Chapter 1 Review

Question 1

With this information explain how to find the:

(1) Axial Stress
(2) Axial Strain
(3) Increase in Length
(4) Lateral Strain
(5) Decrease in Length
(6) Degrease in Cross Sectional Area

Answer 1

(1) Axial Stress; Start using the Axial Stress Equation: σ = P/A

σ = 1500lb / (π/4)(0.25)² = 30,557psi

(2) Axial Strain; Use the equation σ = εE

ε = σ/E = 30,557psi / 25,000*10³psi
ε = 0.001222

(3) Increase in Length; Use the Equation for the Change in length using our new axial strain value: δ = εL

δ = (0.001222)(0.25in) = 0.0003056 in

(4) Lateral Strain; Start by using the Poisson’s Ratio and our current ε_axial value: 𝓥 = - ε_lateral / ε_axial

ε_lateral = -(0.001222)(0.32) = -0.00039104

(5) Decrease in Diameter; We can use our new ε_lateral value in the equation: δ = (ε_lateral)d

δ = (-0.00039104)(0.25 in) = -0.00009776 in

(6) Decrease in Cross Sectional Area; We calculate the first cross sectional area then calculate the new cross sectional area after the change in diameter.

Question 2

Explain how to solve for:

(a) Bearing Stress in the Rivets

Answer 2

(a) Bearing Stress; This will be stress in the Rivets and since there are three layers we will have to divide the force P by two.

Using Bearing Stress Equation: σ = P / 2(dₚ)(t)

where t = thickness of walls

σ = (5010³N) / 2(2010⁻³m)(16*10⁻³m) = 78 MPa

Question 3

Explain how to solve for

(b) Shear Stress on the Rivets

Answer 3

(b) Shear Stress; we start by creating this FBD and solving for V Values in our Shear Equation: Շ = V/A

Summing the Forces we have V = P/4 we can now plug this into the Shear Equation and solve for the Shear Stresses in the Rivets:

Շ = (P/4) / A → P = (4)(180 MPa)(20mm)²(π/4) = 226kN

Question 4

How to find the Reactions at A and C
How to find internal V,N,M @ x = 3ft

Answer 4

(A) External Reactions: This is simply found by analyzing joint BC where we can take moment about B:

↺ΣMc = 0; Cy(9ft) = -500lbft
Cy = 55.56lb
Ay = -55.56lb

Then solve the distributed load: 90lbft(12ft)/2 = 540lb @ 8ft

→ΣFx = 0;
Ax = -540lb

(B) Internal Reactions: Now cut joint AB and set an all positive set of internal forces and moments at the cut.

The distributed load at will be reevaluated using similar of triangles. our new load heigh will be (y/3ft) = (90lb/ft / 12ft); y = 22.5lb/ft

Our new load will be: (22.5lb/ft)(3ft)/2 = 33.75lb @ 2ft

→ΣFx = 0; Vx - Ax + 33.75 = 0
Vx = 506.25lb

↑ΣFy = 0;
Nx = 55.56lb

↺ΣMx = 0; Mc - Ax(3ft) +Ma + 33.75lbft
Mc = -2734lbft

Question 5

(a) Explain how to find the increase in inner diameter of BC.

(b) Explain how to solve for Poisson Ratio of the Brass

(c) Explain how to solve for change in Thickness and Diameter of Top Portion

Answer 5

(a) Increase in inner diameter We can leverage the change in length formula δ = (ε_lateral)L to find the Strain in the axial direction.

ε_lateral = 200x10⁻⁶ in / 0.375 in = 5.33*10⁻⁴

with our new value for ε_lateral, now we may solve for the change in diameter using δ = (ε_lateral)(d_inner)

δ = 5.3310⁻⁴(2.25in - 2(0.375in)) = 8*10⁻⁴in

(b) Poisson’s Ratio: Since both parts are brass we can focus on BC still. We will leverage

We need to first solve for the new Axial Strain: ε_axial
Stress Equation [σ = ε_axial/E = P/A]

We will use the new thickness to solve for our new Area value

t_BC = 0.3752;

A =(π/4)(2.25²in² - (2.25in - 2(0.3752in))² = 2.21in²

now we can use the equation: σ = ε_axial*E = P/A

ε_axial = P/EA = -(26000lbs + 22000lbs) / (14000*10³psi)(2.21in²)

ε_axial = -1.568*10⁻³ in/in

Now we can get Poisson’s Ratio using 𝓥 = ε_lateral/ε_axial

𝓥 = 5.3310⁻⁴/ -1.56810⁻³
𝓥 = 0.34

(c) Change in Thickness and InnerDiameter:

Start by solving for the new ε_axial using σ = ε_axial/E = P/A

σ = ε_axial*E = P/A

ε_axial = P/(E*A)

26000lbs / (π/4)(1.25²in² - (1.25in - 2(0.5in))²(1400*10³psi)

ε_axial = 1.61x10⁻³

Using Poisson’s Ratio:

ε_lateral = 𝓥ε_axial = 0.341.61x10⁻³

ε_lateral = 5.47x10⁻⁴

We can now multiply this into our original Thickness and InnerDiameter using the Delta Formula:

δ = t_ab(ε_lateral) = (5.47x10⁻⁴)(0.5in) = 2.74x10⁻³in
δ = d_ab(ε_lateral) = ( 5.47x10⁻⁴)(0.25in) = 1.36x10⁻³in²

Question 6

Explain the Solution to Part (a)

Answer 6

This problem is lengthy.

