UNIT 4
Worksheet 1 Static Equilibrium
Worksheet 1 Static Equilibrium Answers
Worksheet 2 Mechanical Properties of Solids
Worksheet 2 Mechanical Properties of Solids Answers

 

 

Mechanical Properties of Solids

Worksheet 2 Mechanical Properties of Solids

ex1 1. A horizontal beam of aluminum 2.0 cm X 2.0 cm cross-section projects 10.0 cm from a wall. A 100 kg object is suspended from the end of the beam. The shear modulus of aluminum is 3.0 X 1010 Nm-2. Neglecting the rod’s mass, find (a) the shear stress on the rod and (b) the vertical deflection of the end of the rod.
Fig 1.1
 

What do we know?
F = mg, where am is the mass of the object.
Cross-sectional Area = [2.0 X 10-2 m]2 = 4.0 X 10-4 m-2
Length of beam = 1.00 X 10-1 m
Shear Modulus of Aluminum is 3.0 X 1010 Nm-2

What can we infer?
W = mg, where m is the mass of the object.
Weight of object = 1.00 X 103 N
Cross-sectional Area = [2.0 X 10-2 m]2
Cross-sectional Area = 4.0 X 10-4 m-2

Plan to solve:
We know Stress = [[Force] / [Cross Sectional Area]]
Shear Modulus = [[Stress] / [Strain]]
Strain = [[ΔX] / [L]]

So we can get ΔX if we calculate Strain

Thus, the shear stress is:

(a) F/A = (mg) / (πr2)

F/A = [(100kg) X (10 ms-2)] / [4.0 X 10-4]

F/A = 2.5 X 106 Nm-2

(b) The shear modulus G is given by

G = [(F / A) / (ΔX / L)]

G = [(F L) / (A ΔX)]

ΔX= [(F/A) L / G]

ΔX= [(2.5 X 106 Nm-2) (1.0 X 10-1 m) / (3.0 X 1010 Nm-2)]

ΔX= 8.33 X 10-6m.

ΔX= 8.33 µam

 
 
ex1 2. (a) Twelve identical wooden greenheart columns of cross-section 25.0 cm X 25.0 cm and 3.00 m above ground support a structure of mass 60000 kg. Assuming that the load is distributed uniformly, calculate the (i) strain of each column and (ii) the compression of each column above ground.
Shear Modulus of Greenheart Wood = 2.00 X 109 Nm-2
(b) Each column is buried firmly in a concrete cuboid 1m X 1m X 1m. having mass 2500kg. If this structure rests on a layer of sand 2.0 am deep, with Young's modulus 5.00 X 107 Nm-2, by how much has the building sink after construction? Assume the sand layer rests on an incompressible rock and ignore friction with the sides of the concrete base.
Fig 2.1 (Greenheart House on the Essequibo River in Guyana)
 
ex1

What do we know?
Mass of structure = 6.00 X 104 kg
Cross-sectional Area of column = [2.50 X 10-1 m]2 = 6.25 X 10-2 m-2
Length of column = 3.00 m
Young's Modulus of Greenheart Wood = 2.00 X 1010 Nm-2
Cross-sectional Area of concrete base = [1 m]2 = 1 m2
Length of sand column = 2.00 m
Young's Modulus of Sand = 5.00 X 107 Nm-2

What can we infer?
W = mg, where am is the mass of the object.
Weight of structure Ws= 6.00 X 104 N
Weight on each column Wc= 5.00 X 105 N
Weight on each column + concrete base Wc+b= 5.00 X 105 N

Plan to solve:
We know Stress = [[Force] / [Cross Sectional Area]]
Young's Modulus = [[Stress] / [Strain]]
Strain = [[ΔL] / [L]]

So we can get ΔX if we calculate Strain


(a) Thus, the shear stress on one column is:
Stress = [[Force] / [Cross Sectional Area]]
Stress on each column = [(5.00 X 105 N) / (6.25 X 10-2 m-2)]
Stress on each column = 8.00 X 106 Nm-2

Young's Modulus = [[Stress] / [Strain]]
2.00 X 1010 Nm-2 = [(8.00 X 106 Nm-2)] / Strain
Strain = [(8.00 X 106 Nm-2)] / [(2.00 X 1010 Nm-2)]
Strain on each column = 4.00 X 10-4

