(SEM V) THEORY EXAMINATION 2023-24 STRENGTH OF MATERIAL

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.

1. 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.

1. 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​,L1​δ1​​=L2​δ2​​

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}}PR​1​=PE​1​+PC​1​

where PC=σcAP_C = σ_c APC​=σc​A 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