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1 Seria : 0. T_ME_(+B)_Strength of Materia_9078 Dehi Noida Bhopa Hyderabad Jaipur Luckno Indore une Bhubanesar Kokata atna Web: E-mai: info@madeeasy.in h: 0-56 CLSS TEST 08-9 MECHNICL ENGINEERING Subject : Strength of Materia Date of test : 9/07/08 nser Key. (b) 7. (c). (c) 9. (b) 5. (d). (d) 8. (a). (b) 0. (a) 6. (a). (b) 9. (c) 5. (b). (c) 7. (b). (b) 0. (b) 6. (c). (a) 8. (a) 5. (c). (b) 7. (b). (a) 9. (c) 6. (b). (a) 8. (b). (a) 0. (c)

2 CT-08 ME Strength of Materia 7 Detaied Epanations. (b) kn Q R S m m m R R R F y 0 R + R R kn M R 0 + R R 0 R R kn R 0 M 0 M Q 0 M R kn-m M S + 0 Q R S BMD. (d). Q S R T. Q R S T B. Q R S T. Q R S T C. Q R S T. Q R S T D. Q R S T 5. Q R S T

3 8 Mechanica Engineering. (b). (b) Stiffness of spring z and z z z ( ) ( ) ( ) 7 0 N/cm cm Gd 8Dn 8 0 cm 5 00 N/cm I y ma I y ma m m 0. (z section moduus) times (c) s per maimum shear stress theory, σ bsoute τ ma σ σ σ Ma of,, 6. (b) σyt ( σ yt yied point stress) 80 σ yt 80 Ma T, J τ R Here T and τ are same, so J R shoud be same i.e. poar section moduus i be same. σ yt 7. (c) Radius of Mohr s circe ( ) σ σ + τ y y Diameter of Mohr s circe ( ) σ σ + τ y y Ma 8. (a) It is a case of pure shear stress, τ Ma σ Ma σ σ +00 Ma 00 Ma τ 0 σ σ

4 CT-08 ME Strength of Materia 9 9. (c) Stress induced in the bar α T E Ma 0. (b) Load () 5 kn 5 0 N stress 00 Ma 00 N/mm We kno that σ impact σ static Impact factor For suddeny appied oad, Impact factor Here for safety σ impact shoud not eceed 00 Ma. σimpact 00 σ static Impact factor 50 N/mm π d d d.8 mm. (b) FBD of BC BC 0 kn FBD of B B 80 kn ( ) Here on B, 0 kn is on upper haf and 0kN is on oer haf BC 0 kn, B 80 kn B (0.) m (compete c/s area of B i be taken because of additiona 0 kn) BC 0.785(0.075).56 0 m U B L B B E st + BCLBC BC E br ( 80 0 ) (.5) J ( 0 0 ) ( 0.5) The above resut is vaid ony if σ < σ y B σ B B Ma <(σ y ) st 50 Ma (O.K.) σ BC BC BC Ma <(σ y ) br 0 Ma (O.K.)

5 0 Mechanica Engineering. (a) E G( + ν) ( + ν) ν 0. ν δd d δ mm (c) N ( ) N ( ) (N/m) C D B From the symmetry of the figure, R C R D + Bending moment at mid point, M RC gives. (b) 8 B C D 50 kn 80 kn 0 kn 0 kn 600 mm 000 mm 00 mm art B: The section of the bar in this part is subjected to a tension of 50 kn Etension of B 5 mm mm (etension) E art BC: The section of the bar in this part is subjected to a compression of kn Contraction of BC 5 mm mm (contraction) E art CD: The section of the bar in this part is subjected to a compression of 0 kn. Contraction of BC mm 0. mm (contraction) E Change in ength of the bar i.e., decrease in ength of the bar is 0. mm.

6 CT-08 ME Strength of Materia 5. (b) d 0 mm, f Hz, 5 kw πft π T 000 T.57 Nm 6T Maimum shear stress, τ s N/mm πd π N/mm 6.59 Ma 6. (c) ccording to the given conditions 50 Ma 60 Ma 60 Ma 50 Ma σ, σ ( σ +σ ) ± ( σ σ ) + τ y y y ( σ +σ ) + ( σ σ ) + τ y y y 5 τ y Ma (60 50) + ( ) + τ y 7. (b) Critica oint T 500 Nm τ s σ a + σ b σ a + σ b σ a σ b M e Ma π 50 My I π (50) N Ma τ s 6T Ma πd π (50)

7 Mechanica Engineering 66. Ma 66. Ma ccording to MDET, 0.7 Ma y σ + τ S yt N N (b) B / C By use of superposition principe C / δ δ δ We kno that δ 8EI For case () B.M.D δ δ + 6EI 8EI ( δ δ ) 8EI 96EI 8EI 8 8E I 8E I

8 CT-08 ME Strength of Materia 9. (b) Cross-section remains pane and undistorted for circuar shaft ony but not for non-circuar shaft. 0. (a) ε 0.00 ε 0.00 E 0 5 ; µ 0. σ E ( ) ( ε ) + µε µ σ σ [ ] 00 Ma 00 Ma τ ma σ σ 00Ma. (c) Etension of a tapering circuar bar is given by L πdd E. (a) π mm For bar of uniform diameter L πd E π (.875) % error % mm For soid shaft, T For hoo shaft, T D i π 6 D τ 0 Di π ( D ) τ 6 D 0 D D0 D mm

9 Mechanica Engineering. (a) 0. mm 00 mm σ E π 00 E E 00 0 N/mm E 00 kn/mm Dia. 0 mm. (a) Force diagram W W R R _ + _ + + SFD nd degree curve BMD st degree curve 5. (d) b b C E B D 6. (a) R V E R 0 R B L L M E b+ + + b 0 π Weight of ater ρvg d π 000 ( 0.5) N nd this is uniformy distributed, M maimum N-mm 8 8

10 CT-08 ME Strength of Materia 5 σ maimum M ma Z π d t Ma 8 π ( ) 7. (b) We kno that, 8. (a) V V σ µ E [ ] V V 90 mm [ ( 0.) ] 0 Loading diagram We kno that dv d V dm d / / π 0 sin M 0 π cos + C π 0 π V cos + C π 0 Using boundary conditions, When, 0, M 0, M 0 We get, C 0 and C 0 M π t, π 0 sin π + C + C...(i) π sin ( ) M π 0 9. (c) σ sinφ σ sinφcos φ τ...(i)

11 6 Mechanica Engineering σ cos φ σ n...(ii) Dividing (i) by (ii), e get tanφ τ 00 σ 00 φ 8.5 utting in equation (ii), e get σ cos σ. Ma 0. (c) ΣM B 0 R R 0 kn (upard) R B kn (upard) Making shear force diagram 0 kn n C D E B 8 kn 8 kn Shear force diagram Maimum shear force 68 kn 68 kn

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