Strength Of Materials PDF

S K Mondal’s Strength of Materials Contents Chapter – 1: Stress and Strain Chapter - 2 : Principal Stress and Strain Chapter - 3 : Moment of Inertia and Centroid Chapter - 4 : Bending Moment and Shear Force Diagram Chapter - 5 : Deflection of Beam Chapter - 6 : Bending Stress in Beam Chapter - 7 : Shear Stress in Beam Chapter - 8 : Fixed and Continuous Beam Chapter - 9 : Torsion Chapter-10 : Thin Cylinder Chapter-11 : Thick Cylinder Chapter-12 : Spring Chapter-13 : Theories of Column Chapter-14 : Strain Energy Method Chapter-15 : Theories of Failure Chapter-16 : Riveted and Welded Joint Er. S K Mondal IES Officer (Railway), GATE topper, NTPC ET-2003 batch, 12 years teaching experienced, Author of Hydro Power Familiarization (NTPC Ltd) Page 1 of 429 Note “Asked Objective Questions” is the total collection of questions from:- 20 yrs IES (2010-1992) [Engineering Service Examination] 21 yrs. GATE (2011-1992) and 14 yrs. IAS (Prelim.) [Civil Service Preliminary] Copyright © 2007 S K Mondal Every effort has been made to see that there are no errors (typographical or otherwise) in the material presented. However, it is still possible that there are a few errors (serious or otherwise). I would be thankful to the readers if they are brought to my attention at the following e-mail address: [email protected] S K Mondal Page 2 of 429 1. Stress and Strain Theory at a Glance (for IES, GATE, PSU) 1.1 Stress (ı) When a material is subjected to an external force, a resisting force is set up within the component. The internal resistance force per unit area acting on a material or intensity of the forces distributed over a given section is called the stress at a point. x It uses original cross section area of the specimen and also known as engineering stress or conventional stress. P Therefore, T A x P is expressed in Newton (N) and A, original area, in square meters (m2), the stress ǔ will be expresses in N/ m2. This unit is called Pascal (Pa). x As Pascal is a small quantity, in practice, multiples of this unit is used. 1 kPa = 103 Pa = 103 N/ m2 (kPa = Kilo Pascal) 1 MPa = 106 Pa = 106 N/ m2 =1 N/mm2 (MPa = Mega Pascal) 1 GPa = 109 Pa = 109 N/ m2 (GPa = Giga Pascal) Let us take an example: A rod 10 mm q 10 mm cross-section is carrying an axial tensile load 10 kN. In this rod the tensile stress developed is given by P 10 kN 10q103 N Tt 2 100N/mm2 100MPa A 10 mm q10 mm 100 mm x The resultant of the internal forces for an axially loaded member is normal to a section cut perpendicular to the member axis. x The force intensity on the shown section is defined as the normal stress. %F P T lim and Tavg %Al 0 %A A x Tensile stress (ıt) If ǔ > 0 the stress is tensile. i.e. The fibres of the component tend to elongate due to the external force. A member subjected to an external force tensile P and tensile stress distribution due to the force is shown in the given figure. Page 3 of 429 Chapter-1 Stress and Strain S K Mondal’s x Compressive stress (ıc) If ǔ < 0 the stress is compressive. i.e. The fibres of the component tend to shorten due to the external force. A member subjected to an external compressive force P and compressive stress distribution due to the force is shown in the given figure. x Shear stress ( U ) When forces are transmitted from one part of a body to other, the stresses developed in a plane parallel to the applied force are the shear stress. Shear stress acts parallel to plane of interest. Forces P is applied transversely to the member AB as shown. The corresponding internal forces act in the plane of section C and are called shearing P forces. The corresponding average shear stress U Area 1.2 Strain (İ) The displacement per unit length (dimensionless) is known as strain. x Tensile strain ( F t) The elongation per unit length as shown in the figure is known as tensile strain. ǆt = ƦL/ Lo It is engineering strain or conventional strain. Here we divide the elongation to original length not actual length (Lo + % L) Let us take an example: A rod 100 mm in original length. When we apply an axial tensile load 10 kN the final length of the rod after application of the load is 100.1 mm. So in this rod tensile strain is developed and is given by %L L Lo 100.1mm 100 mm 0.1mm Ft 0.001 (Dimensionless)Tensile Lo Lo 100 mm 100 mm x Compressive strain ( F c) If the applied force is compressive then the reduction of length per unit length is known as compressive strain. It is negative. Then İc = (-ǻL)/ Lo Let us take an example: A rod 100 mm in original length. When we apply an axial compressive load 10 kN the final length of the rod after application of the load is 99 mm. So in this rod a compressive strain is developed and is given by Page 4 of 429 Chapter-1 Stress and Strain S K Mondal’s %L L Lo 99 mm 100 mm 1mm Fc 0.01 (Dimensionless)compressive Lo Lo 100 mm 100 mm x Shear Strain ( H ): When a force P is applied tangentially to the element shown. Its edge displaced to dotted line. Where E is the lateral displacement of the upper face of the element relative to the lower face and L is the distance between these faces. E Then the shear strain is (H ) L Let us take an example: A block 100 mm × 100 mm base and 10 mm height. When we apply a tangential force 10 kN to the upper edge it is displaced 1 mm relative to lower face. Then the direct shear stress in the element 10 kN 10q103 N (U ) 1 N/mm2 1 MPa 100 mmq100 mm 100 mmq100 mm 1mm And shear strain in the element ( H ) = 0.1 Dimensionless 10 mm 1.3 True stress and True Strain The true stress is defined as the ratio of the load to the cross section area at any instant. load T 1 F TT Instantaneous area Where T and F is the engineering stress and engineering strain respectively. x True strain L dl L¬ A ¬ d ¬ FT ¨ ln ln 1 F ln o 2ln o Lo l Lo ® A ® d ® or engineering strain ( F ) = eFT -1 The volume of the specimen is assumed to be constant during plastic deformation. [ ' Ao Lo AL ] It is valid till the neck formation. x Comparison of engineering and the true stress-strain curves shown below Page 5 of 429 Chapter-1 Stress and Strain S K Mondal’s x The true stress-strain curve is also known as the flow curve. x True stress-strain curve gives a true indication of deformation characteristics because it is based on the instantaneous dimension of the specimen. x In engineering stress-strain curve, stress drops down after necking since it is based on the original area. x In true stress-strain curve, the stress however increases after necking since the cross- sectional area of the specimen decreases rapidly after necking. x The flow curve of many metals in the region of uniform plastic deformation can be expressed by the simple power law. ǔT = K(ǆT)n Where K is the strength coefficient n is the strain hardening exponent n = 0 perfectly plastic solid n = 1 elastic solid For most metals, 0.1< n < 0.5 x Relation between the ultimate tensile strength and true stress at maximum load Pmax The ultimate tensile strength Tu Ao Pmax The true stress at maximum load Tu T A A ¬ Ao And true strain at maximum load F ln o or e FT T A ® A Pmax Pmax Ao Eliminating Pmax we get , Tu T q Tu e FT A Ao A Where Pmax = maximum force and Ao = Original cross section area A = Instantaneous cross section area Let us take two examples: (I.) Only elongation no neck formation In the tension test of a rod shown initially it was Ao = 50 mm2 and Lo = 100 mm. After the application of load it’s A = 40 mm2 and L = 125 mm. Determine the true strain using changes in both length and area. Answer: First of all we have to check that does the (If no neck formation member forms neck or not? For that check Ao Lo