ME-304 Strength of Material - Dec 2023 Question Paper

Under each question, How to answer lists the points an answer should cover, outlined by CampusPrep from the syllabus. Check the details against your textbook.

1. 7 Marks
   aA simply supported beam of length $8\text{ m}$ rests on supports $6\text{ m}$ apart, the right-hand end is overhanging by $2\text{ m}$. The beam carries a uniformly distributed load of $1500\text{ N/m}$ over the entire length. Draw S.F. and B.M. diagrams and find the point of contraflexure, if any.
   How to answer

   • Calculate reactions at supports by equilibrium equations
   • Determine shear force values along the span and draw SFD
   • Determine bending moment values, find maximum positive BM and BM at support
   • Equate BM equation to zero to locate the point of contraflexure
   • Draw neat and scaled SFD and BMD
2. 7 Marks
   bAnswer the following: i) Derive strain energy equation for gradual loading. ii) What are the assumptions of simple bending?
   How to answer

   • Part (i): Define strain energy for gradual loading; plot load vs deformation; work done $W = \frac{1}{2} P \delta$; derive $U = \frac{\sigma^2}{2E} \times \text{Volume}$
   • Part (ii): State key assumptions of pure bending (material is homogeneous and isotropic, transverse sections remain plane after bending, Hooke's law obeyed, radius of curvature large compared to dimensions, etc.)
3. 7 Marks
   aA straight circular rod tapering from diameter ‘D’ at one end to a diameter ‘d’ at the other end is subjected to an axial load ‘P’. Obtain an expression for the elongation of the rod.
   How to answer

   • Draw diagram of linearly tapering circular bar showing diameters D, d, and length L
   • Express diameter at an elemental section of thickness $dx$ as $d_x = d + \frac{D-d}{L}x$
   • Write elongation of element: $d(\delta) = \frac{P dx}{A_x E} = \frac{4P dx}{\pi d_x^2 E}$
   • Integrate over the entire length from $0$ to $L$
   • Arrive at final formula $\delta = \frac{4PL}{\pi E D d}$
4. 7 Marks
   bA timber cantilever $200\text{ mm}$ wide and $300\text{ mm}$ deep is $3\text{ m}$ long. It is loaded with a U.D.L of $3\text{ kN/m}$ over the entire length. A point load of $2.7\text{ kN}$ is placed at the free end of the cantilever. Find the maximum bending stress produced.
   How to answer

   • Calculate maximum bending moment at the fixed support: $M_{\max} = (2.7 \times 3) + \frac{3 \times 3^2}{2} = 8.1 + 13.5 = 21.6\text{ kN}\cdot\text{m} = 21.6 \times 10^6\text{ N}\cdot\text{mm}$
   • Calculate section modulus: $Z = \frac{b d^2}{6} = \frac{200 \times 300^2}{6} = 3 \times 10^6\text{ mm}^3$
   • Determine maximum bending stress: $\sigma_{\max} = \frac{M_{\max}}{Z} = \frac{21.6 \times 10^6}{3 \times 10^6} = 7.2\text{ N/mm}^2$
5. 14 Marks
   Derive the stress distribution for circular section and plot shear stress distribution.
   How to answer

   • State the general shear stress formula $q = \frac{F A \bar{y}}{I b}$
   • Consider a circular cross section of radius $R$ and a level at distance $y$ from neutral axis
   • Derive expression for $A\bar{y}$ and width $b = 2\sqrt{R^2 - y^2}$
   • Obtain $q = \frac{4F}{3\pi R^4}(R^2 - y^2)$ showing parabolic distribution
   • Calculate maximum shear stress at neutral axis $q_{\max} = \frac{4}{3} q_{\text{avg}}$
   • Plot the parabolic shear stress distribution across the circular depth
6. 7 Marks
   aFind the expression for the slope and deflection of a cantilever of length L which carries a uniformly distributed load over a length 'a' from the fixed end by Moment area method.
   How to answer

