Mechanics of Solids (Solved Numerical Problems)

A stepped circular bar 150 mm long with diameters 20, 15, and 10 mm along length AB = 40 mm, BC - 45 mm, and CD = 65 mm, respectively is subjected to various forces as shown in Fig. P-14.12(a). Determine the change in its length if E = 200 $kN/m{m^2}$. (AMIE Winter 2023)

Solution

Considering portion AB, a tensile force of 20 kN is to act to maintain equilibrium.

For portion BC, if it is considered to be subjected to a tensile force of 5 kN, then the net force at section B will be 15 kN as shown. Similarly, for the portion CD, a compressive force of 10 kN would make a net force of 15 kN at section C.

Area of cross sections

$\begin{array}{l}{A_1} = \frac{\pi }{4}{(20)^2} = 314.16\,m{m^2}\\{A_2} = \frac{\pi }{4}{(15)^2} = 176.7\,m{m^2}\\{A_3} = \frac{\pi }{4}{(10)^2} = 78.54\,m{m^2}\end{array}$

Extension of AB

$\begin{array}{l}\delta {L_1} = \frac{{20x40}}{{314.16E}}\\ = \frac{{20x40}}{{314.16x200}}\\ = 1.273x{10^{ - 2}}\,mm\end{array}$

Extension of BC

$\begin{array}{l}\delta {L_2} = \frac{{5x45}}{{176.7E}}\\ = \frac{{5x45}}{{176.7x200}}\\ = 0.636\,x\,{10^{ - 2}}\,mm\end{array}$

Contraction in portion CD

$\begin{array}{l}\delta {L_3} = \frac{{10x65}}{{78.54E}}\\ = \frac{{10x65}}{{78.54x200}}\\ = 4.138x{10^{ - 2}}\,mm\end{array}$

Change in length of bar

$\begin{array}{l}\delta L = (1.273 + 0.636 - 4.138)x{10^{ - 2}}\,mm\\ =  - 2.229\,x\,{10^{ - 2}}\,mm\end{array}$

A cast iron water pipe 500 mm outer diameter with 30 mm thickness is supported over a span of 12 m. Find the maximum stress in cast iron when the pipe is running full of water.
Density of cast iron = 7200 $kg/{m^3}$
Density of water = 1000 $kg/{m^3}$

(AMIE Winter 2023)

Solution

D = 0.5 m

d = 0.44 m

${A_c} = \frac{\pi }{4}\left[ {{{0.5}^2} - {{0.44}^2}} \right] = 0.044\,{m^2}$

Consider 1 m length

Weight of CI pipe per unit length

= 0.044 ${m^2}$  x 72 $kN/{m^3}$ = 3.9 KN/m 

Water

Area of cross section

= $\frac{\pi }{4}{(0.44)^2} = 0.16\,{m^2}$

Consider 1 m length

Weight of water in pipe per unit length

= 0.16 ${m^2}$ x 100 $kN/{m^3}$

= 1.6 kN/m

Total UDL of water

w = 3.19 + 1.6 = 4.79 kN/m

Length of pipe = 12 m

$\begin{array}{l}{M_{\max }} = \frac{{w{l^2}}}{8} = \frac{{4.79{{(12)}^2}}}{8}\\ = 86.22\,kNm\end{array}$

Cast iron pipe

Moment of inertia

$\begin{array}{l}I = \frac{\pi }{{64}}({D^4} - {d^4})\\ = \frac{\pi }{{64}}[{0.5^4} - {0.44^4}]\\ = 0.0012\,{m^4}\end{array}$

${y_{\max }} = 0.5/2 = 0.25\,m$

Maximum stress

$\begin{array}{l}{\sigma _{\max }} = \frac{M}{I}{y_{\max }}\\ = \frac{{86.22}}{{0.0012}}(0.25)\\ = 17962\,kN/{m^2}\end{array}$