AL occurs both area and or not? gauge length can be used Here 50 × 100 = 40 × 125 so no neck formation is for a strain calculation.) there. Therefore true strain Page 6 of 429 Chapter-1 Stress and Strain S K Mondal’s L dl 125 ¬ FT ¨ ln 0.223 l 100 ® Lo A ¬ 50 ¬ FT ln o ln 0.223 A ® 40 ® (II.) Elongation with neck formation A ductile material is tested such and necking occurs then the final gauge length is L=140 mm and the final minimum cross sectional area is A = 35 mm2. Though the rod shown initially it was Ao = 50 mm2 and Lo = 100 mm. Determine the true strain using changes in both length and area. Answer: First of all we have to check that does the (After necking, gauge member forms neck or not? For that check Ao Lo AL length gives error but or not? area and diameter can Here AoLo = 50 × 100 = 5000 mm3 and AL=35 × 140 be used for the = 4200 mm3. So neck formation is there. Note here calculation of true strain AoLo > AL. at fracture and before Therefore true strain fracture also.) A ¬ 50 ¬ FT ln o ln 0.357 A ® 35 ® L dl 140 ¬ But not FT ¨ ln 0.336 (it is wrong) l 100 ® Lo 1.4 Hook’s law According to Hook’s law the stress is directly proportional to strain i.e. normal stress (ǔ) B normal strain (ǆ) and shearing stress ( U ) B shearing strain ( H ). ǔ = Eǆ and U GH The co-efficient E is called the modulus of elasticity i.e. its resistance to elastic strain. The co- efficient G is called the shear modulus of elasticity or modulus of rigidity. 1.5 Volumetric strain Fv A relationship similar to that for length changes holds for three-dimensional (volume) change. For P volumetric strain, Fv , the relationship is Fv = (V-V0)/V0 or Fv = ƦV/V0 K x Where V is the final volume, V0 is the original volume, and ƦV is the volume change. x Volumetric strain is a ratio of values with the same units, so it also is a dimensionless quantity. Page 7 of 429 Chapter-1 Stress and Strain S K Mondal’s x ƦV/V= volumetric strain = ǆx +ǆy + ǆz = ǆ1 +ǆ2 + ǆ3 x Dilation: The hydrostatic component of the total stress contributes to deformation by changing the area (or volume, in three dimensions) of an object. Area or volume change is called dilation and is positive or negative, as the volume increases or decreases, p respectively. e Where p is pressure. K PL ı 1.6 Young’s modulus or Modulus of elasticity (E) = = Aį W PL 1.7 Modulus of rigidity or Shear modulus of elasticity (G) = = J AG 'p 'p 1.8 Bulk Modulus or Volume modulus of elasticity (K) = 'v 'R v R 1.10 Relationship between the elastic constants E, G, K, μ 9KG E 2G 1 P 3K 1 2P 3K G [VIMP] Where K = Bulk Modulus, N = Poisson’s Ratio, E= Young’s modulus, G= Modulus of rigidity x For a linearly elastic, isotropic and homogeneous material, the number of elastic constants required to relate stress and strain is two. i.e. any two of the four must be known. x If the material is non-isotropic (i.e. anisotropic), then the elastic modulii will vary with additional stresses appearing since there is a coupling between shear stresses and normal stresses for an anisotropic material. Let us take an example: The modulus of elasticity and rigidity of a material are 200 GPa and 80 GPa, respectively. Find all other elastic modulus. 9KG Answer: Using the relation E 2G 1 P 3K 1 2 P we may find all other elastic modulus 3K G easily E E 200 Poisson’s Ratio ( P ) : 1 P P 1 1 0.25 2G 2G 2 u 80 E E 200 Bulk Modulus (K) : 3K K 133.33GPa 1 2P 3 1 2P 3 1 2 u 0.25 1.11 Poisson’s Ratio (μ) Transverse strain or lateral strain y = = Longitudinal strain x (Under unidirectional stress in x-direction) x The theory of isotropic elasticity allows Poisson's ratios in the range from -1 to 1/2. x Poisson's ratio in various materials Page 8 of 429 Chapter-1 Stress and Strain S K Mondal’s Material Poisson's ratio Material Poisson's ratio Steel 0.25 – 0.33 Rubber 0.48 – 0.5 C.I 0.23 – 0.27 Cork Nearly zero Concrete 0.2 Novel foam negative x We use cork in a bottle as the cork easily inserted and removed, yet it also withstand the pressure from within the bottle. Cork with a Poisson's ratio of nearly zero, is ideal in this application. 1.12 For bi-axial stretching of sheet §L · 1 ln ¨ f 1 ¸ Lo Original length © Lo1 ¹ §L · 2 ln ¨ f 2 ¸ L f -Final length © Lo 2 ¹ Initial thickness(t o ) Final thickness (tf) = e1 u e2 1.13 Elongation x A prismatic bar loaded in tension by an axial force P For a prismatic bar loaded in tension by an axial force P. The elongation of the bar can be determined as PL G AE Let us take an example: A Mild Steel wire 5 mm in diameter and 1 m long. If the wire is subjected to an axial tensile load 10 kN find its extension of the rod. (E = 200 GPa) PL Answer: We know that G AE Here given, Force (P) 10 kN 10 u 1000N Length(L) 1 m 2 Sd2 S u 0.005 Area(A) m2 1.963 u 105 m2 4 4 Modulous of Elasticity (E ) 200 GPa 200 u 109 N/m2 PL 10 u 1000 u 1 Therefore Elongation(G ) m AE 1.963 u 105 u 200 u 109 2.55 u 103 m 2.55 mm x Elongation of composite body Elongation of a bar of varying cross section A1, A2,----------,An of lengths l1, l2,--------ln respectively. P ª l1 l2 l3 l º G « n » E ¬ A1 A2 A3 An ¼ Page 9 of 429 Chapter-1 Stress and Strain S K Mondal’s Let us take an example: A composite rod is 1000 mm long, its two ends are 40 mm2 and 30 mm2 in area and length are 300 mm and 200 mm respectively. The middle portion of the rod is 20 mm2 in area and 500 mm long. If the rod is subjected to an axial tensile load of 1000 N, find its total elongation. (E = 200 GPa). Answer: Consider the following figure Given, Load (P) =1000 N Area; (A1) = 40 mm2, A2 = 20 mm2, A3 = 30 mm2 Length; (l1) = 300 mm, l2 = 500 mm, l3 = 200 mm E = 200 GPa = 200 u 109 N/m2 = 200 u 103 N/mm2 Therefore Total extension of the rod P ª l1 l 2 l 3 º G « » E ¬ A1 A2 A3 ¼ 1000 N ª 300 mm 500 mm 200 mm º u« » 200 u 10 N / mm ¬ 40 mm 2 20 mm 2 30 mm 2 ¼ 3 2 0.196mm x Elongation of a tapered body Elongation of a tapering rod of length ‘L’ due to load ‘P’ at the end 4PL į= (d1 and d2 are the diameters of smaller & larger ends) S Ed1 d 2 PL PL You may remember this in this way, į= i.e. §S · EA eq E ¨ d1 d 2 ¸ ©4 ¹ Let us take an example: A round bar, of length L, tapers uniformly from small diameter d1 at one end to bigger diameter d2 at the other end. Show that the extension produced by a tensile axial load 4PL P is į =. S d1 d 2 E If d2 = 2d1, compare this extension with that of a uniform cylindrical bar having a diameter equal to the mean diameter of the tapered bar. Answer: Consider the figure below d1 be the radius at the smaller end. Then at a X cross section XX located at a distance × from the smaller end, the value of diameter ‘dx’ is equal to Page 10 of 429 Chapter-1 Stress and Strain S K Mondal’s dx d1 x § d 2 d1 · ¨ ¸ 2 2 L© 2 2 ¹ x or d x d1 d 2 d1 L d 2 d1 1 d1 1 kx Where k u L d1 We now taking a small strip of diameter 'd x 'and length 'd x 'at section XX. Elongation of this section 'd x ' length PL P.dx 4P.dx d G 2 AE § S d x2 · S.