NGINRS CDMY xially oaded Members Junior ngineer QUSTION BNK 1. The stretch in a steel rod of circular section, having a length subjected to a tensile load P and tapering uniformly from a diameter d 1 at one end to a diameter d at the other end, is given by (a) P d d 1 P d d 1 (b) (d) P d d 1 P d d 1. The total extension of the bar loaded as shown in the figure is = area of cross-section = modulus of elasticity 10T 3T T 9T (a) 10 30/ 9 30/ 10mm 10mm 10mm (b) 6 10/ (d) 30 / 3. bar of uniform cross-section of one sq. cm is subjected to a set of five forces as shown in the given figure, resulting in its equilibrium. That maximum tensile stress (in kgf/cm ) produced in the bar is. 10 cm long and 5 cm diameter steel rod fits snugly between two rigid walls 10 cm apart at room temperature. Young s modulus of elasticity and coefficient of linear expansion of steel are 10 6 kgf/cm and 1 10 6 / C respectively. The stress developed in the rod due to a 100 C rise in temperature wall be (a) 6 10 10 kgf/cm (b) 6 10 9 kgf/cm. 10 3 kgf/cm (d). 10 kgf/cm 5. For a composite bar consisting of a bar enclosed inside a tube of another material and 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 refer to bar and tube respectively) (a) W 1 + W = W (b) W 1 + W = Constant (d) W W 1 1 1 W 1 1 1 6. tapering bar (diameter of end sectisons being, d 1 and d ) and a bar of uniform cross-section d haved thesame length and are subjected the same axial pull. Both the bars will have the same extension if d is equal to 1 3 (a) d d 1 (b) d1d 11kgf kgf 1kgf 5kgf 5kgf B C D 1 3 (a) 1 (b) 10 (d) 11 # 100-10, Ram Nagar, Bambala Puliya d d 1 (d) d d 1 7. 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 (a) the same half (b) one fourth (d) double mail : info @ engineersacademy.org
NGINRS CDMY C : Strength of Materials xially oaded Members 3 8. slender bar of 100 mm cross-section is subjected to loading as shown in the figure below. If the modulus of elaasticity is taken as 00 10 9 Pa, then the elongation produced in the bar will be (a) 1.3 N/mm (b) 16.0 N/mm 10.7 N/mm (d) Zero 1. The axial movement of top surface of stepped column as shown in figure is 100kN (a) 10 mm 1 mm 00kN 00kN 0.5m 1.0m 0.5m (b) 5 mm (d) nil # 100-10, Ram Nagar, Bambala Puliya 100kN 9. If the rod fitted snugly between the supports as shown in the figure below, is heated, the stress induced in it due to 0 C rise in temperature will be ( = 1.5 10 6 / C) and = 00 GPa) 10 mm 0.5m (a) 0.0795 MPa 0.0397 MPa k =50kN/m (b) 0.0795 MPa (d) 0.0397 MPa 10. rod of material with = 00 10 3 MPa and = 10 3 mm/mm C is fixed at both the ends. It is uniformly heated such that the increase in temperature is 30 C. The stress developed in the rod is (a) 6000 N/mm (tensile) (b) 6000 N/mm (compressive) 000 N/mm (tensile) (d) 000 N/mm (compressive) 11. link is under a pull which lies on one of the faces as shown in the figure below. The magnitude of maximum compressive stress in the link would be kn 50 mm 15 mm 500 mm kn (a).5 P/ (b) 3 P/ 1.5 P/ (d) P/ 13. The principle of superposition is made use of in structural computations when: (a) The geometry of the structure changes by a finite amount during the application of the loads (b) The changes in the geometry of the structure during the application of the loads is too small and the strains in the structure are directly proportional to the corresponding stresses The strains in the structure are not directly proportional to the corresponding stresses, even though the effect of changes in geometry can be neglected. (d) None of the above conditions are met. 1. cantilever beam of tubular section consists of materials copper as outer cylinder and steel as inner cylinder. It is subjected to a temperature rise of 0 C and copper > steel. The stresses developed in the tubes will be (a) Compression is steel and tension in copper (b) Tension in steel and compression in copper No stress in both (d) Tension in both the materials P mail : info @ engineersacademy.org
