Strength of Materials Test 1

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Strength of Materials Test 1

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Question 1
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Modulus of rigidity and Poisson’s ratio are 0.4 × 10⁵ MPa and 0.34 respectively. The Young’s modulus of elasticity is

A
2.1 × 10⁵ MPa

B
1.1 × 10³ MPa

C
1.1 × 10⁵ MPa

D
0.78 × 10⁵ MPa

Solution

Concept :

E = Young's Modulus of Rigidity = Stress / strain

G = Shear Modulus or Modulus of rigidity = Shear stress / Shear strain

ν = Poisson’s ratio = - lateral strain / longitudinal strain

K = Bulk Modulus of elasticity = Direct stress / Volumetric strain

Relation between E, K and ν

E = 2G (1 + ν)

E = 3K (1 - 2ν)

Calculation :

G = 0.4 × 10⁵ MPa, ν = 0.34

E = 2G (1 + ν) = 2 × 0.4 × 10⁵× (1 + 0.34) ≈ 1.1 × 10⁵ MPa
Question 2
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The bending moment distribution in a beam is a function of distance x and it is given by M = (5x²+ 20x - 7) Nm. Find the shear stress at x = 2m if the shearing area is 2 mm².

A
10 MPa

B
20 MPa

C
30 MPa

D
40 MPa

Solution

Concept:

\({\rm{shear\;force}},{\rm{\;F}} = {\rm{}}\frac{{{\rm{dM}}}}{{{\rm{dx}}}}\)

Calculation:

Bending moment (M) = (5x²+ 20x - 7)

Now,

\({\rm{shear\;force}},{\rm{\;F}} = {\rm{}}\frac{{{\rm{dM}}}}{{{\rm{dx}}}}\)

∴ F = 10x + 20

Now,

(F)_{(x = 2 m)} = 10(2) + 20

∴ F = 40 N

Now,

shear stress \({\rm({\tau })} = {\rm{}}\frac{{\rm{F}}}{{\rm{A}}} = {\rm{}}\frac{{40}}{2}\)

∴ τ = 20 MPa
Question 3
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A steel rod 3 m long with a cross sectional area of 250 mm² is stretched between two fixed points. The tensile force is 12000 N at 40°C. Using E = 200 × 10³ N/mm² and α = 11.7 × 10^-6 mm/mm°C.

Calculate the temperature at which the tensile stress in the bar will be 100 N/mm².

A
22.2°C

B
62.2°C

C
17.8°C

D
34.5°C

Solution

Explanation:

The rod is stretched so there will be initial tensile stress.

Initial stress \(= \frac{\sigma }{A} = \frac{{12000}}{{250}} = 48\;N/m{m^2}\)

Final stress = 100 N/mm²

Additional stress required due to temperature change:

σ_th = σ_fin = σ_ini = 100 – 48 = 52 N/mm²

Thermal stress:

σ_th = αΔTE

52 = 11.7 × 10^-6 × ΔT × 200 × 10³

ΔT = T₂ – T_i = 22.22 °C

Note: - (Temperature will be reduced as tensile stress needs to be generated)

40°– T2 = 22.22 ⇒ T₂ = 17.8°C
Question 4
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A thin cylindrical shell 2.5 m long and a diameter of 1.25 m has a thickness of 20 mm. Then the change in dimension of diameter of the shell which is subjected to measure of 1.5 MPa is ______mm (Take Young's Modulus E = 200 GPa, and Poisson's Ratio μ = 0.3)

A
0.24-0.26

B
0.244-0.261

C
0.247-0.266

D
0.249-0.262

Solution

Concept:

For a thin cylinder:

Longitudinal stress: \({\sigma _L} = \frac{{pd}}{{4t}}\)

Hoop stress: \({\sigma _h} = \frac{{pd}}{{2t}} = 2{\sigma _L}\)

Longitudinal Strain:

\({\epsilon_L} = \frac{1}{E}\left( {{\sigma _L} - \mu {\sigma _H}} \right) = \frac{{{\sigma _L}}}{E}\left( {1 - 2\mu } \right) = \frac{{pd}}{{4tE}}\left( {1 - 2\mu } \right)\)

Circumferential or hoop strain:

\({\epsilon_H} = \frac{1}{E}\left( {{\sigma _H} - \mu {\sigma _L}} \right) = \frac{{{\sigma _L}}}{E}\left( {2 - \mu } \right) = \frac{{pd}}{{4tE}}\left( {2 - \mu } \right)\)

Calculation:

Given, l = 2.5 m = 2500 mm, d = 1.25 m = 1250 mm, t = 20 mm

P = 1.5 MPa, E = 200 GPa = 200 × 10³ MPa, μ = 0.3

In cylindrical shell Hoop strain \({{\epsilon }_{H}}=\frac{\delta d}{d}=\frac{Pd}{4tE}~\left( 2-\mu \right)\)

\(\Rightarrow \delta D=\frac{P{{d}^{2}}}{4tE}\left( 2-\mu \right)\)

\(\delta D=\frac{1.5\times {{1250}^{2}}}{4\times 20\times 200\times {{10}^{3}}}\left( 2-0.3 \right)\)

δD = 0.249 mm
Question 5
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Direct stresses of 120 N/mm² (tensile) and 90 N/mm² (compressive) exist on two perpendicular planes at a certain point in a body. They are also accompanied by shear stresses on the planes. The greater principal stress at the point due to these is 150 N/mm². The shear stress on these planes is equal to

A
85 MPa

B
170 MPa

C
135 MPa

D
42 MPa

Solution

Concept:

\( {\sigma _{1,2}} = \frac{1}{2}\left[ {\left( {{\sigma _x} + {\sigma _y}} \right) \pm \sqrt {{{\left( {{\sigma _x} - {\sigma _y}} \right)}^2} + 4\tau _{xy}^2} } \right]\\ {\tau _{max}} = \frac{{{\sigma _1} - {\sigma _2}}}{2} \)

Calculation:

Given: σ_x = 120 N/mm₂, σ_y = -90 N/mm², σ₁ = 150 N/mm²

\( {\sigma _{1,2}} = \frac{1}{2}\left[ {\left( {{\sigma _x} + {\sigma _y}} \right) \pm \sqrt {{{\left( {{\sigma _x} - {\sigma _y}} \right)}^2} + 4\tau _{xy}^2} } \right] \)

\({\sigma _1} = \frac{1}{2}\left[ {\left( {120 - 90} \right) + \sqrt {{{\left( {120 + 90} \right)}^2} + 4{{\left( \tau \right)}^2}} } \right] = 150\)

\( \Rightarrow 30 + \sqrt {{{\left( {210} \right)}^2} + 4{{\left( \tau \right)}^2}} = 300\)

⇒ (210)² + 4(τ)² = (270)²

τ = 84.85 N/mm²

Shear stress on these plane = 84 MPa

Confusion Point:

Sum of normal stress = sum of principal stress

σ_x + σ_y = σ₁ + σ₂

120 – 90 = 150 + σ₂

σ₂ = -120 N/mm²

\({\tau _{max}} = \frac{{{\sigma _1} - {\sigma _2}}}{2} = \frac{{150 - \left( { - 120} \right)}}{2} = 135N/m{m^2}\)

This is maximum shear stress, not shear stress on the oblique plane.