Answer:
1561.84 MPa
Explanation:
L=20 cm
d1=0.21 cm
d2=0.25 cm
F=5500 N
a) σ= F/A1= 5000/(π/4×(0.0025)^2)= 1018.5916 MPa
lateral strain= Δd/d1= (0.0021-0.0025)/0.0025= -0.16
longitudinal strain (ε_l)= -lateral strain/ν = -(-0.16)/0.3
(assuming a poisson's ration of 0.3)
ε_l =0.16/0.3 = 0.5333
b) σ_true= σ(1+ ε_l)= 1018.5916( 1+0.5333)
σ_true = 1561.84 MPa
ε_true = ln( 1+ε_l)= ln(1+0.5333)
ε_true= 0.4274222
The engineering stress on the rod when it is loaded with a 5500 N weight is 1561.84 MPa.
Complete Question
Complete Question is attached below.
Answer:

Explanation:
From the question we are told that:
Diameter 
Power 
Head loss 




Generally the equation for Power is mathematically given by

Therefore



Since

Where


Therefore


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
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