If you’re designing a new pair of shoes, what are some things you’ll have to think about ??
2 answers:
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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:
13.4 mm
Explanation:
Given data :
Load amplitude ( F ) = 22,000 N
factor of safety ( N )= 2.0
Take ( Fatigue limit stress amplitude for this alloy ) б = 310 MPa
<u>calculate the minimum allowable bar diameter to ensure that fatigue failure will not occur</u>
minimum allowable bar diameter = 13.4 * 10^-3 m ≈ 13.4 mm
<em>attached below is a detailed solution</em>