^d1 1 kx ` E ¨ ¸uE © 4 ¹ Therefore total elongation of the taper bar x L 4P dx G ³d G ³ S Ed 2 2 x 0 1 1 kx 4PL S E d1d 2 Comparison: Case-I: Where d2 = 2d1 4PL 2PL Elongation G I S Ed1 u 2d1 S Ed12 Case –II: Where we use Mean diameter d1 d 2 d1 2d1 3 dm d1 2 2 2 PL P.L Elongation of such bar G II 2 AE S §3 · ¨ d1 ¸.E 4©2 ¹ 16PL 9S Ed12 Extension of taper bar 2 9 Extension of uniform bar 16 8 9 Page 11 of 429 Chapter-1 Stress and Strain S K Mondal’s x Elongation of a body due to its self weight (i) Elongation of a uniform rod of length ‘L’ due to its own weight ‘W’ WL į= 2AE The deformation of a bar under its own weight as compared to that when subjected to a direct axial load equal to its own weight will be half. (ii) Total extension produced in rod of length ‘L’ due to its own weight ‘ X ’ per with Z L2 length. į= 2EA (iii) Elongation of a conical bar due to its self weight U gL2 WL į= 6E 2 Amax E 1.14 Structural members or machines must be designed such that the working stresses are less than the ultimate strength of the material. Vy ½ Working stress V w n=1.5 to 2 ° n ° ¾ factor of safety V ult n1 2 to 3 ° n1 °¿ Vp Vp Proof stress n V y or V p or V ult 1.15 Factor of Safety: (n) = Vw 1.16 Thermal or Temperature stress and strain x When a material undergoes a change in temperature, it either elongates or contracts depending upon whether temperature is increased or decreased of the material. x If the elongation or contraction is not restricted, i. e. free then the material does not experience any stress despite the fact that it undergoes a strain. x The strain due to temperature change is called thermal strain and is expressed as, H D 'T x Where ǂ is co-efficient of thermal expansion, a material property, and ƦT is the change in temperature. x The free expansion or contraction of materials, when restrained induces stress in the Page 12 of 429 material and it is referred to as thermal stress. Chapter-1 Stress and Strain S K Mondal’s Vt D E 'T Where, E = Modulus of elasticity x Thermal stress produces the same effect in the material similar to that of mechanical stress. A compressive stress will produce in the material with increase in temperature and the stress developed is tensile stress with decrease in temperature. Let us take an example: A rod consists of two parts that are made of steel and copper as shown in figure below. The elastic modulus and coefficient of thermal expansion for steel are 200 GPa and 11.7 × 10-6 per °C respectively and for copper 70 GPa and 21.6 × 10-6 per °C respectively. If the temperature of the rod is raised by 50°C, determine the forces and stresses acting on the rod. Answer: If we allow this rod to freely expand then free expansion GT D 'T L 11.7 u 106 u 50 u 500 21.6 u 106 u 50 u 750 1.1025 mm Compressive But according to diagram only free expansion is 0.4 mm. Therefore restrained deflection of rod =1.1025 mm – 0.4 mm = 0.7025 mm Let us assume the force required to make their elongation vanish be P which is the reaction force at the ends. § PL · § PL · G ¨ ¸ ¨ ¸ © AE ¹Steel © AE ¹Cu P u 500 P u 750 or 0.7025 S 2½ 9 S 2½ 9 ® u 0.075 ¾ u 200 u 10 ® u 0.050 ¾ u 70 u 10 ¯ 4 ¿ ¯ 4 ¿ or P 116.6 kN Therefore, compressive stress on steel rod P 116.6 u 103 V Steel N/m2 26.39 MPa ASteel S 2 u 0.075 4 And compressive stress on copper rod P 116.6 u 103 V Cu N/m2 59.38 MPa ACu S 2 u 0.050 4 Page 13 of 429 Chapter-1 Stress and Strain S K Mondal’s 1.17 Thermal stress on Brass and Mild steel combination A brass rod placed within a steel tube of exactly same length. The assembly is making in such a way that elongation of the combination will be same. To calculate the stress induced in the brass rod, steel tube when the combination is raised by toC then the following analogy have to do. (a) Original bar before heating. (b) Expanded position if the members are allowed to expand freely and independently after heating. (c) Expanded position of the compound bar i.e. final position after heating. x Compatibility Equation: Assumption: G G st G sf G Bt G Bf 1. L = Ls LB 2. D b ! D s x Equilibrium Equation: 3. Steel Tension V s As V B AB Brass Compression Where, G = Expansion of the compound bar = AD in the above figure. G st = Free expansion of the steel tube due to temperature rise toC = D s L t = AB in the above figure. G sf = Expansion of the steel tube due to internal force developed by the unequal expansion. = BD in the above figure. G Bt = Free expansion of the brass rod due to temperature rise toC = D b L t = AC in the above figure. G Bf = Compression of the brass rod due to internal force developed by the unequal expansion. = BD in the above figure. And in the equilibrium equation Tensile force in the steel tube = Compressive force in the brass rod Where, V s = Tensile stress developed in the steel tube. V B = Compressive stress developed in the brass rod. As = Cross section area of the steel tube. AB = Cross section area of the brass rod. Let us take an example: See the Conventional Question Answer section of this chapter and the question is “Conventional Question IES-2008”Page and14it’s of 429 answer. Chapter-1 Stress and Strain S K Mondal’s 1.18 Maximum stress and elongation due to rotation UZ 2L2 UZ 2L3 (i) V max and G L 8 12E UZ 2L2 UZ 2L3 (ii) V max and G L 2 3E For remember: You will get (ii) by multiplying by 4 of (i) 1.18 Creep When a member is subjected to a constant load over a long period of time it undergoes a slow permanent deformation and this is termed as “creep”. This is dependent on temperature. Usually at elevated temperatures creep is high. x The materials have its own different melting point; each will creep when the homologous Testing temperature temperature > 0.5. Homologous temp = > 0.5 Melting temperature A typical creep curve shows three distinct stages with different creep rates. After an initial rapid elongation ǆo, the creep rate decrease with time until reaching the steady state. 1) Primary creep is a period of transient creep. The creep resistance of the material increases due to material deformation. 2) Secondary creep provides a nearly constant creep rate. The average value of the creep rate during this period is called the minimum creep rate. A stage of balance between competing. Strain hardening and recovery (softening) of the material. 3) Tertiary creep shows a rapid increase in the creep rate due to effectively reduced cross- sectional area of the specimen leading to creep rupture or failure. In this stage intergranular cracking and/or formation of voids and cavities occur. c2 Creep rate =c1 V Creep strain at any time = zero time strain intercept + creep rate ×Time = 0 c1 V c2 u t Where, c1 , c2 are constants V stress 1.19 If a load P is applied suddenly to a bar then the stress & strain induced will be double than those obtained by an equal load applied gradually. Page 15 of 429 1.20 Stress produced by a load P in falling from height ’h’ Chapter-1 Stress and Strain S K Mondal’s ª 2h º ½° Vd V «1 1 »¾ ı, «¬ L »¼ °¿ being stress & strain produced by static load P & L=length of bar. Aª 2 AEh º «1 1 » P¬ PL ¼ 1.21 Loads shared by the materials of a compound bar made of bars x & y due to load W, Ax Ex Px W. Ax Ex Ay E y Ay E y Py W. Ax Ex Ay E y PL 1.22 Elongation of a compound bar, G Ax Ex Ay E y 1.23 Tension Test i) True elastic limit: based on micro-strain measurement at strains on order of 2 × 10-6. Very low value and is related to the motion of a few hundred dislocations. ii) Proportional limit: the highest stress at which stress is directly proportional to strain. iii) Elastic limit: is the greatest stress the material can withstand without any measurable permanent strain after unloading. Elastic limit > proportional limit. iv) Yield strength is the stress required to produce a small specific amount of deformation. The offset yield strength can be determined by the stress corresponding to the intersection of the stress-strain curve and a line parallel to the elastic line offset by a strain of 0.2 or 0.1%. ( H = 0.002 or 0.001). Page 16 of 429 Chapter-1 Stress and Strain S K Mondal’s x The offset yield stress is referred to proof stress either at 0.1 or 0.5% strain used for design and specification purposes to avoid the practical difficulties of measuring the elastic limit or proportional limit. v) Tensile strength or ultimate tensile strength (UTS) V u is the