STRENGTH/MECHANICS OF MATERIALS
SIMPLE STRESSES, SHEAR, STRAIN, BENDING, TORSION, BUCKLING, ETC.

A rigid body has a Poisson's ratio of:

  1. 0
  2. 1
  3. Greater than 1
  4. Less than 1

The ratio of stress and strain is known as _____

  1. Modulus of Elasticity
  2. Young's Modulus
  3. Both A and B
  4. None of the above

A 50 lb load is attached at the end of a 1/2" diameter rod. If the cross sectional area of the rod is 0.196 in2, the amount of Uniformly Distributed Normal Stress in psi is most nearly:

  1. 100
  2. 255
  3. 2400
  4. 4300

Also known as Normal or Unit Stress (S) is defined as the amount of force per unit area: S=P/A = 50 lb/0.196 in2 = 255.1 psi

A 50 lb load is attached at the end of a rod. If the cross sectional area of the rod is 126.451 mm2, the amount of Normal Stress in psi is most nearly:

  1. 150
  2. 4001
  3. 3300
  4. 255

It is important to have a few unit conversions memorized: (50 lb/126.451 mm2)(645.16 mm2/in2) = 255.1 psi

The actual breaking stress in stress-strain diagram is the ratio of ______

  1. load at breaking point and original cross-sectional area
  2. load at breaking point and reduced cross-sectional area
  3. maximum load and original cross-sectional area
  4. yield load and original cross-sectional area

Two parallel, equal and opposite forces acting tangentially to the surface of the body is called as _____

  1. Complementary stress
  2. Compressive stress
  3. Shear stress
  4. Tensile stress

Modulus of rigidity (Shear Modulus) is the ratio of ______

  1. Shear stress and shear strain
  2. Shear strain and shear stress
  3. Lateral strain and linear strain
  4. Linear stress and lateral strain

Bulk Modulus (K or B) is the ratio of ______

  1. Hydrostatic pressure stress and volumetric strain
  2. Shear strain and shear stress
  3. Lateral strain and linear strain
  4. Linear stress and lateral strain

What is the bulk modulus of a material, if a cube of 100 mm changes its volume to 4000 mm3 when subjected to compressive force of 2.5 x 106 N?

  1. 62.5 GPa
  2. 65 GPa
  3. 67.5 GPa
  4. 70 GPa

Given: Compressive force = 2.5 x 106 N, change in volume = 4000 mm3, Side of cube = 100 mm

Formula: 1) σ = P/A

Stress is defined as the force per unit area.

2) K = σ / ( δV/V)

here, K is bulk modulus , δV/V is the volumetric strain

Solution:

1) σ = P/A

Area of cube = (100)2 and Compressive force = 2.5 x 106 N

σ = 2.5 x 106 / (100)2 = 250 N/mm2

2) Bulk modulus K = σ / ( δV/V)

K = 250 / (4000 /1003)

K = 62500 Mpa = 62.5 GPa

The relation between modulus of elasticity (E), modulus of rigidity (G) and bulk modulus (K) is given as ________

  1. K+G / (3K+ G)
  2. 3 KG / (3K+ G)
  3. 3 KG / (9K+ G)
  4. 9 KG / (3K+ G)

Bulk Modulus (K) and Rigidity Modulus (G) are defined as follows:

K = \frac{E}{3(1-2 \mu)}, G = \frac{E}{2(1+ \mu)} \,\mathrm{solving \, for \,}\mu:\,  9KG = E(3K+G)

\mathrm{solving \, for \,}E:\,  E = \frac{9KG}{(3K+G)}

Robert Hooke (1676) observed the linear relationship between stress and strain of some materials. Which equation represent Hooke's Law?

  1. 2E u_{0}= \sigma^{2}
  2. \epsilon_{long} L_{0}= \Delta L
  3. \sigma A_{0}= P
  4. \sigma =E \epsilon
  5. \epsilon_{long} \nu= -\epsilon_{lat}

The property called ________ (in.lb) is the potential energy stored up in a deformed body. Or the ability of a material to absorb energy without any permanent damage to the material.

  1. Modulus of Elasticity
  2. Young's Modulus
  3. Bulk Modulus
  4. Resilience

Energy stored in a body within an elastic limit is called as _____

  1. resilience
  2. strain energy
  3. both a. and b.
  4. none of the above

Resilience is called many things in practice (e.g. Proof resilience, Strain Energy). It is the area under the elastic zone of the stress-strain curve multiplied by the volume of the material:

U = \frac{\sigma}{2} \epsilon V\,\,\mathrm{because\, of\, Hooke's\, Law}: U = \frac{\sigma^{2}}{2 E}V

Therefore, the units are in Joules (J) or N-m or lb-in.
Now, without the volume you only have the Modulus or Unit Resilience:

U_{p} = \frac{\sigma}{2} \epsilon

Which has units of N-m/m3 (J/m3 ) because σ is (N/m2) and ε is (m/m).

