About this previous year paper
Subject: Strength of Material
Subject Code: KME502
Branch: Mechanical Engineering
Maximum Marks: 100 | Duration: 3 Hours
SECTION A – Short Answer Questions (10 × 2 = 20 Marks)
Students must attempt all ten questions in brief. These test fundamental definitions, concepts, and relations used in mechanics of solids.
Define Poisson’s Ratio.
Poisson’s Ratio (ν) is the ratio of lateral strain to longitudinal strain within the elastic limit of a material when subjected to uniaxial stress.
- ν=Lateral StrainLongitudinal Strainν = \frac{\text{Lateral Strain}}{\text{Longitudinal Strain}}ν=Longitudinal StrainLateral Strain
It represents the material’s ability to contract laterally when stretched or expand laterally when compressed.
Define Bulk Modulus.
Bulk modulus (K) is the ratio of volumetric stress to volumetric strain. It measures a material’s resistance to uniform compression.
- K=Volumetric StressVolumetric Strain=pΔV/VK = \frac{\text{Volumetric Stress}}{\text{Volumetric Strain}} = \frac{p}{\Delta V / V}K=Volumetric StrainVolumetric Stress=ΔV/Vp
Define Impact Load.
Impact load is a sudden or dynamic load applied in a very short time, producing stress greater than the static load of equal magnitude. Example: a falling weight striking a beam.
Define Composite Beam.
A composite beam consists of two or more materials rigidly joined so that they act together as one unit when loaded (e.g., steel–concrete beam).
Define Spring.
A spring is an elastic member designed to absorb energy, store potential energy, or resist external load by deformation.
Define Eccentric Loading.
When the line of action of a load does not pass through the centroidal axis of the section, it is called eccentric loading. It causes both direct and bending stresses.
Define Shrinkage Allowance.
When manufacturing parts like compound cylinders or shrink fits, one component is heated and expanded to fit over another. The allowance given for contraction on cooling is shrinkage allowance.
Define Unsymmetrical Bending.
When a beam is loaded in a plane other than one containing a principal axis of the cross-section, bending is unsymmetrical. The neutral axis does not coincide with the centroidal axis.
Define Shear Centre.
Shear centre is the point in a cross-section through which the applied transverse load passes without causing twisting of the beam.
Define Principal Stress.
Principal stresses are the maximum and minimum normal stresses acting on mutually perpendicular planes where shear stress is zero.
SECTION B – Descriptive Questions (3 × 10 = 30 Marks)
Attempt any three questions. Each question carries 10 marks and checks derivation and concept clarity.
Q2(a) Define Hooke’s Law and derive the relationship between Elastic Moduli:
E=2G(1+ν)E = 2G(1 + ν)E=2G(1+ν)
and
E=3K(1−2ν)E = 3K(1 - 2ν)E=3K(1−2ν)
Where E = Young’s Modulus, G = Shear Modulus, K = Bulk Modulus, ν = Poisson’s Ratio.
Explanation:
Hooke’s Law states that stress is directly proportional to strain within the elastic limit:
σ=Eεσ = Eεσ=Eε
The relationship between the moduli is derived by analyzing strain energy relationships for an isotropic body under triaxial loading.
Q2(b) Composite Bar under Axial Load
When two materials are joined in series or parallel (e.g., steel and copper), they share load proportionally to their stiffness.
P=P1+P2,δ1L1=δ2L2P = P_1 + P_2, \quad \frac{δ_1}{L_1} = \frac{δ_2}{L_2}P=P1+P2,L1δ1=L2δ2
Used for determining stresses in composite members like reinforced columns or compound rods.
Q3(a) Torsion Equation
Derive:
TJ=τr=GθL\frac{T}{J} = \frac{τ}{r} = \frac{Gθ}{L}JT=rτ=LGθ
Where
TTT: Torque
JJJ: Polar moment of inertia
τττ: Shear stress
rrr: Radius
GGG: Modulus of rigidity
θθθ: Angle of twist
LLL: Length of shaft
Assumptions: material homogeneous, uniform twist, circular cross-section, plane sections remain plane.
Q3(b) Power Transmitted by Shaft
P=2πNT60P = \frac{2πNT}{60}P=602πNT
Where NNN = speed in rpm and TTT = torque (Nm).
Q4 Euler’s Column Theory
Derive critical load for a long column:
Pcr=π2EI(KL)2P_{cr} = \frac{π^2EI}{(KL)^2}Pcr=(KL)2π2EI
where KKK is the effective length factor depending on end conditions (1 for both ends hinged, 0.5 for fixed, etc.).
This relation gives the buckling load at which the column fails due to instability.