maximum load Pmax divided by the original cross-sectional area Ao of the specimen. Lf Lo vi) % Elongation, , is chiefly influenced by uniform elongation, which is dependent on the Lo strain-hardening capacity of the material. Ao Af vii) Reduction of Area: q Ao x Reduction of area is more a measure of the deformation required to produce failure and its chief contribution results from the necking process. x Because of the complicated state of stress state in the neck, values of reduction of area are dependent on specimen geometry, and deformation behaviour, and they should not be taken as true material properties. x RA is the most structure-sensitive ductility parameter and is useful in detecting quality changes in the materials. viii) Stress-strain response 1.24 Elastic strain and Plastic strain The strain present in the material after unloading is called the residual strain or plastic strain and the strain disappears during unloading is termed as recoverable or elastic strain. Equation of the straight line CB is given by V total uE Plastic uE Elastic uE Carefully observe the following figures and understand which one is Elastic strain and which one is Plastic strain Page 17 of 429 Chapter-1 Stress and Strain S K Mondal’s Let us take an example: A 10 mm diameter tensile specimen has a 50 mm gauge length. The load corresponding to the 0.2% offset is 55 kN and the maximum load is 70 kN. Fracture occurs at 60 kN. The diameter after fracture is 8 mm and the gauge length at fracture is 65 mm. Calculate the following properties of the material from the tension test. (i) % Elongation (ii) Reduction of Area (RA) % (iii) Tensile strength or ultimate tensile strength (UTS) (iv) Yield strength (v) Fracture strength (vi) If E = 200 GPa, the elastic recoverable strain at maximum load (vii) If the elongation at maximum load (the uniform elongation) is 20%, what is the plastic strain at maximum load? S 2 Answer: Given, Original area A0 u 0.010 m2 7.854 u 105 m2 4 S 2 Area at fracture Af u 0.008 m2 5.027 u 105 m2 4 Original gauge length (L0) = 50 mm Gauge length at fracture (L) = 65 mm Therefore L L0 65 50 (i) % Elongation u 100% u 100 30% L0 50 A0 Af 7.854 5.027 (ii) Reduction of area (RA) = q u 100% u 100% 36% A0 7.854 Pmax 70 u 103 (iii) Tensile strength or Ultimate tensile strength (UTS), V u 5 N/m2 891 MPa Ao 7.854 u 10 Py 55 u 103 (iv) Yield strength V y 5 N/m2 700 MPa Ao 7.854 u 10 PFracture 60 u 103 (v) Fracture strength V F 5 N/m2 764MPa Ao 7.854 u 10 Pmax / Ao 891u 106 (vi) Elastic recoverable strain at maximum load H E 9 0.0045 Page 18Eof 429 200 u 10 Chapter-1 Stress and Strain S K Mondal’s (vii) Plastic strain H P H total H E 0.2000 0.0045 0.1955 1.25 Elasticity This is the property of a material to regain its original shape after deformation when the external forces are removed. When the material is in elastic region the strain disappears completely after removal of the load, The stress-strain relationship in elastic region need not be linear and can be non-linear (example rubber). The maximum stress value below which the strain is fully recoverable is called the elastic limit. It is represented by point A in figure. All materials are elastic to some extent but the degree varies, for example, both mild steel and rubber are elastic materials but steel is more elastic than rubber. 1.26 Plasticity When the stress in the material exceeds the elastic limit, the material enters into plastic phase where the strain can no longer be completely removed. Under plastic conditions materials ideally deform without any increase in stress. A typical stress strain diagram for an elastic-perfectly plastic material is shown in the figure. Mises-Henky criterion gives a good starting point for plasticity analysis. 1.27 Strain hardening If the material is reloaded from point C, it will follow the previous unloading path and line CB becomes its new elastic region with elastic limit defined by point B. Though the new elastic region CB resembles that of the initial elastic region OA, the internal structure of the material in the new state has changed. The change in the microstructure of the material is clear from the fact that the ductility of the material has come down due to strain hardening. When the material is reloaded, it follows the same path as that of a virgin material and fails on reaching the ultimate strength which remains unaltered due to the intermediate loading and unloading process. Page 19 of 429 Chapter-1 Stress and Strain S K Mondal’s 1.28 Stress reversal and stress-strain hysteresis loop We know that fatigue failure begins at a local discontinuity and when the stress at the discontinuity exceeds elastic limit there is plastic strain. The cyclic plastic strain results crack propagation and fracture. When we plot the experimental data with reversed loading and the true stress strain hysteresis loops is found as shown below. True stress-strain plot with a number of stress reversals Due to cyclic strain the elastic limit increases for annealed steel and decreases for cold drawn steel. Here the stress range is Ʀǔ. Ʀǆp and Ʀǆe are the plastic and elastic strain ranges, the total strain range being Ʀǆ. Considering that the total strain amplitude can be given as Ʀǆ = Ʀǆp+ Ʀǆe Page 20 of 429 Chapter-1 Stress and Strain S K Mondal’s OBJECTIVE QUESTIONS (GATE, IES, IAS) Previous 20-Years GATE Questions Stress in a bar due to self-weight GATE-1. Two identical circular rods of same diameter and same length are subjected to same magnitude of axial tensile force. One of the rods is made out of mild steel having the modulus of elasticity of 206 GPa. The other rod is made out of cast iron having the modulus of elasticity of 100 GPa. Assume both the materials to be homogeneous and isotropic and the axial force causes the same amount of uniform stress in both the rods. The stresses developed are within the proportional limit of the respective materials. Which of the following observations is correct? [GATE-2003] (a) Both rods elongate by the same amount (b) Mild steel rod elongates more than the cast iron rod (c) Cast iron rod elongates more than the mild steel rod (d) As the stresses are equal strains are also equal in both the rods PL 1 GATE-1. Ans. (c) G L or G L f [AsP, L and A is same] AE E G L mild steel ECI 100 ? G L CI ! G L MS G L C.I EMS 206 GATE-2. A steel bar of 40 mm × 40 mm square cross-section is subjected to an axial compressive load of 200 kN. If the length of the bar is 2 m and E = 200 GPa, the elongation of the bar will be: [GATE-2006] (a) 1.25 mm (b) 2.70 mm (c) 4.05 mm (d) 5.40 mm PL 200 u 1000 u 2 GATE-2. Ans. (a) G L m 1.25mm AE 0.04 u 0.04 u 200 u 109 True stress and true strain GATE-3. The ultimate tensile strength of a material is 400 MPa and the elongation up to maximum load is 35%. If the material obeys power law of hardening, then the true stress-true strain relation (stress in MPa) in the plastic deformation range is: [GATE-2006] 0.30 0.30 0.35 (a) V 540H (b) V 775H (c) V 540H (d) V 775H 0.35 GATE-3. Ans. (c) A true stress – true strain curve in tension V kH n k = Strength co-efficient = 400 × (1.35) = 540 MPa n = Strain – hardening exponent = 0.35 Elasticity and Plasticity GATE-4. An axial residual compressive stress due to a manufacturing process is present on the outer surface of a rotating shaft subjected to bending. Under a given Page 21 of 429 Chapter-1 Stress and Strain S K Mondal’s bending load, the fatigue life of the shaft in the presence of the residual compressive stress is: [GATE-2008] (a) Decreased (b) Increased or decreased, depending on the external bending load (c) Neither decreased nor