NGINRS CDMY xially oaded Members Junior ngineer 15. In a linear elastic structural element (a) Stiffness is directly proportional to flexibility (b) Stiffness is inversely proportional to flexibility Stiffness is equal to flexibility (d) Stiffness and flexibility are not related 16. The total elongation of the structural element fixed, at one end, free at the other end, and of varying cross-section as shown in the figure when subjected to a force p at free end is given by (a) P/ (b) 3 P/.5 P/ (d) P/ 17. large uniform plate containing a rivet-hole is subjected to uniform uniaxial tension of 95 MPa. The maximum stress in the plate is P P (a) T1 T Pal Pbl T,T a b a b (b) 1 Pbl Pal T,T a b a b 1 Pal Pbl T,T a b a b 1 19. rod of length I and cross-section area 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 is the angular velocity of rotation of the rod.) (a) wl g wl g 3 (b) wl 3g 3 3g (d) 3 wl 95MPa 10cm # 100-10, Ram Nagar, Bambala Puliya 5cm (a) 100 MPa (b) 85 MPa 190 MPa (d) Indeterminate 18. Below Fig. shows a rigid bar hinged at and supported in a horizontal position by two vertical identical steel wires. Neglect the weight of the beam. The tension Tl and T induced in these wires by a vertical load P applied as shown are T l b P T 1 l a 0. 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) (a) y x y x (b) (d) 1. free bar of length is uniformly heated form 0 C to a temperature t C. is the coefficient of linear expansion and the modulus of elasticity. The stress in the bar is (a) t (b) t/ zero (d) none of the above y x y x mail : info @ engineersacademy.org
# 100-10, Ram Nagar, Bambala Puliya NGINRS CDMY C : Strength of Materials xially oaded Members 5. Which one of the following pairs is NOT correctly matched? ( = Young s modulus, a = Coefficient of linear expansion, T = Termperature rise, = rea of cross-section, 1 = Original length) (a) Temperature strain with permitted expansion... ( Tl ) (b) Temperature stress... T Temperature thrust... T (d) Temperature stress with permitted expansion ( Tl )... l 3. The reactions at the rigid supports at and B for the bar loaded as shown in the figur e are respectively. 1m C (a) 0/3 kn, 10/3 KN (b) 10/3 kn, 0/3 kn 5 kn, 5 kn (d) 6 kn, kn 10 kn m. steel rod of diameter 1 cm and 1 m long is heated from 0 C to 10 C. Its = 1 10 6 /K and = 00 GN/m. If the rod is free to expand, the thermal stress developed in it is: (a) 1 10 N/m (b) 0 kn/m zero (d) infinity 5. steel rod 10 mm in diameter and 1m long is heated from 0 C to 10 C, = 00 GPa and a = 1 10-6 per C. If the rod is not free to expand, the thermal stress developed is: (a) 10 MPa (tensile) (b) 0 MPa (tensile) 10 MPa (compressive) (d) 0 MPa (compressive) B 6. heavy uniform rod of length and material density is hung vertically with its top end rigidly fixed. How is the total elongation of the bar under its own weight expressed? (a) g g (b) (d) g g 7. Given that for an element in a body of homogeneous isotropic material subjected to plane stresses. If x, y and z are normal strains in x, y and z direction respectively and is the poisson s ratio, the magnitude of unit volume change of the element is given by (a) x + y + z (b) x + ( y + z ) ( x + y + z ) (d) 1 1 1 x y z 8. solid metal bar of uniform diameter D and length is hung vertically from a ceiling. If the density of the material of the bar is and the modulus of elasticity is, then the total elongation of the bar due to its own weight is (a) (b) (d) 9. bar of circular cross-section varies uniformly from a cross-section