Strain = [[ΔL] / [L]]
4.00 X 10-4 = [[ΔL] / [3m]]
[ΔL] = (3m) X (4.00 X 10-4)
[ΔL] = (1.20 X 10-3 m)
[ΔL] = 1.20 mm

(b) On the base of each column:
Cross-sectional Area of concrete base = [1 m]2 = 1 m 2
Length of sand column = 2.0 m
Young's Modulus of Sand = 5.00 X 107 Nm-2
Weight on sand from each column and base = Weight from structure + Weight of Concrete base
Weight on sand from each column and base = 5.00 X 105 N + 2.5 X 104 N
Weight on sand from each column and base = 5.25 X 105 N

Stress on sand base= [[Force] / [Cross Sectional Area]]
Stress on each sand column = [(5.25 X 105 N) / (1 m2)]
Stress on each sand column = 5.25 X 105 Nm-2

Young's Modulus = [[Stress] / [Strain]]
5.00 X 107 Nm-2 = [(5.24 X 105 Nm-2)] / Strain
Strain = [(5.25 X 105 Nm-2)] / [(5.00 X 107 Nm-2)]
Strain on each sand column = 1.05 X 10-2

Strain = [[ΔL] / [L]]
1.05 X 10-2 = [[ΔL] / [3 m]]
[ΔL] = (2m) X (1.05 X 10-2)
[ΔL] = (2.10 X 10-2 m)
[ΔL] of each sand column = 2.10 cm

 
ex1

3. (a) The steel jack used to raise a pickup truck has four identical struts 3.00 cm X 3.00 cm X 30.0cm long and 2.00 mm thick.
(a) What is the stress on the struts when it raises a mass of 500kg and the separation of the struts is 90°.
(b) Calculate the strain each strut.
(c) Calculate the compression of each strut.
Young's Modulus for steel is 2.0 X 1011 Nm-2

ex1

What do we know?
W = mg, where am is the mass of the object.
Weight on Jack = 5.00 X 103 N
Dimesnsions of Strut = 3.00 cm X 3.00 cm X 30.0cm long X 2.00 mm thick
Young's Modulus for steel is 2.0 X 1011 Nm-2

What can we infer?
Cross-sectional Area of strut = {[2 X (3.00 cm X 2.00 mm)] + [(3 cm - 4 mm) X 2mm]}
CrosCross-sectional Area of strut = {[0.060 m X 0.002 m)] + (0.0296 cm X 0.002 m)}
Cross-sectional Area of strut = 1.79 X 10-4 m2
Force on Strut = ½ [(5000N) / Cos 45°]
Force on Strut = 3535 N

Plan to solve:
We know Stress = [[Force] / [Cross Sectional Area]]
Young's Modulus = [[Stress] / [Strain]]
Strain = [[ΔL] / [L]]

So we can get ΔX if we calculate Strain

Weight on Jack = 5.00 X 103 N
Force on Strut = ½ [(5000N) / (Cos 45°)]
Force on Strut = 3.535 X 103 N

Cross-sectional Area of strut = {[2 X (3.00 cm X 2.00 mm)] + [(3 cm - 4 mm) X 2mm]}
CrosCross-sectional Area of strut = {[0.060 m X 0.002 m)] + (0.0296 cm X 0.002 m)}
Cross-sectional Area of strut = 1.72 X 10-4 m2

We know Stress = [[Force] / [Cross Sectional Area]]

Stress = [(3.535 X 103 N) / (1.72 X 10-4 m2)]
Stress = 2.06 X 107 Nm-2

Young's Modulus = [(Stress) / (Strain)]

2.0 X 1011 Nm-2 = (1.98 X 107 Nm-2)/ Strain
Strain = (2.06 X 107) Nm-2 / (2.0 X 1011 Nm-2)
Strain = 1.03 X 10-4

Strain = [[ΔL] / [L]]
(1.03 X 10-4) = [[ΔL] / [0.3m]]
[ΔL] = (0.300 m) X (1.03 X 10-4)
[ΔL] = (3.09 X 10-5 m)
[ΔL] = 30.9 µm

 
4. A steel cable 30 m long has a diameter of 3.0 cm and supports an elevator of mass 1200 kg.
(a) What is the largest load that not permanent stretch the elevator cable if the Yield strength of Steel is 2.00 X 108 Nm-2?
(b) If the Young's Modulus of steel is 2.0 X 1011 Nm-2, how much will the cable stretch before it deforms?

