameter.
decrease in dia.=0.01578mm
new diameter=11-0.01578= 10.984mm
