Chapter 2 Review

Question 1

Two wires, one copper and the other steel, of equal length stretch the same amount under an applied load P. The moduli of
elasticity for each is: Eₛ=210 GPa, E꜀ =120 GPa. Determine the ratio of the diameter of the copper wire to that of the steel wire.

Answer 1

Since the two wires have the same change in length, we can use out Delta Equations and Solve for the ratios between the two diameters:

PL/E₁A₁ = PL/E₂A₂ → 1/E₁(π/4)(d₁)² = 1/E₂(π/4)(d₂)²

We can now solve for the diameter of the two using algebra.

since its copper to steel we need [d₂/d₁]

[d₂/d₁] = √(E₁/E₂) = √(210/120)

[d₂/d₁] = 1.32

Question 2

A plane truss with span length L= 4.5 m is constructed using cast iron pipes (E = 170 GPa) with cross sectional area of
4500 mm². The displacement of joint B cannot exceed 2.7 mm. Yield stress is 290 MPa* in tension. Determine the maximum
value of loads P and max stress. Is it going to yield?

Answer 2

Since there is only one leftward force acting on the Joint AB, we can use our Elongation Equation with the value of P as our N

δ = PL/EA

Solving for P_max: P_max = EAδ/L

P_max = (170GPa)(4500 mm²)(2.7 m) / 4.5 m
P_max = 459 kN

Comparing with the given yield:

σ = p_max / A → σ = 459 kN / 4500 mm²

** σ = 102 MPa**

Since this is less than our original value it WILL NOT yield.

Question 3

A monel shell (Em = 170 GPa, d3 = 12 mm, d2 = 8 mm) encloses a brass core (Eb = 96 GPa, d1 =6 mm). Initially, both shell and core are of length 100 mm. A load P is applied to both shell and core through a cap plate. Determine the load P required to compress both shell and core by 0.10 mm.

Answer 3

Start by finding the area of the two objects. ₁₂

Since the length should change the same for both objects we can start by setting the Delta Equations of both objects equal to each other: δ = PL/EA

We also know that the force P is P₁ + P₂ = P

δ₁ = δ₂ → P₁L/E₁A₁ = P₂L/E₂A₂ → P₁ = P₂E₁A₁/E₂A₂

then we can leverage that

P₂ = E₂A₂/L

Now we plug into our P equations and solve:

P = P₂(1 + E₁A₁/E₂A₂)

P = (E₂A₂/L)(1 + E₁A₁/E₂A₂)

Question 4

How to solve for the Max Shear Stress in the AB section ?

P = 200 kN
A - 3970 mm²

Answer 4

We will start by finding the reaction forces in the Bar’s Axial Direction

Ax = 2P = 400 kN

Then we will leverage the Շ_max = σ/2 equation:

σ = 400x10³ N / 0.003970 m² = 100.76 MPa

Շ_max = 100.76 MPa / 2

Շ_max = 50.4 MPa

Question 5

How to find Max Shear Stress in the Rod

P = 11.5 kN
d = 10 mm

Answer 5

We can leverage the equation Շ_max = σ/2 equation

Շ_max = 11.5*10³ kN / 2(π/4)(10x10⁻³ m)²

Շ_max = 73 MPa

Question 6

(A) How to solve for Theta

(B) How to solve for P_max

σ = 5 MPa
Շ = 3 MPa
A = 225 mm²

Answer 6

(A) Solving for Theta: We start by leveraging the two Angles Stress equations

σ = σₓcos²(θ) → σₓ = σ /cos²(θ)
Շ = σₓsinθcosθ → σₓ = Շ /sinθcosθ

Now set these equations equal and solve for θ:

σ /cos²(θ) = Շ/sinθcosθ → σsinθcosθ/cos²(θ) = Շ → tanθ = Շ/σ

σ = tan⁻¹[3/5] = 30.96

(B) Solving for P_max We will again leverage the Angle Stress Equation
σₓ = σ /cos²(θ) along with σₓ = P/A

P = Aσₓ

P = (225 mm²)(5 MPa) /cos²(30.96)

P = 1.75 kN

Question 7

A Brass WIre (d = 2 mm, E = 110GPa) is pretensioned to T = 85 N. The coefficient of Thermal Expansion for the wire is 19.5x10⁻⁶/°C

Find the change in Temperature.

Answer 7

We start by leveraging our equations for Stress: σ = T/A = EαΔT

ΔT = T/EαA = (85 N) / (110x10⁹ Pa)(19.5x10⁻⁶/°C)(π/4)(2x10⁻³ m)²

ΔT = 12.61°C

Question 8

Explain the solution

Answer 8

This problem is simple. We use the Stress and Shear equations with the given values to solve for when P is allowed to be the lowest.

σ = P_allow/A → σA = P_allow₁
Շ = P_allow/A → ՇA = P_allow₂

P_allow₁ = (100 MPa)(38 mm)(50 mm) = 190 kN
P_allow₂ = (48 MPa)(38 mm)(50 mm) = 91.2 kN

P_allow: 91.2 kN