Seria : 0 GH1_ME Strength of Materia_1019 Dehi Noida hopa Hyderabad Jaipur Lucknow Indore une hubaneswar Kokata atna Web: E-mai: info@madeeasy.in h: 011-5161 LSS TEST 019-00 MEHNIL ENGINEERING Subject : Strength of Materia Date of test : 1/0/019 nswer Key 1. (c) 7. (c) 1. (d) 19. (a) 5. (d). (b) 8. (c) 1. (b) 0. (b) 6. (c). (d) 9. (c) 15. (b) 1. (d) 7. (c). (a) 10. (c) 16. (a). (c) 8. (c) 5. (b) 11. (b) 17. (b). (d) 9. (a) 6. (c) 1. (b) 18. (a). (b) 0. (d)
T-019 ME Strength of Materia 7 Detaied Expanations 1. (c) M Loading diagram Shear force ( ) diagram M/L M/L. (b) Load () 5 kn 5 10 N Stress 100 Ma 100 N/mm We know that, σ impact σ static Impact factor For suddeny appied oad, Impact factor Here for safety σ impact shoud not exceed 100 Ma. σ static 50 σ Impact factor 50 N/mm Impact 100 π d d 11.8 mm. (d) 5 10 666.1977 d. Q S R T. Q R S T. Q R S T. Q R S T. Q R S T. Q R S T D. Q R S T 5. Q R S T
8 Mechanica Engineering. (a) kx T x T θ k x Td x GJ 1 1 Td k x x dx GJ GJ 0 k GJ x k GJ 0 5. (b) M e (σ b ) My 5e (tensie) I I xx xx Tota stress at (σ a ) (compressive) 5e (compressive) I xx 6. (c) For beam of uniform strength, maximum bending stress remains constant throughout. 7. (c) ccording to the given conditions 50 Ma 60 Ma 60 Ma 50 Ma σ 1, σ 1 1 ( σ +σ ) ± ( σ σ ) + τ x y x y xy 1 ( σ +σ ) + ( σ σ ) + τ x y x y xy
T-019 ME Strength of Materia 9 15 1 (60 50) + (60 + 50) + τ τ xy 106.65 Ma xy 8. (c) d t σ per (1750 16) 50 16 5 Interna pressure, 50 16 1.67 Ma (1750 ) 5 9. (c) π E I e ( L ) L e Effective ength of coumn For both end fixed, L e e e π EI 10. (c) verage force, F d dt mv t ( u) ( ) 0000 50 10 1000 11111.11 N or 11.11 kn 0 600 11. (b) ross-section remains pane and undistorted for circuar shaft ony but not for non-circuar shaft. 1. (b) ritica oint T 500 Nm τ s σ a + σ b σ a + σ b M e σ a σ b 50 10 5.6Ma π 50 My I π (50) N 50 10 10 5 6 0.7 Ma
10 Mechanica Engineering τ s 16T 16 500 10 0.7Ma π (50) 66. Ma 66. Ma ccording to MDET, x x y yt σ + τ S N N.67 0.7 Ma 1. (d) R 1 R, 1 F K K R D R F or R D R + R D F δ 1 + δ 0 1 + 0 K K 1 R ( R F) + 0 K K R F U 1 1 1 F 1 1δ 1 K1 1 F F F K 8K 1. (b) w y use of superposition principe / 1
T-019 ME Strength of Materia 11 w w / δ 1 δ δ We know that δ w 8E I For case ().M.D δ w w + 6EI 8EI w w + 96EI 18EI δ 1 w w w ( δ δ ) 8EI 96EI 18EI 8w w w 8E I 1w 8E I 15. (b) 16 T e 6 τ aowabe 16 0 10 50 d 16.7 mm...(i) M e and σ aowabe 6 1 10 100 d 11.56 mm...(ii) From (i) and (ii), we get d 16.7 mm 16. (a) 17. (b)
1 Mechanica Engineering diameter (d) 10 mm radius (r) 5 mm θ π radian τ 0 N/mm G.7 10 N/mm We know that, τ r Gθ L L Gθ r π τ 0.7 10 5 105.75 mm 1.05 m 1.1 m 18. (a) The extension of the rod is given by, δ πed d 1 6000 00 δ 5 π 10 60 0 6.66 µm 6.7 µm 19. (a) Given: E 00 10 N/mm y max t y max 1 mm R 000 Maximum bending stress: 1000 mm R Ey (σ b ) max max Eymax Eymax R t 1 R R + t R >>> 00 10 1 1000 00 N/mm 0. (b) hange in voume i.e. diatation, σ x + σ y + σz v ( 1 µ ) E Diatation (sum of norma stresses) Diatation 1 E E G(1 + µ) Reationship between E and G µ 0.1 : for concretes
T-019 ME Strength of Materia 1 1. (d) µ 0.5 0.5 : for eastic materias µ 0.5 : rubber Largest possibe vaue of µ is 0.5. E G 1 ( +µ ) ; s (1 + µ) > 0, G < E y σ x 80 N/mm. (c) For zero strain in, s y 0 σ y E D σ µ x E σ 0. 80 E E σ N/mm for auminium, µ 0. for rubber, µ 0.8 to 0.5. (d) Let us draw Free ody Diagram of and separatey : µ < µ r, auminium resists atera deformation more effectivey than rubber. 900 N x 1. m 1.8 m y y ΣM 0, 900 1. y y 900 1. 60 N
1 Mechanica Engineering xy x 1 100 00 N m x M y Sum of a moments at for entire frame: 8 M + 60 900 1. 00 0 M 1080 + 1080 + 600 M 600 Nm. (b) σ 1 100 N/mm, σ 50 N/mm, σ 0 ccording to maximum principa stress theory, σy σ 1 FOS 5. (d) FOS 50.5 100 ccording to maximum shear stress theory, σy τ max, absoute FOS FOS.5.5 kn x 5. kn 0.5m 0.5m 1 m 1.5 m 1.5 m x M x.5 0.5 + 5.(1.5 + 1.0 + 0.5) M x.5 + 16. M x 18.5 knm
T-019 ME Strength of Materia 15 6. (c) 7. (c) U 1 W δ W U 7000 δ 60 00 N 8 00 1000 8 δ 8000 60 D 1 mm 1. mm Maximum shear stress, τ max 8. (c) 8WD 8 00 1 π ( 1.) 6.8 Ma L L L L L rea of bending moment, 1 L L L 8 L L x + x 5 L 6 δ x E I L 5L 1 5L 8 6 EI 8EI 9. (a) Given, σ 1 100 Ma σ 50 Ma, σ 5 Ma S yt 0 Ma, For maximum shear strain energy theory,
16 Mechanica Engineering S yt (σ 1 σ ) + (σ σ ) + (σ σ 1 ) N [Where, N factor of safety] (100 50) + (50 5) + (5 100) 0 N fter soving, Factor of safety, N.6 ~. 0. (d) 1 55 φ 1 1 Shear strain, φ tan 55 1.8 0.5 radian τ ab 16000 150 5 0.7 Ma Shear moduus, G τ φ 0.7 0.5 1.896 1.90 Ma