   • Sketch the beam and draw the $\frac{M}{EI}$ diagram for cantilever with UDL over length $a$ from fixed end
   • Apply Mohr's 1st Theorem: $\theta = \frac{1}{EI} \times \text{Area of } M\text{-diagram}$ to find slope at free end
   • Calculate area of parabolic bending moment diagram of length $a$: $\text{Area} = \frac{1}{3} \left(\frac{w a^2}{2}\right) a = \frac{w a^3}{6}$
   • Apply Mohr's 2nd Theorem to compute deflection: $y = \frac{1}{EI} (\text{Area} \times \bar{x})$ where $\bar{x}$ is measured from free end
   • Conclude with expressions for maximum slope and maximum deflection
7. 7 Marks
   bA cast iron pipe of external diameter $60\text{mm}$, internal diameter of $40\text{mm}$ and of length $5\text{ m}$ is supported at its ends. Calculate the maximum bending stress induced in the pipe if it carries a point load of $100\text{ N}$ at its centre.
   How to answer

   • Calculate maximum bending moment: $M_{\max} = \frac{W L}{4} = \frac{100 \times 5000}{4} = 125,000\text{ N}\cdot\text{mm}$
   • Compute section modulus $Z = \frac{\pi}{32} \frac{D^4 - d^4}{D} = \frac{\pi}{32} \frac{60^4 - 40^4}{60} \approx 16,886\text{ mm}^3$
   • Calculate maximum bending stress: $\sigma = \frac{M_{\max}}{Z} = \frac{125,000}{16,886} \approx 7.40\text{ N/mm}^2$
8. 7 Marks
   aAn elemental cube is subjected to tensile stresses of $30\text{ N/mm}^2$ and $10\text{ N/mm}^2$ acting on two mutually perpendicular planes and a shear stress of $10\text{ N/mm}^2$ on these planes. Draw the Mohr's circle of stresses and hence or otherwise determine the magnitudes and directions of principal stresses and also the greatest shear stress.
   How to answer

   • State center and radius of Mohr's circle: $C = \left(\frac{30+10}{2}, 0\right) = (20, 0)$ and $R = \sqrt{10^2 + 10^2} = 10\sqrt{2} \approx 14.14\text{ N/mm}^2$
   • Draw Mohr's circle using proper scale
   • Calculate major and minor principal stresses: $\sigma_1 = 20 + 14.14 = 34.14\text{ N/mm}^2$, $\sigma_2 = 20 - 14.14 = 5.86\text{ N/mm}^2$
   • Determine principal plane direction: $\tan 2\theta_p = \frac{2(10)}{30-10} = 1 \implies 2\theta_p = 45^\circ \implies \theta_p = 22.5^\circ, 112.5^\circ$
   • Calculate greatest shear stress: $\tau_{\max} = R = 14.14\text{ N/mm}^2$
9. 7 Marks
   bCalculate the modulus of rigidity and bulk modulus of a cylindrical bar of diameter $30\text{ mm}$ and of length $1.5\text{ m}$ if the longitudinal strain in a bar during a tensile stress is four times the lateral strain. Find the change in volume, when the bar is subjected to a hydrostatic pressure of $100\text{ N/mm}^2$. Take $E = 1 \times 10^5\text{ N/mm}^2$.
   How to answer

   • Identify Poisson's ratio: $\mu = \frac{\text{lateral strain}}{\text{longitudinal strain}} = \frac{1}{4} = 0.25$
   • Compute Modulus of Rigidity: $G = \frac{E}{2(1+\mu)} = \frac{10^5}{2(1+0.25)} = 40,000\text{ N/mm}^2$
   • Compute Bulk Modulus: $K = \frac{E}{3(1-2\mu)} = \frac{10^5}{3(1 - 2(0.25))} = \frac{10^5}{1.5} \approx 66,666.67\text{ N/mm}^2$
   • Calculate initial volume $V = \frac{\pi}{4}(30)^2 \times 1500 = 1.0603 \times 10^6\text{ mm}^3$
   • Compute change in volume under hydrostatic pressure: $\Delta V = -\frac{p V}{K} = -\frac{100 \times 1.0603 \times 10^6}{66,666.67} \approx -1590.4\text{ mm}^3$
10. 14 Marks
    A weight of $10\text{ kN}$ falls by $30\text{ mm}$ on a collar rigidly attached to a vertical bar $4\text{ m}$ long and $1000\text{ mm}^2$ in section. Find the instantaneous expansion of the bar. Take $E = 210\text{ GPa}$. Derive the formula you use.
    How to answer