Modulus of resilience is the ratio of ______

  1. minimum strain energy and unit volume
  2. maximum stress energy and unit volume
  3. proof resilience and unit volume
  4. resilience and unit area

What is the proof resilience of a square bar of 2500 mm2 and 200 mm long, when a load of 150 kN is induced gradually?
(Take E = 150 x 103 Mpa)

  1. 45 J
  2. 8 J
  3. 5.3 J
  4. 6 J

Proof resilience: It is the maximum strain energy a material is capable to store up to its elastic limit. It is also called as strain energy.

Given: Area of square bar = 2500 mm2 , gradual load applied = 150 kN, length of bar 200 mm, E = 150 x 103 Mpa

Formula: Proof resilience:

u= \frac{\sigma^{2} V}{2E}

or in this case: (σ2 / 2E) x AL

Solution:

1) Stress = Applied load / Area = 150 x 103 / 2500 = 60 N/mm2

2) Proof resilience = (σ2 / 2E) x AL

Substituting the given values, we get

= (602 / 2 x 150 x 103) x (2500 x 200)

= 6000 N-mm

= 6 Joule

Proof resilience of a square bar is 6 Joule.

Strain energy stored in a uniform bar is given as ______

  1. (σ E/ 2A)
  2. (σ L/ 2AE)
  3. (σ2 AL/ 4E)
  4. (σ2 AL/ 2E)

In the diagram shown below, a tensile load of 50 kN is applied axially. What is the increase in length of the rectangular tapered plate of width 10 mm. (Assume E = 220 GPa)

  1. 0.100 mm
  2. 0.161 mm
  3. 0.126 mm
  4. 0.560 mm

Given: Tensile load (P)= 50 kN, width of tapered plate (w) = 10 mm, d1 = 100 mm, d2 = 50 mm,
force (P) = 50 x 103 N, E = 220 GPa = 220 x 103 MPa = 220 x 103 N/mm2

Formula: increase in length δL = [(PL) / ((d1- d2) w E)] [loge (d1/d2)]

Substituting the given values,

δL = [(50 x 103 x 400) / ((100 - 50 ) 10 x 220 x 103)] [loge (100/50)]

= 0.126 mm

Increase in length of the rectangular tapered plate of width 10 mm = 0.126 mm

When a rectangular bar is uniaxially loaded, the volumetric strain (ev) is given as ________

  1. σx / E(1- μ)
  2. σx / E(1+ μ)
  3. σx / E(1- 2μ)
  4. σx / E(1+2μ)

Volumetric strain (δV/V) or (ev) is given by Hooke's Law as

(δV/V) = (1/E)( σz+ σx +σy)[1 - 2μ]

here, E = Modulus of elasticity, σx = strain in X-direction, σy = strain in Y-direction, σz = strain in Z-direction

- When a rectangular bar is uniaxially loaded, σz= σx= 0, therefore, the volumetric strain (ev) is given as (σx/ E)(1- 2μ)

- When a rectangular bar is biaxially loaded σz=0, therefore the volumetric strain (ev) is given as (1/ E)(σx -σy) [1- 2μ]

- When a rectangular bar is triaxially loaded, the volumetric strain (ev) is given as [(σz+ σx +σy)/E] [1 - 2μ]

A rectangular bar has volume of 1.5 x 106 mm3. What is the change in volume, if stresses in x, y and z direction are 100 Mpa, 150 Mpa and 160 Mpa respectively. (Assume E = 2 x 105 N/mm2 & μ = 0.3)

  1. 1000 mm3
  2. 1230 mm3
  3. 1500 mm3
  4. 2000 mm3

Given: Volume = 1.5 x 106 mm3, Poisson's ratio (μ) = 0.3, (Note that E = 2 x 105 N/mm2 = 2 x 105 Mpa)

Formula:

1) ev = [σx / E + σy / E + σz / E ] x [1 - 2μ], where ev is the volumetric strain

2) Change in volume: δV = ev x V

Solution:

1) ev = [(100 / 2 x 105) + (150 / 2 x 105) + (160 / 2 x 105)] x [1 - 2μ]