D to D. If extension of the bar is calculated treating it as a bar of average diameter, then the perentage error will be (a) 10 (b) 5 33.33 (d) 50 30. The length, coefficient of thermal expansion and young s modulus of bar are twice that of bar B. If the temperature of both bars is increased by the same amount while preventing any expansion, then the ratio of stress developed in bar to that in bar B will be (a) (b) 8 (d) 16 mail : info @ engineersacademy.org
NGINRS CDMY 6 xially oaded Members Junior ngineer 31. The side D of the square block BCD as shown in the given figure is fixed at the base and it is under a stage of simple shear causing shear stress and shear strain. where = Modulus of Rigidity (G) The distored shape is B C D. The diagonal strain (linear) will be B B C C D (a) / (b) / (d) 3. The lists given below refer to a bar of length cross sectional area, Young s modulus, poisson s ratio and subjected to axial stress p. Match ist-i with ist-ii and select the correct answer using the codes given below the lists: ist-i ist-ii. Volumetric strain 1. (1 + ) B. Strain energy per. 3(1 ) unit volume C. Ratio of young s 3. p (1 ) 33. If all the dimensions of a prismatic bar of square cross-section suspended freely from the ceiling of a roof are doubled then the total elongation produced by its own weight will increase (a) eight times three times (b) four times (d) two times 3. sseriton (): The amount of elastic deformation at a certain point, which an elastic body undergoes, under given stress is the same irrespective of the stresses being tensile or compressive. Reason (R) : The modulus of elasticity and Poisson s ratio are assumed to be the same in tension as well as compression. (a) Both and R are true and R is the correct explanation of (b) Both and R are true but R is not a correct explanation of is true but R is false (d) is false but R is true 35. If the loads and reactions of the beam shown are as given in the following figure..t 6.98T 3.T 3.6TT.T.T C D B 5.87T 7.3T 1m 1.5m m 1.5m The thrust diagram on the section of the beam, taking tension positive, will be modulus to bulk modulus D. Ratio of young s. modulus to modulus of rigidity Codes: B C D (a) 3 1 (b) 5 1 5 1 (d) 3 1 5 # 100-10, Ram Nagar, Bambala Puliya p 5. (1 ) (a) 3. + C D + 0. B (b) 3. 3. (d) + 0. B C D C D + 0.B 0. + C D B 3. mail : info @ engineersacademy.org
NGINRS CDMY C : Strength of Materials xially oaded Members 7 36. copper bar of 5 cm length is fixed by means of supports at its ends. Supports can yield (total) by 0.01 cm. If the temperature of the bar is raised by 100 C, then the stress induced in the bar for c = 0 10 6 C and c = 1 10 6 kg/ cm wil be (a) 10 kg/cm (b) 10 kg/cm 8 10 kg/cm (d) 16 10 kg/cm 37. Two wires of equal length are suspended vertically at a distance of 0 cm as shown in the figure below. Their upper ends are fixed to the ceiling while their lower ends support a rigid horizontal bar which carries a central load of 1t midway between the wires. Details of the wires are given below: Wire rea No. (cm ) 0 cm 1 m 1 t Modulous of Material lasticity longation 6 (kg/cm ) 1 Copper 110 Steel 10 The ratio of the elongation of the two wires, c / s is (a) 0.05 (b) 0.5 (d) 1 38. composite section shown in the figure below was formed at 0 C and was made of two materials and B. If the coefficient of thermal expansion of is greater than that of B and the composite section is heated to 0 C, then and B will 6 B Rigid bar c s (a) be in tension and compression respectively (b) both be in compression both be in tension (d) be in compression and tension respectively 39. mild steel bar is in two parts having equal lengths. The area of cross-section of part-1 is double that of Part-. If the bar carries an axial load