Strength Of Materials : Practice Questions and Answers

CONCEPT: Tension is a kind of Force applied to a string or wire along the length from both ends to stretch it. SI unit of Tension is Newton. EXPLANATION: When Tension on a wire is applied, due to forces from both ends the distance between its atoms increases.

Explanation: As per the stress-strain curve, The occurrence of the following points are given below: Proportional limit: It is the highest stress at which stress and strain are directly proportional so that the stress-strain graph is a straight line and follows the Hooke's law. Elastic limit: It is the greatest stress that can be applied to it without causing plastic (permanent) deformation. When a material is stressed to a point below its elastic limit, it will return to its original length once the stress is removed.

Concept: Deflection of Helical Springs of Circular Wire: $\delta = \frac{{8W{D^3}n}}{{Gd^4}}$ Where D = Mean diameter of the spring coil, d = Diameter of the spring wire, n = Number of active coils, G = Modulus of rigidity for the spring material, W = Axial load on the spring Calculation: It is given that D2 = 2D1 d2 = 2d1 ${\delta 1} = \frac{{8\;W\;D1^3n}}{{G\;{d^41}}}$ ${\delta 2} = \frac{{8\;W\;D2^3n}}{{G\;{d^42}}}$ $\delta = \frac{{8W{D^3}n}}{{Gd^4}} \Rightarrow \delta \propto \frac{D^3}{d^4}$ (rest other parameters are same) $\frac{{{\delta 2}}}{{{\delta 1}}} = (\frac{{{D2}}}{{{D1}}})^3 \times(\frac{d1}{d2})^4= \frac{{{2^3}}}{{{2^4}}} = \frac{1}{2}$ ${\delta 2} = \frac{1}{2}{\delta 1}$ Therefore If d and D is doubled, δ will be halved.

Explanation Shear stress at any point in a cross-section is given by the following formula: ${\rm{f}} = \frac{{{\rm{VQ}}}}{{{\rm{Ib}}}}$ Where, f = shear stress Q = First moment of area about the neutral axis of the area above the point where we want to calculate the shear stress Q = A × y V is Shear force y is the centroid of the area above the point where we want to calculate the shear stress measured from the neutral axis. A = Cross-sectional area above the point where we want to calculate the shear stress I = moment of inertia about the neutral axis b= width of the section where shear stress is to be calculated. Shear stress is zero at extreme fibres For a symmetric ‘I’ section, the intensity of shear stress is maximum at the centroid of the section which is neutral axis as shown in the figure.

Explanation: Stiffness (k): It is defined as the ability of the material to resist the deformation of an object. Stiffness is also defined as the ratio of load applied to the deflection. $k= \frac{P}{\delta}$ Deformation of axially loaded members: The deformation of a structural member with known geometry and subjected to an axial load can be determined by the equation of the form, $\delta = \frac{{PL}}{{AE}}$ $\therefore \delta = \frac{P}{{\frac{{AE}}{L}}}$ ${\frac{{AE}}{L}} = \frac{P}{{\delta}} =k$ The term $\frac{{AE}}{L}$ is defined as the stiffness of the bar.

Concept: Net weight of the UDL = wL = W RA + RB = wL Due to symmetry, ${RA} = \frac{{wL}}{2};{RB} = \frac{{wL}}{2}$ Taking a moment about B $M = {RA}x - \frac{{w{x^2}}}{{2}} = \frac{{wL}}{2}x - \frac{{w{x^2}}}{2}$ For maximum B.M, $\frac{{dM}}{{dx}} = 0\;i.e.\;\frac{{wL}}{2} - \frac{{w.2x}}{2} = 0$ $x = \frac{L}{2}$ ${M{max}} = \frac{{wL}}{2} \times \frac{L}{2} - \frac{{w{L^2}}}{8} = \frac{{w{L^2}}}{8}$

Explanation: Following are the assumptions made in the theory of Pure Bending: 1. The material of the beam is homogenous and isotropic. 2.

Concept: The torsion equation is given by: $\frac{T}{J} = \frac{τ }{R} = \frac{{G\theta }}{L}$ Where, T is the torque applied, J = polar moment of inertia, τ = shear strength of material, G = modulus of rigidity, θ = angular deflection of shaft (in radians). $\frac{T}{J} = \frac{τ }{r}$ $T = \frac{\tau J}{r} =\frac{\tau}{r} \times \frac{\pi d^4}{32}=\frac{\pi \tau d^3}{16}$ Calculation: τ = 48 N/cm2, d = 3 cm $T =\frac{\pi \tau d^3}{16}=\frac{\pi \times 48 \times3^3}{16}=81\pi~Ncm$

Explanation: Bending equation states that: $\frac{\sigma }{y} = \frac{M}{I} = \frac{E}{R}$ Various terms in above bending equation are: σ = Bending stress y = Distance from neutral axis M = Bending moment I = Area moment of inertia E = Modulus of elasticity R = Radius of curvature Bending Stress distribution for I section: Bending stress is directly proportional to y (Distance from the neutral axis). Bending stress will be zero at neutral axis. The maximum stress always occurs at the outermost fibre.

Explanation: Different type of support: Fixed support: The fixed support does not allow the rotation and translation motion to the structural members. It offers vertical, horizontal, and moment reactions. Pinned or hinge support: It resists the horizontal and vertical loads but is unable to resist the moment.

Explanation: Shear Force: Shear force at any section is the summation of all transverse loading, either to the left or to the right of any section. Bending Moment: Bending moment at any section is a resultant algebraic sum of moments caused by transverse forces either from the left or right of the section. Relationship between bending moment, shear force, and loading: The rate of change of shear force at any section is equal to load intensity at that section.

Explanation: Bending moment Diagram: Bending moment diagrams are simply plots of the bending moment (on the y-axis) versus the position of various points along the beam (on the x-axis). For UDL(Uniformly Distributed Load) Relationship between Shear force, Bending moment and Loading rate $-(\frac {ds}{dx}) =w$ i.e. negative slope of Shear force diagram at any section will be equal to the load intensity at that section $\frac {dM}{dx} = ~S$ i.e.