= 2.05 x 10-3 [1-2 (0.3)]

= 8.2 x 10-4

2) Change in volume: δV = ev x V

Substituting the value of ev = 8.2 x 10-4 and V = 1.5 x 106 mm3

= (8.2 x 10-4)(1.5 x 106 mm3)

= 1230 mm3

Change in volume = 1230 mm3

What is the strain energy stored in a cube of 50 mm, when it is subjected to shear stress of 200 MPa. (G = 100 GPa)

  1. 25 Nm
  2. 75 Nm
  3. 125 Nm
  4. 150 Nm

For a square volume element the following strain-energy relation applies:

u_{cube}= \frac{\tau \gamma V}{2} = \frac{\tau^{2} V}{2G}

Then for Shear (τ) 200 MPa, Volume (50 mm)3 = 125,000 mm3 and Modulus of Rigidity (Shear Modulus) G = 100E3 MPa.

ucube = (200 N/mm2)2(125,000 mm3)(1/2)(mm2/100E3 N) = 25,000 N.mm

Since 1 m = 1000 mm, then ucube = 25 N.m

What is the strain energy stored in a simply supported beam due to bending moment?

  1. ∫ (M2/EI)
  2. ∫ (M2/2EI)
  3. ∫ (M/2EI)
  4. ∫ (2M/EI)

Normal stress on a bean undergoing bending is represented:

\sigma = \frac{My}{I}

When combined with the Resilience (Strain-Energy) equation:

U = \int_V \frac{\sigma^{2}}{2E} \mathrm{d}V = \int_V \frac{1}{2E} \left(\frac{My}{I}\right)^{2} \mathrm{d}A\mathrm{d}x = \int_0^L \frac{M^{2}}{2EI^{2}} \left(\int_A y^{2} \mathrm{d}A\right) \mathrm{d}x

U =\int_0^L \frac{M^{2}}{2EI} \mathrm{d}x

Stress on an object due to sudden load is _____ the stress induced when the load is applied gradually.

  1. equal to
  2. twice
  3. half
  4. thrice

Dynamic, Impact or Sudden loads and deflections at the point of impact are found from the static stress (σstatic) and static deflection (ΔLstatic):

\sigma_{impact} = n \sigma_{static},  \Delta L_{impact} = n \Delta L_{static}

Factor n represents the magnification of a statically applied load so that it can be treated dynamically, n can be calculated for any member that has a linear relationship between load and deflection.

n = 1+ \sqrt{1+2\left(\frac{h}{\Delta L_{static}}\right)}

The equation above is the result of using the conservation of energy to analyse elastic deflection of a load impacting a spring and h being the height at which the load is dropped from. Provided the block does not rebound off the spring, it will continue to vibrate until the motion dampens out and the block assumes the static position, (ΔLstatic). Note also that if the block is held just above the spring, h = 0, and released then n = 2, therefore:

\sigma_{impact} = 2 \sigma_{static},  \Delta L_{impact} = 2 \Delta L_{static}

Impact stresses and displacements have twice (2) the magnitude of those attained by applying a load gradually.

What is the maximum stress induced in a bar 2500 mm2, when a load of 2000 kN is applied suddenly?

  1. 400 N/mm2
  2. 800 N/mm2
  3. 1600 N/mm2
  4. Insufficient data

Given F = 2000 kN and A = 2500 mm2:

\sigma_{impact} = 2 \sigma_{static} = 2 \left(\frac{F}{A}\right)

σimpact = 2(2000 x 103 N/2500 mm2) = 1600 N/mm2

What is the strain energy caused due to self weight in a cylindrical bar?

  1. (W2 L)/(2 AE)
  2. (W L)/(8 AE)
  3. (τ2/ 2G)V
  4. (τ2/ G)V

The strain in the bar is given by Hooke's Law:

\epsilon = \frac{\sigma}{E} = \frac{W}{AE}  where \sigma = \frac{W}{A}

Then we substitute ε into the Strain-Energy equation:

u_{bar} = \frac{\sigma \epsilon V}{2} = \frac{W \epsilon V}{2A} = \frac{W}{2A} \frac{W V}{AE} = \frac{W}{2A} \frac{W}{AE} \frac{AL}{1}

u_{bar} = \frac{W^{2} L}{2AE}

Load is gradually applied on a steel plate shown below, what is the modulus of resilience, if 130 N/mm2 is the maximum stress applied? (Assume E = 100 x 103 Mpa)

  1. 0.845 N/mm2
  2. 0.0845 N/mm2
  3. 0.00845 N/mm2
  4. Insufficient data

Proof resilience: It is the maximum strain energy which is stored within elastic limit. It is also called as strain energy.

