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
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Answer:
Explanation:
a) On the verge of tipping over, reaction acts at the corner A
When slippage occurs,
Block moves w/ const. velocity equilibrium
Three-force member: reaction at A must pass through B
tan b/2h, h b/ 2 θ µ = = ∴= k k ( µ )
b) When slippage occurs,
Block moves w/ const. velocity equilibrium
Three-force member: reaction at C must pass through G
k tanθ µ =
tan x/ H/2 , x H/2
