melinda is using a rectangular brass bar in a sculpture she is creating. the brass bar has a length that is 4 more than 3 times
2 answers:
You might be interested in
The modulus of elasticity is 28.6 X 10³ ksi
<u>Explanation:</u>
Given -
Length, l = 5in
Force, P = 8000lb
Area, A = 0.7in²
δ = 0.002in
Modulus of elasticity, E = ?
We know,
Modulus of elasticity, E = σ / ε
Where,
σ is normal stress
ε is normal strain
Normal stress can be calculated as:
σ = P/A
Where,
P is the force applied
A is the area of cross-section
By plugging in the values, we get
σ =
σ = 11.43ksi
To calculate the normal strain we use the formula,
ε = δ / L
By plugging in the values we get,
ε =
ε = 0.0004 in/in
Therefore, modulus of elasticity would be:
Thus, modulus of elasticity is 28.6 X 10³ ksi
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step and very detailed solution of the given problem
Answer:
The shear strain is 0.05797 rad.
Explanation:
Shear strain is the ratio of change in dimension along the shearing load direction to the height of the plate under application of shear load. Width of the plate remains same. Length of the plate slides under shear load.
Step1
Given:
Height of the pad is 1.38 in.
Deformation at the top of the pad is 0.08 in.
Calculation:
Step2
Shear strain is calculated as follows:
For small angle of ,
can take as
.
Thus, the shear strain is 0.05797 rad.