increased (d) Increased GATE-4. Ans. (d) A cantilever-loaded rotating beam, showing the normal distribution of surface stresses. (i.e., tension at the top and compression at the bottom) The residual compressive stresses induced. Net stress pattern obtained when loading a surface treated beam. The reduced magnitude of the tensile stresses contributes to increased fatigue life. GATE-5. A static load is mounted at the centre of a shaft rotating at uniform angular velocity. This shaft will be designed for [GATE-2002] (a) The maximum compressive stress (static) (b) The maximum tensile stress (static) (c) The maximum bending moment (static) (d) Fatigue loading GATE-5. Ans. (d) GATE-6. Fatigue strength of a rod subjected to cyclic axial force is less than that of a rotating beam of the same dimensions subjected to steady lateral force because (a) Axial stiffness is less than bending stiffness [GATE-1992] (b) Of absence of centrifugal effects in the rod (c) The number of discontinuities vulnerable to fatigue are more in the rod (d) At a particular time the rod has only one type of stress whereas the beam has both the tensile and compressive stresses. GATE-6. Ans. (d) Relation between the Elastic Modulii GATE-7. A rod of length L and diameter D is subjected to a tensile load P. Which of the following is sufficient to calculate the resulting change in diameter? (a) Young's modulus (b) Shear modulus [GATE-2008] (c) Poisson's ratio (d) Both Young's modulus and shear modulus Page 22 of 429 Chapter-1 Stress and Strain S K Mondal’s GATE-7. Ans. (d) For longitudinal strain we need Young's modulus and for calculating transverse strain we need Poisson's ratio. We may calculate Poisson's ratio from E 2G (1 P ) for that we need Shear modulus. GATE-8. In terms of Poisson's ratio (μ) the ratio of Young's Modulus (E) to Shear Modulus (G) of elastic materials is [GATE-2004] 1 1 (a) 2(1 P ) (b) 2(1 P ) (c) (1 P ) (d ) (1 P ) 2 2 GATE-8. Ans. (a) GATE-9. The relationship between Young's modulus (E), Bulk modulus (K) and Poisson's ratio (μ) is given by: [GATE-2002] (a) E 3 K 1 2P (b) K 3 E 1 2P (c) E 3 K 1 P (d) K 3 E 1 P 9KG GATE-9. Ans. (a) Remember E 2G 1 P 3K 1 2P 3K G Stresses in compound strut GATE-10. In a bolted joint two members are connected with an axial tightening force of 2200 N. If the bolt used has metric threads of 4 mm pitch, then torque required for achieving the tightening force is (a) 0.7Nm (b) 1.0 Nm (c) 1.4Nm (d) 2.8Nm [GATE-2004] 0.004 GATE-10. Ans. (c) T Fur 2200 u Nm 1.4Nm 2S GATE-11. The figure below shows a steel rod of 25 mm2 cross sectional area. It is loaded at four points, K, L, M and N. [GATE-2004, IES 1995, 1997, 1998] Assume Esteel = 200 GPa. The total change in length of the rod due to loading is: (a) 1 μm (b) -10 μm (c) 16 μm (d) -20 μm GATE-11. Ans. (b) First draw FBD of all parts separately then PL Total change in length = AE GATE-12. A bar having a cross-sectional area of 700mm2 is subjected to axial loads at the positions indicated. The value of stress in the segment QR is: [GATE-2006] Page 23 of 429 Chapter-1 Stress and Strain S K Mondal’s P Q R S (a) 40 MPa (b) 50 MPa (c) 70 MPa (d) 120 MPa GATE-12. Ans. (a) F.B.D P 28000 V QR MPa 40MPa A 700 GATE-13. An ejector mechanism consists of a helical compression spring having a spring constant of K = 981 × 103 N/m. It is pre-compressed by 100 mm from its free state. If it is used to eject a mass of 100 kg held on it, the mass will move up through a distance of (a) 100mm (b) 500mm (c) 981 mm (d) 1000mm [GATE-2004] GATE-13. Ans. (a) No calculation needed it is pre- compressed by 100 mm from its free state. So it can’t move more than 100 mm. choice (b), (c) and (d) out. GATE-14. The figure shows a pair of pin-jointed gripper-tongs holding an object weighing 2000 N. The co-efficient of friction (μ) at the gripping surface is 0.1 XX is the line of action of the input force and YY is the line of application of gripping force. If the pin-joint is assumed to be frictionless, then magnitude of force F required to hold the weight is: (a) 1000 N (b) 2000 N (c) 2500 N (d) 5000 N [GATE-2004] GATE-14. Ans. (d) Frictional force required = 2000 N 2000 Force needed to produce 2000N frictional force at Y-Y section = 20000N 0.1 Page 24 of 429 So for each side we need (Fy) = 10000 N force Chapter-1 Stress and Strain S K Mondal’s Taking moment about PIN Fy u 50 10000 u 50 Fy u 50 F u 100 or F 5000N 100 100 GATE-15. A uniform, slender cylindrical rod is made of a homogeneous and isotropic material. The rod rests on a frictionless surface. The rod is heated uniformly. If the radial and longitudinal thermal stresses are represented by ǔr and ǔz, respectively, then [GATE-2005] (a) V r 0, V z 0 (b) V r z 0, V z 0 (c ) V r 0, V z z 0 (d ) V r z 0, V z z 0 GATE-15. Ans. (a) Thermal stress will develop only when you prevent the material to contrast/elongate. As here it is free no thermal stress will develop. Tensile Test GATE-16. A test specimen is stressed slightly beyond the yield point and then unloaded. Its yield strength will [GATE-1995] (a) Decrease (b) Increase (c) Remains same (d) Becomes equal to ultimate tensile strength GATE-16. Ans. (b) GATE-17. Under repeated loading a material has the stress-strain curve shown in figure, which of the following statements is true? (a) The smaller the shaded area, the better the material damping (b) The larger the shaded area, the better the material damping (c) Material damping is an independent material property and does not depend on this curve [GATE-1999] (d) None of these GATE-17. Ans. (a) Previous 20-Years IES Questions Stress in a bar due to self-weight IES-1. A solid uniform metal bar of diameter D and length L is hanging vertically from its upper end. The elongation of the bar due to self weight is: [IES-2005] (a) Proportional to L and inversely proportional to D2 (b) Proportional to L2 and inversely proportional to D2 (c) Proportional of L but independent of D (d) Proportional of U but independent of D Page 25 of 429 Chapter-1 Stress and Strain S K Mondal’s WL WL 1 IES-1. Ans. (a) G ?G u L & Gu 2AE S D2 D2 2u uE 4 IES-2. The deformation of a bar under its own weight as compared to that when subjected to a direct axial load equal to its own weight will be: [IES-1998] (a) The same (b) One-fourth (c) Half (d) Double IES-2. Ans. (c) IES-3. A rigid beam of negligible weight is supported in a horizontal position by two rods of steel and aluminum, 2 m and 1 m long having values of cross - sectional areas 1 cm2 and 2 cm2 and E of 200 GPa and 100 GPa respectively. A load P is applied as shown in the figure [IES-2002] If the rigid beam is to remain horizontal then (a) The forces on both sides should be equal (b) The force on aluminum rod should be twice the force on steel (c) The force on the steel rod should be twice the force on aluminum (d) The force P must be applied at the centre of the beam IES-3. Ans. (b) Bar of uniform strength IES-4. Which one of the following statements is correct? [IES 2007] A beam is said to be of uniform strength, if (a) The bending moment is the same throughout the beam (b) The shear stress is the same throughout the beam (c) The deflection is the same throughout the beam (d) The bending stress is the same at every section along its longitudinal axis IES-4. Ans. (d) IES-5. Which one of the following statements is correct? [IES-2006] Beams of uniform strength vary in section such that (a) bending moment remains constant (b) deflection remains constant (c) maximum bending stress remains constant (d) shear force remains constant IES-5. Ans. (c) IES-6. For bolts of uniform strength, the shank diameter is made