P, then the ratio of elongation in Part-1 to that in Part- will be (a) (b) 1/ (d) 1/ 0. sseriton (): bar tapers from a diameter of d 1 to a diameter of d over its length and is subjected to a tensile force P. If extension is calculated based on treating it as a bar of average diameter, the calculated extension will be more than the actual extension. Reason (R): The actual extension in such bars P is given by, =. d d 1 (a) Both and R are true and R is the correct explanation of (b) Both and R are true but R is not a correct explanation of istrue but R is false (d) is false but R is true 1. round bar made of same material consists of 3 parts each of 100 mm length having diameters of 0 mm, 50 mm and 60 mm, respectively. If the bar is subjected to an axial load of 10 kn, the total elongation of the bar in mm would be ( is the modulus of elasticity in kn/mm ) (a) 0. 1 1 1 16 5 36 (b) 1 1 1 16 5 36 1 1 1 16 5 36 (d) 0 1 1 1 16 5 36 # 100-10, Ram Nagar, Bambala Puliya mail : info @ engineersacademy.org
NGINRS CDMY 8 xially oaded Members Junior ngineer. If a member is subjectd to tensile stress of p x, compressive stress of p y and tensile stress of p z, along the X, Y and Z directions respectively, then the resultant strain e x along the X direction would be ( is Young s modulus of elasticity, is Poisson s ratio) 1 (p p p ) (a) x y z 1 (p p p ) (b) x y z 1 (p p p ) x y z 3. steel bar 300 mm long and having mm diameter, is turned down to 18 mm daimeter for one third of its length. It is heated 30 C above room temperature, clamped at both ends and than allowed to cool to room temperature. If the distance betwen the clamps is unchanged, the maximum stress in the bar ( = 1.5 10 6 per C and = 00 GN/m ) is (a) 5 MN/m (b) 50 MN/m 75 MN/m (d) 100 MN/m 1 (p p p ) (d) x y z # 100-10, Ram Nagar, Bambala Puliya mail : info @ engineersacademy.org
NGINRS CDMY C : Strength of Materials xially oaded Members 9 1. ns. (d) NSWRS ND XPNTIONS Deflection of circular tapering rod subjected to tensile load P. is. ns. (b) = P d d 1 10T 10T 9T 9T = 3. ns. (d) F.B.D 7T 7T ( l) = ( l) + ( l) + ( l) total 1 3 1010 7 10 910 = 610 11 11 10 10 9 9 5 5 Maximum tensile force = 11 kg.f Max. tensile stress = = 11 kg/cm 11 11 5. ns. (Strain) 1 = (strain) 6. ns. (b) l = W W 1 1 = 1 pl d d 1 pl pl d1d rea rea = d eq 1 7. ns. (d ) d d d d eq 1 Deformation under own weight l dx. ns. 5 cm l = 10 cm = 10 6 kgf/cm = 1 10 6 / c Strain is prevented stress will be induced in steel rod. It is satically indeterminate. So we used one equation of compatibility. T = P = T = 1 10 6 10 6 100 =. 10 3 kgf/cm # 100-10, Ram Nagar, Bambala Puliya Consider a small strip dx at a distance x as shown in figure. We shall find change in length for dx and then integrate for whole length. Force exerted by weight below strip dx Force (P) = Volume below dx specific weight = p ( = cross sectional area) longation of the strip ( l) dx ( l)dx load p length dx xdx (l) dx = For total elongation integrate and take limits l l x l xdx o l l o...(i) mail : info @ engineersacademy.org
NGINRS CDMY 50 xially oaded Members Junior ngineer Now elongation due to load (W) Wl l load = own weight (givne) W = l The stress induced is compressive is nature as a compressive force is exerted by the supports which prevents increase in length due to increase is temperature. 11. ns. Tension = W l l w # 100-10, Ram Nagar, Bambala Puliya l Now from equation (i) and (ii) Where, l l w 1...