CE-305 Strength of Materials - Dec 2025 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 square steel rod $20 \text{mm} \times 20 \text{mm}$ in section is to carry an axial load (compressive) of $100 \text{ KN}$. Calculate the shortening in a length of $50 \text{ mm}$. $E = 2.14 \times 10^{8} \text{ KN/m}^{2}$.
   How to answer

   • Axial stress = P/A compressive
   • Axial strain = stress / E with consistent units
   • Shortening delta L = P*L/(A*E)
   • Substitute values and compute result in mm
2. 7 Marks
   bDerive an expression for the thermal stress developed in a bar due to a temperature change.
   How to answer

   • Free expansion delta_T = alpha*T*L if unrestrained
   • Prevented expansion induces compressive strain
   • Compatibility: thermal strain = mechanical strain
   • Derive sigma = alpha*T*E
3. 7 Marks
   aExplain the distribution of shear stress across a rectangular section with neat sketches.
   How to answer

   • Shear stress formula $\tau = VQ/It$
   • Zero at top/bottom fibres, parabolic variation
   • Maximum at neutral axis = 1.5 times average
   • Neat cross-section and parabolic stress diagram
4. 7 Marks
   bA rectangular beam $200 \text{mm}$ deep and $300 \text{ mm}$ wide is simply supported over a span of $8 \text{m}$. What uniformly distributed load per metre the beam is carrying if the bending stress is not to exceed $120 \text{N/mm}^{2}$.
   How to answer

   • Section modulus Z = b*d^2/6 for rectangle
   • Maximum bending moment for simply supported UDL: M = w*L^2/8
   • Bending formula sigma = M/Z
   • Solve for w with sigma_max = 120 N/mm2
5. 7 Marks
   aA simply supported beam of span $L$ carries a uniformly distributed load $w$ over the entire span. Using the double integration method, determine:

   1. (a))Slope at the supports
   2. (b))Maximum deflection
   How to answer

   • Bending moment equation for UDL and EI*d2y/dx2 = M(x)
   • Integrate twice and apply boundary conditions y=0 at supports
   • Symmetry gives slope equal and opposite at supports
   • Obtain slope = wL^3/24EI and max deflection = 5wL^4/384EI at midspan
6. 7 Marks
   bExplain Macaulay's method for finding slope and deflection of beams.
   How to answer

   • Macaulay's method uses single moment equation with Macaulay brackets
   • Integration rules for concentrated and distributed loads
   • Application of boundary conditions to find constants
   • Advantages for discontinuous loading
7. 7 Marks
   aA hollow mild steel tube $6 \text{m}$ long $4 \text{cm}$ internal diameter and $6 \text{ mm}$ thick is used as a strut with both ends hinged. Find the crippling load and safe load taking factor of safety as $3$. Take $E = 2 \times 10^{5} \text{ N/mm}^{2}$.
   How to answer

   • Compute outer diameter, area and moment of inertia of hollow tube
   • Effective length Le = L for both ends hinged
   • Euler crippling load Pcr = pi^2*E*I/Le^2
   • Safe load = Pcr / factor of safety 3
8. 7 Marks
   bDerive Euler's formula for long columns with different end conditions.
   How to answer

   • Differential equation EI*d2y/dx2 + P*y = 0 for buckled column
   • Solution for pinned ends gives Euler load pi^2EI/L^2
   • Introduce effective length for other end conditions
   • List Le = 2L, L, L/sqrt(2), L/2 for free, pinned, fixed-pinned, fixed
9. 7 Marks
   aA hollow circular shaft of external diameter $50 \text{mm}$ and internal diameter $40 \text{mm}$ transmit a torque of $10 \text{ kNm}$. Find the maximum shear induced in the shaft.
   How to answer

   • Polar moment of inertia J = pi/32*(D^4-d^4)
   • Torsion formula tau = T*r/J
   • Maximum shear at outer radius
   • Substitute T = 10 kNm to get value
10. 7 Marks
    bExplain the torsion of hollow circular shaft and compare their strength and stiffness with solid shaft.
    How to answer

    • Torsion formula for hollow shaft and shear stress variation
    • Torque capacity ratio hollow vs solid for same weight/length
    • Stiffness ratio via polar moments of inertia
    • Conclusion on material economy of hollow shaft
11. 7 Marks
    aExplain how principal stresses, principal planes and maximum shear stresses are obtained using Mohr's circle.
    How to answer

    • Construction of Mohr's circle from biaxial stresses
    • Principal planes where shear is zero from circle intersections
    • Principal stresses as max/min normal stresses on circle
    • Maximum shear stress as radius of circle
12. 7 Marks
    bExplain shear stress distribution across rectangular and circular sections.
    How to answer

    • Shear stress formula and parabolic distribution for rectangle
    • Distribution for circular section with max at neutral axis
    • Comparison of max to mean values (1.5 for rectangle, 1.33 for circle)
    • Sketches of both distributions
13. 7 Marks
    aExplain strain energy method and Castigliano's theorem.
    How to answer

    • Definition of strain energy and strain energy density
    • Castigliano's first and second theorems statements
    • Application to deflection and load determination
    • Example of beam or truss application
14. 7 Marks
    bA column of timber section $15 \text{cm} \times 20 \text{ cm}$ is $6 \text{ metre}$ long both ends being fixed. If the Young's modulus for timber $= 17.5 \text{kN/mm}^{2}$ determine:

    1. (i))Crippling load and
    2. (ii))Safe load for the column if factor of safety = 3.
    How to answer

    • Minimum I for rectangular timber section governs buckling
    • Effective length Le = L/2 for both ends fixed
    • Euler crippling load Pcr = pi^2*E*I_min/Le^2
    • Safe load = Pcr/3
15. 7 Marks
    aDraw the shear force and bending moment diagram for a simply supported beam of length $9 \text{m}$ and carrying a uniformiy distributed load of $10 \text{ kN/m}$ for a distance of $6 \text{ m}$ from the left end. Also calculate the maximum B.M. on the section.
    How to answer

    • Calculate reactions with partial UDL over 6 m of 9 m span
    • Shear force diagram with linear drop under UDL
    • Bending moment diagram parabolic under loaded length
    • Locate zero shear for maximum B.M. and compute value
16. 7 Marks
    bExplain Rankine's formula and its limitations.
    How to answer

    • Rankine's formula combines Euler and crushing loads
    • Expression 1/PR = 1/PE + 1/PC
    • Applicability to short, medium and long columns
    • Limitations: empirical constant, material dependence