equal to [IES-2003] (a) Major diameter of threads (b) Pitch diameter of threads (c) Minor diameter of threads (d) Nominal diameter of threads IES-6. Ans. (c) IES-7. A bolt of uniform strength can be developed by [IES-1995] (a) Keeping the core diameter of threads equal to the diameter of unthreaded portion of the bolt Page 26 of 429 (b) Keeping the core diameter smaller than the diameter of the unthreaded portion Chapter-1 Stress and Strain S K Mondal’s (c) Keeping the nominal diameter of threads equal the diameter of unthreaded portion of the bolt (d) One end fixed and the other end free IES-7. Ans. (a) Elongation of a Taper Rod IES-8. Two tapering bars of the same material are subjected to a tensile load P. The lengths of both the bars are the same. The larger diameter of each of the bars is D. The diameter of the bar A at its smaller end is D/2 and that of the bar B is D/3. What is the ratio of elongation of the bar A to that of the bar B? [IES-2006] (a) 3 : 2 (b) 2: 3 (c) 4 : 9 (d) 1: 3 PL IES-8. Ans. (b) Elongation of a taper rod G l S ddE 4 1 2 Gl A d2 B § D / 3 · 2 or ¨ ¸ Gl B d2 A © D / 2 ¹ 3 IES-9. A bar of length L tapers uniformly from diameter 1.1 D at one end to 0.9 D at the other end. The elongation due to axial pull is computed using mean diameter D. What is the approximate error in computed elongation? [IES-2004] (a) 10% (b) 5% (c) 1% (d) 0.5% PL PL IES-9. Ans. (c) Actual elongation of the bar G l act §S · §S · ¨ 4 d1d2 ¸ E ¨ 4 u 1.1D u 0.9D ¸ E © ¹ © ¹ PL Calculated elongation of the bar G l Cal S D2 uE 4 G l act G l cal § D2 · ? Error % u 100 ¨ 1¸ u 100% 1% G l cal © 1.1D u 0.9D ¹ IES-10. The stretch in a steel rod of circular section, having a length 'l' subjected to a tensile load' P' and tapering uniformly from a diameter d1 at one end to a diameter d2 at the other end, is given [IES-1995] Pl pl.S pl.S 4 pl (a) (b) (c) (d) 4 Ed1d 2 Ed1d 2 4 Ed1d 2 S Ed1d 2 PL IES-10. Ans. (d) Actual elongation of the bar G l act §S · ¨ d1d2 ¸ E ©4 ¹ IES-11. A tapering bar (diameters of end sections being d1 and d2 a bar of uniform cross-section ’d’ have the same length and are subjected the same axial pull. Both the bars will have the same extension if’d’ is equal to [IES-1998] d1 d 2 d1d 2 d1 d 2 a b d1d 2 c d 2 2 2 IES-11. Ans. (b) Poisson’s ratio IES-12. In the case of an engineering material under unidirectional stress in the x- direction, the Poisson's ratio is equal to (symbols have the usual meanings) [IAS 1994, IES-2000] Page 27 of 429 Chapter-1 Stress and Strain S K Mondal’s Hy Hy Vy Vy (a) (b) (c) (d) Hx Vx Vx Hx IES-12. Ans. (a) IES-13. Which one of the following is correct in respect of Poisson's ratio (v) limits for an isotropic elastic solid? [IES-2004] (a) f dQ d f (b) 1/ 4 dQ d1/ 3 (c) 1dQ d1/ 2 (d) 1/ 2 dQ d1/ 2 IES-13. Ans. (c) Theoretically 1 P 1/ 2 but practically 0 P 1/ 2 IES-14. Match List-I (Elastic properties of an isotropic elastic material) with List-II (Nature of strain produced) and select the correct answer using the codes given below the Lists: [IES-1997] List-I List-II A. Young's modulus 1. Shear strain B. Modulus of rigidity 2. Normal strain C. Bulk modulus 3. Transverse strain D. Poisson's ratio 4. Volumetric strain Codes: A B C D A B C D (a) 1 2 3 4 (b) 2 1 3 4 (c) 2 1 4 3 (d) 1 2 4 3 IES-14. Ans. (c) IES-15. If the value of Poisson's ratio is zero, then it means that [IES-1994] (a) The material is rigid. (b) The material is perfectly plastic. (c) There is no longitudinal strain in the material (d) The longitudinal strain in the material is infinite. IES-15. Ans. (a) If Poisson's ratio is zero, then material is rigid. IES-16. Which of the following is true (μ= Poisson's ratio) [IES-1992] (a) 0 P 1/ 2 (b) 1 P 0 (c) 1 P 1 (d) f P f IES-16. Ans. (a) Elasticity and Plasticity IES-17. If the area of cross-section of a wire is circular and if the radius of this circle decreases to half its original value due to the stretch of the wire by a load, then the modulus of elasticity of the wire be: [IES-1993] (a) One-fourth of its original value (b) Halved (c) Doubled (d) Unaffected IES-17. Ans. (d) Note: Modulus of elasticity is the property of material. It will remain same. IES-18. The relationship between the Lame’s constant ‘ǌ’, Young’s modulus ‘E’ and the Poisson’s ratio ‘Ǎ’ [IES-1997] EP EP EP EP a O (b)O c O d O 1 P 1 2P 1 2P 1 P 1 P 1 P IES-18. Ans. (a) IES-19. Which of the following pairs are correctly matched? [IES-1994] 1. Resilience…………… Resistance to deformation. 2. Malleability …………..Shape change. 3. Creep........................ Progressive deformation. 4. Plasticity.... ………….Permanent deformation. Select the correct answer using the codes given below: Codes: (a) 2, 3 and 4 (b) 1, 2 and 3 (c) 1, 2 and 4 (d) 1, 3 and 4 IES-19. Ans. (a) Strain energy stored by a body within elastic limit is known as resilience. Page 28 of 429 Chapter-1 Stress and Strain S K Mondal’s Creep and fatigue IES-20. What is the phenomenon of progressive extension of the material i.e., strain increasing with the time at a constant load, called? [IES 2007] (a) Plasticity (b) Yielding (b) Creeping (d) Breaking IES-20. Ans. (c) IES-21. The correct sequence of creep deformation in a creep curve in order of their elongation is: [IES-2001] (a) Steady state, transient, accelerated (b) Transient, steady state, accelerated (c) Transient, accelerated, steady state (d) Accelerated, steady state, transient IES-21. Ans. (b) IES-22. The highest stress that a material can withstand for a specified length of time without excessive deformation is called [IES-1997] (a) Fatigue strength (b) Endurance strength (c) Creep strength (d) Creep rupture strength IES-22. Ans. (c) IES-23. Which one of the following features improves the fatigue strength of a metallic material? [IES-2000] (a) Increasing the temperature (b) Scratching the surface (c) Overstressing (d) Under stressing IES-23. Ans. (d) IES-24. Consider the following statements: [IES-1993] For increasing the fatigue strength of welded joints it is necessary to employ 1. Grinding 2. Coating 3. Hammer peening Of the above statements (a) 1 and 2 are correct (b) 2 and 3 are correct (c) 1 and 3 are correct (d) 1, 2 and 3 are correct IES-24. Ans. (c) A polished surface by grinding can take more number of cycles than a part with rough surface. In Hammer peening residual compressive stress lower the peak tensile stress Relation between the Elastic Modulii IES-25. For a linearly elastic, isotropic and homogeneous material, the number of elastic constants required to relate stress and strain is: [IAS 1994; IES-1998] (a) Two (b) Three (c) Four (d) Six IES-25. Ans. (a) IES-26. E, G, K and Ǎ represent the elastic modulus, shear modulus, bulk modulus and Poisson's ratio respectively of a linearly elastic, isotropic and homogeneous material. To express the stress-strain relations completely for this material, at least [IES-2006] (a) E, G and Ǎ must be known (b) E, K and Ǎ must be known (c) Any two of the four must be known (d) All the four must be known IES-26. Ans. (c) IES-27. The number of elastic constants for a completely anisotropic elastic material which follows Hooke's law is: [IES-1999] (a) 3 (b) 4 (c) 21 (d) 25 IES-27. Ans. (c) IES-28. What are the materials which show direction dependent properties, called? (a) Homogeneous materials (b) Viscoelastic materials [IES 2007] (c) Isotropic materials (d) Anisotropic materials IES-28. Ans. (d) Page 29 of 429 IES-29. An orthotropic material, under plane stress condition will have: [IES-2006] Chapter-1 Stress and Strain S K Mondal’s (a) 15 independent elastic constants (b) 4 independent elastic constants (c) 5 independent elastic constants (d) 9 independent elastic constants IES-29. Ans. (d) IES-30. Match List-I (Properties) with List-II (Units) and select the correct answer using the codes given below the lists: [IES-2001] List I List II A. Dynamic viscosity 1. Pa B. Kinematic viscosity 2. m2/s C. Torsional stiffness 3. Ns/m2 D. Modulus of rigidity 4. N/m Codes: A B C D A B C D (a) 3 2 4 1 (b) 5 2 4 3 (b) 3 4 2 3 (d) 5 4 2 1 IES-30. Ans. (a) IES-31. Young's modulus of elasticity and Poisson's ratio of a material are 1.25 × 105 MPa and 0.34 respectively. The modulus of rigidity of the material is: [IAS 1994, IES-1995, 2001, 2002, 2007] (a) 0.4025 ×105 Mpa (b) 0.4664 × 105 Mpa (c) 0.8375 × 105 MPa (d) 0.9469 × 105 MPa IES-31. Ans.