A metal bar 4 m long and 100 mm × 200 mm in cross section is subjected to a pull of 50 kN. If Young’s modulus of the material of the bar is 200 $kN/m{m^2}$, find: (i) stress in the bar (ii) Strain produced (iii) elongation of the bar (iv) work done. (AMIE Winter 2023)

Solution

Stress in the bar

$\sigma  = \frac{P}{A} = \frac{{50}}{{100\,x\,200}} = 2.5\,x\,{10^{ - 3}}\,kN/m{m^2}$

Strain

$e = \frac{\sigma }{E} = \frac{{2.5\,x\,{{10}^{ - 3}}}}{{200}} = 1.25\,x\,{10^{ - 5}}$

Elongation

$\begin{array}{l}\delta L = \frac{{PL}}{{AE}}\\ = \frac{{50(4000)}}{{100\,x\,200\,x\,200}}\\ = 0.05\,mm\end{array}$

Work done

W = Force x change in length

= (50 x 1000) N x (0.05/1000) m

= 2.5 N-m = 2.5 J

Note: In the question, value of E is given as 2000 $kN/m{m^2}$. Which is incorrect, it should be 200 $kN/m{m^2}$. We have taken corrected value in the question.

A steel wire, 10 mm in diameter, is used for hoisting purposes during the construction of a building. If 150 m of the wire is hanging vertically, and a load of 800 N is being lifted at the lower end of the wire, determine the total elongation of the wire.
Given: specific weight of steel = $7.8\,x\,{10^4}\,N/{m^3}$ and  E= 200 $GN/{m^2}$

(AMIE Winter 2023)

Solution

Specific weight of steel = $7.8\,x\,{10^4}\,N/{m^3}$ 

= $0.078\,x\,{10^{ - 3}}\,N/m{m^3}$

$E = 200\,x\,1000\,N/m{m^2}$

d = 10 mm

$A = \frac{\pi }{4}{(10)^2} = 78.5\,m{m^2}$

l = 150 m = 150 x 1000 mm

W = 800 N

Elongation due to weight W

$\begin{array}{l}\delta {l_1} = \frac{{Wl}}{{AE}}\\ = \frac{{800(150\,x\,1000)}}{{78.5\,x\,(200\,x\,1000)}}\\ = 7.64\,mm\end{array}$

Elongation due to self weight

$\begin{array}{l}\delta {l_2} = \frac{{w{l^2}}}{{2E}}\\ = \frac{{(0.078\,x\,{{10}^{ - 3}}){{(150\,x\,1000)}^2}}}{{2(200\,x\,1000)}}\\ = 4.39\,mm\end{array}$

Total elongation

= 7.64 + 4.39

= 12 mm

A hollow steel shaft 3 m long must transmit a torque of 25 kN-m. The total angle of twist in this length is not to exceed 2.5°, and the allowable shearing stress is 90 MPa. Determine the inside and outside diameter of the shaft if G = 85 GPa. (AMIE Winter 2023)

Solution

$\begin{array}{l}{2.5^0}\left( {\frac{{rad}}{{57.3\,\deg }}} \right) = \frac{{(250000(3)}}{{(85x{{10}^9})(\pi /32)({d_o}^4 - {d_i}^4)}}\\ \Rightarrow \,{d_o}^4 - {d_i}^4 = 206x{10^{ - 6}}\end{array}$

The maximum shear force will occur at outer fibre at $\rho  = {d_o}/2$

Using the equation  $\tau  = \frac{{T\rho }}{J}$

$\begin{array}{l}90\,x\,{10^6} = \frac{{(250000({d_o}/2)}}{{(\pi /32)({d_o}^4 - {d_i}^4}}\\ \Rightarrow {d_o}^4 - {d_i}^4 = (1414{d_o})({10^{ - 6}})\end{array}$

Comparing equations of  ${d_o}^4 - {d_i}^4$

$\begin{array}{l}206\,x\,{10^{ - 6}} = 1414{d_o}({10^{ - 6}})\\ \Rightarrow {d_o} = 0.145\,m = 145\,mm\\From\,this,{d_i} = 0.125\,m = 125\,mm\end{array}$

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