We start by using the Delta Formula: δ = PL/EA which we can apply to each section.

Bar #1: δ = (1400kN)(5m) / (110GPa)(π/4)(80mm)²

δ₁ = 12.67mm

Bar #2 and #3 require sectioning the bar in order to get values for each part. Working with section 1 & 3:

Bar #2: δ₁₃ = (2)(1400kN)(25m) / (5)(110GPa)(π/4)(280mm)²

δ₁₃ = 2.53mm

δ₂ = (1400kN)(1m) / (110GPa)(π/4)(80mm)²

δ₂ = 2.53mm

the summation of these two length will give us the total change in length of the rods

Question 7

(A) Find the Average Shear Stress in the Pin

Answer 7

(A) Average Shear Stress in the Pin

We can leverage the Shear Stress Formula: Շ = V/A

We can get the value of V just from a FBD of the pin shown. Summing the Forces:

∑F = 0; (P/2) = 2V

V = P/4

Շ = 160x10³ lbs / (4)(π/4)(2in)²

Շ = 12.73 ksi

Question 8

(B) Find the Bearing Stress between the flange plates and the pin and also between the gusset plates and the pin.

Answer 8

(B1) Between the Flange Plates:

Simply use the diameter of the pins and the thickness of the Flange Plates to solve for the Bearing Stress using: σ = P/(d*t)

σ = [(160x10³ lbs)/2] / 2(2in)(1in)

σ = 20 ksi

(B2) Between Gusset Plate:

Use the same equation however for Gusset Plate

σ = [(160x10³ lbs)/2] / (2in)(1.5in)

σ = 26.7 ksi

Question 9

Explain how to solve for:

(A) The Allowable Tensile Force P_allow considering tension in the tubes.

Answer 9

(A) Allowable Tension (P_allow):

Start by calculating the area of action on the two tube. this will require subtracting the area of the four pins.

A₁ = (π/4)(d₁² - (d₁ - 2t₁)²) - 4(pin)(t₁)
A₁ = (π/4)(41²mm² - (41mm - 2(6.5mm))²) - 4(11mm)(6.5mm)
A₁ = 418.5mm²

A₂ = (π/4)(d₂² - (d₂ - 2t₂)²) - 4(pin)(t₂)
A₂ = (π/4)(28²mm² - (28mm - 2(7.5mm)²) - 4(11mm)(7.5mm)
A₂ = 153mm²

We can now solve for the Allowable Tension by using the area which the force is acting on and our equation for Stress: σ = P(Fs) / A

We will factor in the two safety conditions as well to get an immediate value:

P_ₐᵧ = (200MPa)(153mm²) / (3.5)
P_ₐᵧ = 8.74 kN

P_ₐᵤ = (340MPa)(153mm²) / (4.5)
P_ₐᵤ = 11.56 kN

The lowest value P_ₐᵧ = 8.74 kN
is our best Tensile Force

Question 10

(B) Allowable Tensile for the given Shears in Pins:

Yield: 80 MPa
Ultimate: 140 Mpa

Answer 10

(B) Allowable Tensile for the given Shears in Pins:

Since we are dealing with a pipe that has 4 separate pins we must take them into account when doing our Shear Calculations: Շ = (V)(Fs) / A

Since 4 pin occasions:

∑F = 0; P = 4V
V = P/4

Now we sub in our values

Շ = (P_ₐ)(Fs) / (4)A

P_ₐᵧ = (4)(80MPa)(π/4)(11mm)² / (3.5)
P_ₐᵧ = 8.7 kN

P_ₐᵤ = (4)(140MPa)(π/4)(11mm)² / (4.5)
P_ₐᵤ = 11.8 kN

The P_ₐᵧ will be our best value

Question 11

(C) Allowable Tension for Bearing between pins with Bearing Stress Values:

Yield: 80 MPa

Ultimate: 450 MPa

Answer 11

(C) Allowable Bearing between pins:

For allowable Bearing we will work with this FBD and the diameter of the pins along with the thickness of the plates.

We need to leverage the Bearing Stress Equation:
σ = P(Fs) / (4)(d)(t)

Right Pipe
PRP_ₐᵧ = (260MPa)(4)(11mm)(6.5mm) / (3.5mm) = 21.2 kN
PRP_ₐᵤ = (450MPa)(4)(11mm)(6.5mm) / (4.5mm) = 28.6kN

Left Pipe
PLP_ₐᵧ = (260MPa)(4)(11mm)(7.5mm) / (3.5mm) = 24.5 kN
PLP_ₐᵤ = (450MPa)(4)(11mm)(7.5mm) / (4.5mm) = 33.0 kN