(b) E 2G (1 P ) or 1.25x10 = 2G(1+0.34) or G = 0.4664 × 105 MPa 5 IES-32. In a homogenous, isotropic elastic material, the modulus of elasticity E in terms of G and K is equal to [IAS-1995, IES - 1992] G 3K 3G K 9 KG 9 KG (a) (b) (c) (d) 9 KG 9 KG G 3K K 3G IES-32. Ans. (c) IES-33. What is the relationship between the linear elastic properties Young's modulus (E), rigidity modulus (G) and bulk modulus (K)? [IES-2008] 1 9 3 3 9 1 9 3 1 9 1 3 (a) (b) (c) (d) E K G E K G E K G E K G 9KG IES-33. Ans. (d) E 2G 1 P 3K 1 2 P 3K G IES-34. What is the relationship between the liner elastic properties Young’s modulus (E), rigidity modulus (G) and bulk modulus (K)? [IES-2009] KG 9KG 9 KG 9 KG (a) E (b) E (c) E (d) E 9K G K G K 3G 3K G 9KG IES-34. Ans. (d) E 2G 1 P 3K 1 2P 3K G IES-35. If E, G and K denote Young's modulus, Modulus of rigidity and Bulk Modulus, respectively, for an elastic material, then which one of the following can be possibly true? [IES-2005] (a) G = 2K (b) G = E (c) K = E (d) G = K = E 9KG IES-35. Ans.(c) E 2G 1 P 3K 1 2P 3K G 1 the value of P must be between 0 to 0.5 so E never equal to G but if P then 3 E k so ans. is c IES-36. If a material had a modulus of elasticity of 2.1 × 106 kgf/cm2 and a modulus of rigidity of 0.8 × 106 kgf/cm2 then the approximate value of the Poisson's ratio of the material would be: Page 30 of 429 [IES-1993] (a) 0.26 (b) 0.31 (c) 0.47 (d) 0.5 Chapter-1 Stress and Strain S K Mondal’s IES-36. Ans. (b) Use E 2G 1 P IES-37. The modulus of elasticity for a material is 200 GN/m2 and Poisson's ratio is 0.25. What is the modulus of rigidity? [IES-2004] (a) 80 GN/m2 (b) 125 GN/m2 (c) 250 GN/m2 (d) 320 GN/m2 E 200 IES-37. Ans. (a) E 2G 1 P or G 80GN / m2 2 1 P 2 u 1 0.25 IES-38. Consider the following statements: [IES-2009] 1. Two-dimensional stresses applied to a thin plate in its own plane represent the plane stress condition. 2. Under plane stress condition, the strain in the direction perpendicular to the plane is zero. 3. Normal and shear stresses may occur simultaneously on a plane. Which of the above statements is /are correct? (a) 1 only (b) 1 and 2 (c) 2 and 3 (d) 1 and 3 IES-38. Ans. (d) Under plane stress condition, the strain in the direction perpendicular to the plane is not zero. It has been found experimentally that when a body is stressed within elastic limit, the lateral strain bears a constant ratio to the linear strain. [IES-2009] Stresses in compound strut IES-39. Eight bolts are to be selected for fixing the cover plate of a cylinder subjected to a maximum load of 980·175 kN. If the design stress for the bolt material is 315 N/mm2, what is the diameter of each bolt? [IES-2008] (a) 10 mm (b) 22 mm (c) 30 mm (d) 36 mm S d2 P 980175 IES-39. Ans. (b) Total load P 8 uV u or d 22.25 mm 4 2SV 2S u 315 IES-40. For a composite consisting of a bar enclosed inside a tube of another material when compressed under a load 'w' as a whole through rigid collars at the end of the bar. The equation of compatibility is given by (suffixes 1 and 2) refer to bar and tube respectively [IES-1998] W1 W2 W1 W2 (a) W1 W2 W (b) W1 W2 Const. (c) (d ) A1 E1 A2 E2 A1 E2 A2 E1 IES-40. Ans. (c) Compatibility equation insists that the change in length of the bar must be compatible with the boundary conditions. Here (a) is also correct but it is equilibrium equation. IES-41. When a composite unit consisting of a steel rod surrounded by a cast iron tube is subjected to an axial load. [IES-2000] Assertion (A): The ratio of normal stresses induced in both the materials is equal to the ratio of Young's moduli of respective materials. Reason (R): The composite unit of these two materials is firmly fastened together at the ends to ensure equal deformation in both the materials. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES-41. Ans. (a) IES-42. The figure below shows a steel rod of 25 mm2 cross sectional area. It is loaded at four points, K, L, M and N. [GATE-2004, IES 1995, 1997, 1998] Page 31 of 429 Chapter-1 Stress and Strain S K Mondal’s Assume Esteel = 200 GPa. The total change in length of the rod due to loading is (a) 1 μm (b) -10 μm (c) 16 μm (d) -20 μm IES-42. Ans. (b) First draw FBD of all parts separately then PL Total change in length = AE IES-43. The reactions at the rigid supports at A and B for the bar loaded as shown in the figure are respectively. (a) 20/3 kN,10/3 kN (b) 10/3 kN, 20/3 kN (c) 5 kN, 5 kN (d) 6 kN, 4 kN [IES-2002; IAS- 2003] IES-43. Ans. (a) Elongation in AC = length reduction in CB R A u 1 RB u 2 AE AE And RA + RB = 10 IES-44. Which one of the following is correct? [IES-2008] When a nut is tightened by placing a washer below it, the bolt will be subjected to (a) Compression only (b) Tension (c) Shear only (d) Compression and shear IES-44. Ans. (b) IES-45. Which of the following stresses are associated with the tightening of nut on a bolt? [IES-1998] 1. Tensile stress due to the stretching of bolt 2. Bending stress due to the bending of bolt 3. Crushing and shear stresses in threads 4. Torsional shear stress due to frictional resistance between the nut and the bolt. Select the correct answer using the codes given below Codes: (a) 1, 2 and 4 (b) 1, 2 and 3 (c) 2, 3 and 4 (d) 1, 3 and 4 IES-45. Ans. (d) Thermal effect IES-46. A 100 mm × 5 mm × 5 mm steel bar free to expand is heated from 15°C to 40°C. What shall be developed? [IES-2008] (a) Tensile stress (b) Compressive stress (c) Shear stress (d) No stress IES-46. Ans. (d) If we resist to expand then only stress will develop. IES-47. Which one of the following statements is correct? [GATE-1995; IES 2007] Page If a material expands freely due to 32 of 429 it will develop heating, Chapter-1 Stress and Strain S K Mondal’s (a) Thermal stress (b) Tensile stress (c) Compressive stress (d) No stress IES-47. Ans. (d) IES-48. A cube having each side of length a, is constrained in all directions and is heated uniformly so that the temperature is raised to T°C. If ǂ is the thermal coefficient of expansion of the cube material and E the modulus of elasticity, the stress developed in the cube is: [IES-2003] D TE D TE D TE D TE (a) (b) (c) (d) J 1 2J 2J 1 2J 3 %V T p a3 1 BT a3 IES-48. Ans. (b) V K a3 P Or 3BT E 3 1 2H IES-49. Consider the following statements: [IES-2002] Thermal stress is induced in a component in general, when 1. A temperature gradient exists in the component 2. The component is free from any restraint 3. It is restrained to expand or contract freely Which of the above statements are correct? (a) 1 and 2 (b) 2 and 3 (c) 3 alone (d) 2 alone IES-49. Ans. (c) IES-50. A steel rod 10 mm in diameter and 1m long is heated from 20°C to 120°C, E = 200 GPa and ǂ = 12 × 10-6 per °C. If the rod is not free to expand, the thermal stress developed is: [IAS-2003, IES-1997, 2000, 2006] (a) 120 MPa (tensile) (b) 240 MPa (tensile) (c) 120 MPa (compressive) (d) 240 MPa (compressive) IES-50. Ans. (d) D E't 12 u 106 u 200 u 