    • Derive impact stress formula by equating potential energy loss $W(h + \delta)$ to strain energy stored $\frac{\sigma^2}{2E} A L$
    • Express $\delta = \frac{\sigma L}{E}$ to form quadratic equation in $\sigma$ and solve to obtain $\sigma = \frac{W}{A}\left(1 + \sqrt{1 + \frac{2hAE}{WL}}\right)$
    • Substitute values: $W = 10000\text{ N}$, $A = 1000\text{ mm}^2$, $h = 30\text{ mm}$, $L = 4000\text{ mm}$, $E = 210000\text{ N/mm}^2$
    • Compute static stress $\sigma_{\text{st}} = 10\text{ N/mm}^2$ and impact factor $\left(1 + \sqrt{1 + \frac{2 \times 30 \times 1000 \times 210000}{10000 \times 4000}}\right) = 1 + \sqrt{1 + 315} = 1 + 17.776 = 18.776$
    • Find instantaneous stress $\sigma_i = 187.76\text{ N/mm}^2$ and instantaneous expansion $\delta = \frac{\sigma_i L}{E} = \frac{187.76 \times 4000}{210000} \approx 3.58\text{ mm}$
11. 14 Marks
    A cylindrical shell made of mild steel plate and $1.2\text{ m}$ in diameter is to be subjected to an internal pressure of $1.5\text{ MN/m}^2$. If the material yields at $200\text{ MN/m}^2$, calculate the thickness of the plate on the basis of the following three theories, assuming a factor of safety $3$ in each case: i) maximum principal stress theory ii) maximum shear stress theory iii) maximum shear strain energy theory
    How to answer

    • Find allowable stress: $\sigma_{\text{allow}} = \frac{200}{3} = 66.67\text{ MN/m}^2$
    • Express principal stresses: $\sigma_1 = \frac{pd}{2t} = \frac{1.5 \times 1200}{2t} = \frac{900}{t}$, $\sigma_2 = \frac{pd}{4t} = \frac{450}{t}$, $\sigma_3 = 0$
    • Part (i) Maximum Principal Stress Theory: $\sigma_1 = \sigma_{\text{allow}} \implies \frac{900}{t} = 66.67 \implies t = 13.5\text{ mm}$
    • Part (ii) Maximum Shear Stress Theory: $\tau_{\max} = \frac{\sigma_1 - 0}{2} = \frac{\sigma_{\text{allow}}}{2} \implies \frac{900}{t} = 66.67 \implies t = 13.5\text{ mm}$
    • Part (iii) Distortion Energy Theory: $\sigma_1^2 - \sigma_1\sigma_2 + \sigma_2^2 = \sigma_{\text{allow}}^2 \implies \left(\frac{900}{t}\right)^2 \left(1 - 0.5 + 0.25\right) = 66.67^2 \implies \frac{900\sqrt{0.75}}{t} = 66.67 \implies t = 11.69\text{ mm}$
12. 7 Marks
    aWhat are the assumptions for Euler's formula? When we apply Rankine's formula in place of Euler's formula than comment on results.
    How to answer

    • List assumptions of Euler's formula (initially straight column, perfectly axial load, failure entirely by buckling, stress within elastic limit, negligible self-weight)
    • Explain limitations of Euler's formula: valid only for long/slender columns ($L_e/k >$ critical value) where buckling occurs before crushing
    • Explain when and why Rankine's formula is applied: valid for both short, intermediate, and long columns
    • State Rankine's formula $\frac{1}{P_R} = \frac{1}{P_c} + \frac{1}{P_e}$ and comment on how it smoothly transitions from crushing failure to buckling failure
13. 7 Marks
    bWrite the short note on Area moment method for deflection of beams and its advantage and limitation over double integration method.
    How to answer

    • Introduce Mohr's Moment-Area theorems (Theorem 1 for change in slope, Theorem 2 for tangential deviation)
    • Highlight advantages over double integration: avoids tedious integration, quick calculation for standard loadings and stepped/non-uniform cross-sections
    • Highlight limitations: difficult for continuous beams, requires reference tangent of known zero slope (like at fixed support), complex BMD areas for complex loadings require standard area-centroid knowledge