(ii) l = longation due to self load (w) (l) w = longation externel due to load w l lw / 8. ns. (d) F.B.D. P 100 100 total elongation 100 100 100 100 = 1 100 0.5 100 1 100 0.5 0 9. ns. (b) xpansion of rod = l t = 0.5 1.5 10 6 0 = 1.5 10 m Force will be induced due to spring = 1.5 10 50 1000 = 6.5 F 6.5 stress = (10) 10. ns. (b) Thermla stress () = t = 10 3 00 10 3 30 = 6000 N/mm (compresive) 0.0795MPa KN quivalent figure, Bending stress b = M y I = Compressive 500 mm 15 50 000 5 1 15 (50) 1 KN 000 5 N-mm 3 5 = 16 MPa F 000N Tensile stress, t = 15 50mm Maximum compressive stress 1. ns. = b t = 16 5.33 = 10.67 MPa P P = 1.5P 13. ns.(b) Superposition principle is applicable for structural members with negligible deformation and the loads acting on the member are within elastic limit. 1. ns.(b) Steel is less active due to temperature changes compared to copper, therefore steel is subjected to tension and copper is subjected to compression due to composite action, 15. ns: (b) For any linear elastic member stiffness is reciprocal of flexibility. mail : info @ engineersacademy.org
NGINRS CDMY C : Strength of Materials xially oaded Members 51 16. ns. P P P.5P 17. ns. Maximum stress in plate develops across rivet hole Max stress, = 18. ns. M = 0 P = 95 t 10cm = t (10 5)cm 10 95 190MPa 5 T b T a = P(l)...(1) 1 l 1 b = l a T1 b T = a a T 1 =.T b Sub P 1 in equation (1) a T b T = P() b 3. ns. (a) longation in C = length reduction in CB R 1 R B = and R + R B = 10. ns. Thermal stress will develop only if expansion is restricted. 5. ns. (d) t = (1 10 6 ) (00 10 3 ) (10 0) = 0 MPa It will be compressive as elongation restricted. 6. ns. (d) longation due to self weight = 7. ns. (a) = W = ( g) g 8. ns. (b) b T a b = P T = Pb a b Pa Similarly T = 1 19. ns. (b) 0. ns. (a) 1. ns. # 100-10, Ram Nagar, Bambala Puliya a b bar free to expand due to temperature has no stress. ns. (a) Dimensional analysis gives (a) is wrong. dx x The elongation of bar due to its own weight (W) is Now = W W = = mail : info @ engineersacademy.org
NGINRS CDMY 5 xially oaded Members Junior ngineer 9. ns. (a) 0 = P D D 1 33. ns. (b) The area of bar will become times, and the volume as well as weight will increase 8 times. So increase in elongation P P ctual extension, 0 = (D ) D The average diameter of bar = D D 1.5D pproximate extension, P P = (1.5D).5D rror in calculation = # 100-10, Ram Nagar, Bambala Puliya 1 100 0 = 1 100 = 11.11% 10%.5 30. ns. (b) Temperature Stress = T Stress in bar Stress in bar B 31. ns. (a) 3. ns. (a) = Volumetric strain B B ( ) (1 ) V x y z V p (1 ) B = Strain energy per unit volume 1 p l stress Strain C = = 3K (1 ) 3(1 ), G(1 ) K W 3. ns. (a) 8 will be = times If the material is homogeneous & isotropic, magnitude of deformation will be same if & are same in all direction. x y x ( ) ( z) x y x Magnitude of x in (i) as well as (ii) is same. 35. ns. (a) Section C 3. mail : info @ engineersacademy.org z Thrust (T) CD 3. 3.6 = 0. D 0. +. =.0 B = 0 3. 36. ns. (d) + + C D 0. The elongation in the bar due to increase in temperature = T Yield of support reduces strain in the bar by. = 1 10 6 = 1600 kg/cm = T = T 6 0.01 010 100 5 B
NGINRS CDMY C : Strength of Materials xially oaded Members 53 37. ns. (d) Since load of 1t is applied mid-way so force in both wires will be same and equals to 0.5 t. 38. ns. (b) c = 500 510 m 6 110 500 s = 6 10 = 5 10 m c = 1.0 s Since both the ends of the bar are unyielding. With increase in temperature, both bars will be in compression and the load will be transfered to the supports. 39. ns. 1 = 1 1 1. ns. (d) P 1 1 1 Total elongation = d1 d d3 = 10100 1 1 1 mm 100 16 5 36 = 0 1 1 1 mm 16 5 36. ns. (a) e x = x y z 3. ns. 1 p p p We knowthat temperature stresses do not depend upon properties of cross section like length and area. They only depends upon properties of the material. = T = 1.5 10 6 00 10 3 30 = 75 MN/m Since 1 = and 1 = 0. ns. (d) 1 = 1 1 1 The actual extension is more than approximate extension based on average diameter. # 100-10, Ram Nagar, Bambala Puliya mail : info @ engineersacademy.org