103 u 120 20 240MPa It will be compressive as elongation restricted. IES-51. A cube with a side length of 1 cm is heated uniformly 1° C above the room temperature and all the sides are free to expand. What will be the increase in volume of the cube? (Given coefficient of thermal expansion is ǂ per °C) (a) 3 ǂ cm3 (b) 2 ǂ cm3 (c) ǂ cm3 (d) zero [IES-2004] IES-51. Ans. (a) co-efficient of volume expansion J 3 u co efficient of linear exp ansion D IES-52. A bar of copper and steel form a composite system. [IES-2004] They are heated to a temperature of 40 ° C. What type of stress is induced in the copper bar? (a) Tensile (b) Compressive (c) Both tensile and compressive (d) Shear IES-52. Ans. (b) IES-53. Į =12.5×10-6 / o C, E = 200GPa If the rod fitted strongly between the supports as shown in the figure, is heated, the stress induced in it due to 20oC rise in temperature will be: [IES-1999] (a) 0.07945 MPa (b) -0.07945 MPa (c) -0.03972 MPa (d) 0.03972 MPa Page 33 of 429 Chapter-1 Stress and Strain S K Mondal’s IES-53. Ans. (b) Let compression of the spring = x m Therefore spring force = kx kN Expansion of the rod due to temperature rise = LD't kx u L Reduction in the length due to compression force = AE kx u L Now LD't x AE 0.5 u 12.5 u 106 u 20 Or x 0.125 mm ½ °° 50 u 0.5 °° ®1 2 ¾ ° S u 0.010 u 200 u 106 ° ¯° 4 ¿° kx 50 u 0.125 ? Compressive stress = 0.07945MPa A § S u 0.0102 · ¨ ¸ © 4 ¹ IES-54. The temperature stress is a function of [IES-1992] 1. Coefficient of linear expansion 2. Temperature rise 3. Modulus of elasticity The correct answer is: (a) 1 and 2 only (b) 1 and 3 only (c) 2 and 3 only (d) 1, 2 and 3 IES-54. Ans. (d) Stress in the rod due to temperature rise = D't u E Impact loading IES-55. Assertion (A): Ductile materials generally absorb more impact loading than a brittle material [IES-2004] Reason (R): Ductile materials generally have higher ultimate strength than brittle materials (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES-55. Ans. (c) IES-56. Assertion (A): Specimens for impact testing are never notched. [IES-1999] Reason (R): A notch introduces tri-axial tensile stresses which cause brittle fracture. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES-56. Ans. (d) A is false but R is correct. Page 34 of 429 Chapter-1 Stress and Strain S K Mondal’s Tensile Test IES-57. During tensile-testing of a specimen using a Universal Testing Machine, the parameters actually measured include [IES-1996] (a) True stress and true strain (b) Poisson’s ratio and Young's modulus (c) Engineering stress and engineering strain (d) Load and elongation IES-57. Ans. (d) IES-58. In a tensile test, near the elastic limit zone [IES-2006] (a) Tensile stress increases at a faster rate (b) Tensile stress decreases at a faster rate (c) Tensile stress increases in linear proportion to the stress (d) Tensile stress decreases in linear proportion to the stress IES-58. Ans. (b) IES-59. Match List-I (Types of Tests and Materials) with List-II (Types of Fractures) and select the correct answer using the codes given below the lists: List I List-II [IES-2002; IAS-2004] (Types of Tests and Materials) (Types of Fractures) A. Tensile test on CI 1. Plain fracture on a transverse plane B. Torsion test on MS 2. Granular helecoidal fracture C. Tensile test on MS 3. Plain granular at 45° to the axis D. Torsion test on CI 4. Cup and Cone 5. Granular fracture on a transverse plane Codes: A B C D A B C D (a) 4 2 3 1 (c) 4 1 3 2 (b) 5 1 4 2 (d) 5 2 4 1 IES-59. Ans. (d) IES-60. Which of the following materials generally exhibits a yield point? [IES-2003] (a) Cast iron (b) Annealed and hot-rolled mild steel (c) Soft brass (d) Cold-rolled steel IES-60. Ans. (b) IES-61. For most brittle materials, the ultimate strength in compression is much large then the ultimate strength in tension. The is mainly due to [IES-1992] (a) Presence of flaws and microscopic cracks or cavities (b) Necking in tension (c) Severity of tensile stress as compared to compressive stress (d) Non-linearity of stress-strain diagram IES-61. Ans. (a) IES-62. What is the safe static tensile load for a M36 × 4C bolt of mild steel having yield stress of 280 MPa and a factor of safety 1.5? [IES-2005] (a) 285 kN (b) 190 kN (c) 142.5 kN (d) 95 kN W S d2 IES-62. Ans. (b) V c or W V c u ; S d2 4 4 W V c u S u d2 280 u S u 362 Wsafe N 190kN fos fos u 4 1.5 u 4 IES-63. Which one of the following properties is more sensitive to increase in strain rate? [IES-2000] (a) Yield strength (b) Proportional limit (c) Elastic limit (d) Tensile strength IES-63. Ans. (b) Page 35 of 429 Chapter-1 Stress and Strain S K Mondal’s IES-64. A steel hub of 100 mm internal diameter and uniform thickness of 10 mm was heated to a temperature of 300oC to shrink-fit it on a shaft. On cooling, a crack developed parallel to the direction of the length of the hub. Consider the following factors in this regard: [IES-1994] 1. Tensile hoop stress 2. Tensile radial stress 3. Compressive hoop stress 4. Compressive radial stress The cause of failure is attributable to (a) 1 alone (b) 1 and 3 (c) 1, 2 and 4 (d) 2, 3 and 4 IES-64. Ans. (a) A crack parallel to the direction of length of hub means the failure was due to tensile hoop stress only. IES-65. If failure in shear along 45° planes is to be avoided, then a material subjected to uniaxial tension should have its shear strength equal to at least [IES-1994] (a) Tensile strength (b) Compressive strength (c) Half the difference between the tensile and compressive strengths. (d) Half the tensile strength. IES-65. Ans. (d) IES-66. Select the proper sequence [IES-1992] 1. Proportional Limit 2. Elastic limit 3. Yielding 4. Failure (a) 2, 3, 1, 4 (b) 2, 1, 3, 4 (c) 1, 3, 2, 4 (d) 1, 2, 3, 4 IES-66. Ans. (d) Previous 20-Years IAS Questions Stress in a bar due to self-weight IAS-1. A heavy uniform rod of length 'L' and material density 'ǅ' is hung vertically with its top end rigidly fixed. How is the total elongation of the bar under its own weight expressed? [IAS-2007] 2G L2 g G L2 g G L2 g G L2 g (a) (b) (c) (d) E E 2E 2E WL G ALg L G L2 g IAS-1. Ans. (d) Elongation due to self weight = 2 AE 2 AE 2E IAS-2. A rod of length 'l' and cross-section area ‘A’ rotates about an axis passing through one end of the rod. The extension produced in the rod due to centrifugal forces is (w is the weight of the rod per unit length and Z is the angular velocity of rotation of the rod). [IAS 1994] Zwl 2 Z 2 wl 3 Z 2 wl 3 3 gE (a) (b) (c) (d) gE 3 gE gE Z 2 wl 3 IAS-2. Ans. (b) Page 36 of 429 Chapter-1 Stress and Strain S K Mondal’s Elongation of a Taper Rod IAS-3. A rod of length, " L " tapers uniformly from a diameter ''D1' to a diameter ''D2' and carries an axial tensile load of "P". The extension of the rod is (E represents the modulus of elasticity of the material of the rod) [IAS-1996] 4 P1 4 PE1 S EP1 S P1 (a) (b) (c) (d) S ED1 D2 S D1 D2 4 D1D2 4 ED1D2 Pl IAS-3. Ans. (a) The extension of the taper rod = §S · ¨ 4 D1D2 ¸.E © ¹ Poisson’s ratio IAS-4. In the case of an engineering material under unidirectional stress in the x- direction, the Poisson's ratio is equal to (symbols have the usual meanings) [IAS 1994, IES-2000] Hy Hy Vy Vy (a) (b) (c) (d) Hx Vx Vx Hx IAS-4. Ans. (a) IAS-5. Assertion (A): Poisson's ratio of a material is a measure of its ductility. Reason (R): For every linear strain in the direction of force, Poisson's ratio of the material gives the lateral strain in directions perpendicular to the direction of force. [IAS-1999] (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true IAS-5. ans. (d) IAS-6. Asser

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