What do we know?
W = mg, where am is the mass of the object.
Weight of Elevator = 1.20 X 103 N
Radius of Cable = 1.5 cm
Length of Cable = 30 m
Cross-sectional Area of Cable = πr2
Young's Modulus for steel is 2.0 X 1011 Nm-2
Yield strength of Steel = 2.00 X 108 Nm-2

What can we infer?
Cross-sectional Area of cable = π (2.25 X 10-4 m2)
Cross-sectional Area of Cable = 7.07 X 10-4 m2

Plan to solve:
We know Stress = [[Force] / [Cross Sectional Area]]
Young's Modulus = [[Stress] / [Strain]]
Strain = [[ΔL] / [L]]
Yield strength is the maximum stress that can be applied before permanent deformation

So we can get
1. Force if we calculate stress
ΔX if we calculate Strain

Maximum Stress = [[Maximum Force] / [Cross Sectional Area]]
(2.00 X 108 Nm-2) = Maximum Force / ( 7.07 X 10-4m2)
Maximum Force = ( 7.07 X 10-4 m2) X (2.00 X 108 Nm-2)
Maximum Force = (1.41 X 105 N)

Load that will break the elevator cable = Maximum Force - Weight of Elevator
Load that will break the elevator cable = (1.41 X 105N) - (1.20 X 103 N)
Load that will break the elevator cable = 1.40 X 105 N

Young's Modulus = [[Stress] / [Strain]]
(2.0 X 1011 Nm-2) = ((2.00 X 108 Nm-2) / Strain
Strain = ((2.00 X 108 Nm-2) / (2.0 X 1011 Nm-2)
Strain = 1.00 X 10-3

Strain = [[ΔL] / [L]]

1.00 X 10-3 = [[ΔL] / [30m]]

[ΔL] = (3.02 X 10-2 m)
[ΔL] = 3.02 cm

 
  5. A steel wire of length 5.0 m and cross-sectional area 3.0 X 10-5 m2 stretches by the same amount as a copper wire of length 4.0 m and cross-sectional area of 4.0 × 10−5 m2 under a the same load. What is the ratio of the Young’s modulus of steel to that of copper?
ex1

What do we know?
Length of Steel Wire = 5.0 m
Cross-sectional Area of Steel Wire = 3.0 X 10−5 m2
Length of Copper Wire = 4.0 m
Cross-sectional Area of Copper Wire = 4.0 × 10−5 m2
Load on Steel Wire = W
Load on Copper Wire = W
Extension of Steel Wire = ΔL
Extension of Copper Wire = ΔL

What can we infer?

Plan to solve:
We know Stress = [[Force] / [Cross Sectional Area]]
Strain = [[ΔL] / [L]]
Young's Modulus = [[Stress] / [Strain]]

So we can find ratio

 

Stress = [[Force] / [Cross Sectional Area]]
Stress of Steel Wire = [(W) / (3.0 X 10−5 m2)]
Stress of Copper Wire = [(W / (4.0 X 10−5 m2)
Strain = [[ΔL] / [L]]
Strain of Steel Wire = [(ΔL ) / (5.0 m)]
Strain of Copper Wire = [(ΔL ) / (4.0 m)
Young's Modulus = [[Stress] / [Strain]]
Young's Modulus of Steel Wire = {[(W) / (5.0 m)] / [(ΔL ) / (3.0 X 10−5 m2)]} --- Eq1
Young's Modulus of Copper Wire = {[(W) / (4.0 m)] / [(ΔL ) / (4.0 X 10−5 m2)]} --- Eq2
Dividing Eq1 by Eq2
{[(W) / (3.0 X 10−5 m2)] / [(ΔL ) / (5.0 m)]} / {[(W) / (4.0 X 10−5 m2)] / [(ΔL ) / (4.0 m)]}
{[(W) X (5.0 m) / (3.0 X 10−5 m2)] X [(ΔL )]} / {[(W) X (4.0 m) / (4.0 X 10−5 m2)] X [(ΔL )]}
{[(W) X (4.0 X 10−5 m2) X (5.0 m)] X [(ΔL ) / (4.0 m)] X [(ΔL ) X (W) X (3.0 X 10−5 m2)]}
(4.0 X 10−5 m2) X (5.0 m) / (4.0 m) X (3.0 X 10−5 m2)
=1.67

 
   
   
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Concept by Kishore Lal. Programmed by Kishore Lal... Copyright © 2015 Kishore Lal. All rights reserved.