Modulus of resilience: It is defined as the proof resilience per unit volume. It is given as (σ2 / 2E)

Given: Maximum stress applied = 130 N/mm2, E = 100 x 103 Mpa

Formula: Modulus of resilience = (σ2 / 2E)

Solution: Substituting the given values, we get

Modulus of resilience = (σ2 / 2E)

= (1302) / (2 x 100 x 103)

= 0.0845 N/mm2, which could also be presented as 0.0845 N-mm/mm3 to highlight the Energy/Volume ratio.

Modulus of resilience = 0.0845 N/mm2

Which stress is induced in a member, when expansion or contraction due to temperature variation is prevented?

  1. Compressive stress
  2. Tensile stress
  3. Thermal stress
  4. None of the above

The thermal stress induced in a steel rod is compressive, if temperature _____

  1. decrease
  2. increase
  3. remains constant
  4. none of the above

The deformation of a material caused due temperature variation develops ______

  1. stress
  2. strain
  3. both a. and b.
  4. none of the above

In the diagram shown below, material A and B expand freely as shown in condition (2). If expansion is prevented for equilibrium condition(3) to reach, then

  1. tensile force exerted by material B on A is equal to tensile force exerted by A on B.
  2. tensile force exerted by material B on A is equal to compressive force exerted by A on B.
  3. compressive force exerted by material B on A is equal to tensile force exerted by A on B.
  4. no force acts on the composite member in equilibrium conditions

Force of 100 kN is applied on 900 mm long rod of 20 mm diameter and is simultaneously heated at 70o C. What is the total elongation of the rod? (Assume coefficient of thermal expansion = 10 x 10-6 oC, Young's modulus = 150 Gpa)

  1. 2.54 mm
  2. 2.00 mm
  3. 1.27 mm
  4. 0.63 mm

Given: Diameter of rod = 20 mm, length of rod (L) = 900 mm, heating temperature (t)= 70 °C, coefficient of thermal expansion (α) = 10 x 10-6 /°C, Young's modulus = 150 Gpa

Formula: 1) Total elongation = elongation when heated + elongation due to pull

2) Elongation due to pull (δL) = (PL/AE)

3) Elongation due to heating = α t L

Solution:

1) Elongation due to pull (δL) = (PL/AE)

= [100000 x 900] /[(π/4) x 202 x 150 x 103]

= 1.9098

2) Elongation due to heating = α t L

= 10 x 10-6 x 70 x 900

= 0.63

3) Total elongation = elongation when heated + elongation due to pull

Total elongation = 1.9098+0.63 = 2.54 mm

In a circuit box, a copper rod at room temperature of 25o C has a gap of 0.3 mm between the end of the rod Q and the rigid wall. If temperature of the rod increases to 70o C ,then what is the compressive stress acting on the rod PQ? (Assume Young's modulus = 100 Gpa & α = 15 x 10-6 /oC)

  1. 6 MPa
  2. 7.5 MPa
  3. 8.2 MPa
  4. None of the above

Given: Gap = 0.3 mm, Young's modulus = 100 Gpa & α = 15 x 10-6 /°C,

Solution:

Free expansion of copper rod = α t L = 15 x 10-6 x (70 - 25) x 500

= 0.3375

Actual expansion prevented = Free expansion of copper rod - gap

= 0.3375 - 0.3

= 0.0375

Temperature strain (e) = (δL/L) = (0.0375/ 500) = 7.5 x 10-5

Temperature stress = e E = (7.5 x 10-5 x 100 x 103) = 7.5 Mpa

Compressive stress acting on the rod PQ = 7.5 Mpa

The value of elasticity increases, when temperature _____

  1. increases
  2. decreases
  3. remains constant
  4. None of the above

If a composite bar of copper and steel is heated at 120o C, the stress induced in copper bar is _______

  1. compressive stress
  2. shear stress
  3. tensile stress
  4. thermal stress

Thermal stress is not a function of _____

  1. change in temperature
  2. coefficient of linear expansion
  3. modulus of elasticity
  4. None of the above

What is the value of thermal stress, if a rod of 3 m is heated at 50 oC and is fixed at both the ends?
(Take α = 10 x 10-6 /oC & E= 200 x 103 Mpa)

  1. 25 MPa
  2. 50 MPa
  3. 100 MPa
  4. 150 MPa
Review Prepared by Estiven R. Sierra, 2016