Q5 Thin and Thick Cylinders
(a) Thin Cylinder Stress Analysis
For thin-walled pressure vessels (t/r < 1/10):
σh=pD2t,σl=pD4tσ_h = \frac{pD}{2t}, \quad σ_l = \frac{pD}{4t}σh=2tpD,σl=4tpD
where σhσ_hσh = hoop stress, σlσ_lσl = longitudinal stress, ppp = internal pressure, DDD = diameter, ttt = thickness.
(b) Thick Cylinder (Lame’s Equations)
σr=A−Br2,σt=A+Br2σ_r = A - \frac{B}{r^2}, \quad σ_t = A + \frac{B}{r^2}σr=A−r2B,σt=A+r2B
Constants A, B determined from boundary conditions. Used for gun barrels and hydraulic cylinders.
Q6 Springs and Energy
(a) Closed-Coil Helical Spring**
Shear stress:
τ=16WRπd3τ = \frac{16WR}{πd^3}τ=πd316WR
Deflection:
δ=8WR3nGd4δ = \frac{8WR^3n}{Gd^4}δ=Gd48WR3n
Strain Energy:
U=W2R3nGd4U = \frac{W^2R^3n}{Gd^4}U=Gd4W2R3n
(b) Leaf Spring**
Bending stress:
σ=3WL2nbt2σ = \frac{3WL}{2nbt^2}σ=2nbt23WL
Deflection:
δ=3WL38Enbt3δ = \frac{3WL^3}{8Enbt^3}δ=8Enbt33WL3
Q7 Principal Stress and Strain Energy
Explain Mohr’s Circle for stress transformation.
When a body is subjected to biaxial stress (σx, σy, τxy), the principal stresses are:
σ1,2=σx+σy2±(σx−σy2)2+τxy2σ_{1,2} = \frac{σ_x + σ_y}{2} \pm \sqrt{\left(\frac{σ_x - σ_y}{2}\right)^2 + τ_{xy}^2}σ1,2=2σx+σy±(2σx−σy)2+τxy2
and maximum shear stress:
τmax=(σx−σy2)2+τxy2τ_{max} = \sqrt{\left(\frac{σ_x - σ_y}{2}\right)^2 + τ_{xy}^2}τmax=(2σx−σy)2+τxy2
This is used to locate the failure plane and design safe structures.
SECTION C – Long / Problem-Solving Questions (5 × 10 = 50 Marks)
Students must attempt one part from each question (Q3–Q7).
Q3. Torsion and Power Transmission
A solid circular shaft must transmit 300 kW at 200 rpm with a maximum shear stress of 50 MPa.
Find:
Required diameter.
Angle of twist per meter length if G = 80 GPa.
P=2πNT60⇒T=60P2πNP = \frac{2πNT}{60} \Rightarrow T = \frac{60P}{2πN}P=602πNT⇒T=2πN60P
Then apply torsion formula to find ddd and θθθ.
Q4. Column and Strut
A steel column (L = 3 m, E = 200 GPa, I = 3×10⁶ mm⁴) is hinged at both ends.
Calculate Euler’s buckling load:
Pcr=π2EI(KL)2P_{cr} = \frac{π^2EI}{(KL)^2}Pcr=(KL)2π2EI
For real columns, Rankine’s formula is used:
1PR=1PE+1PC\frac{1}{P_{R}} = \frac{1}{P_{E}} + \frac{1}{P_{C}}PR1=PE1+PC1
where PC=σcAP_C = σ_c APC=σcA is crushing load.
Q5. Thin Cylinder Under Internal Pressure
A cylindrical shell (diameter = 500 mm, thickness = 10 mm) subjected to internal pressure p=3MPap = 3 MPap=3MPa.
Find hoop and longitudinal stresses, and change in diameter and volume using:
σh=pD2t,σl=pD4tσ_h = \frac{pD}{2t}, \quad σ_l = \frac{pD}{4t}σh=2tpD,σl=4tpD ΔDD=1E(σh−νσl)\frac{ΔD}{D} = \frac{1}{E}(σ_h - νσ_l)DΔD=E1(σh−νσl)
Q6. Springs
A helical spring of wire diameter 10 mm, mean coil diameter 80 mm, number of coils 12, carries load 100 N.
Find:
Shear stress
Deflection
τ=16WRπd3,δ=8WR3nGd4τ = \frac{16WR}{πd^3}, \quad δ = \frac{8WR^3n}{Gd^4}τ=πd316WR,δ=Gd48WR3n
Q7. Compound Cylinder
Inner cylinder and outer cylinder assembled by shrink fit.
Given interference and modulus of elasticity, calculate radial pressure at common surface using Lame’s theory and shrinkage equations.
Key Topics Covered
Stress–strain relationships and elastic constants Torsion of shafts
Energy and impact loading Bending and shear stress distribution
Columns and buckling Springs (helical and leaf)
Thin and thick cylinders Compound bars and shrink fits