Mechanics of Materials MENG 270 Fall 2003 Exam 3 Time allowed: 90min. Q.1(a) Q.1 (b) Q.2 Q.3 Q.4 Total

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1 Mechanics of Materials MENG 70 Fall 00 Eam Time allowed: 90min Name. Computer No. Q.(a) Q. (b) Q. Q. Q.4 Total Problem No. (a) [5Points] An air vessel is 500 mm average diameter and 0 mm thickness, the length being meters. Find the stresses induced in the material and change in diameter & length when charged to 0 N/mm internal pressure. Take E 00 kn/ mm and Poisson s ratio is 0. Given: d 50 mm t 0 mm l m p 0 N/ mm E 00 kn/ mm υ 0. δd??? δl??? Solution: Stresses in the material Hoop Stress pr N / mm t 0 ongitudinal stress pr N / mm t 0 Change in the diameter δd ε d ε ν E E 50 5 ε ε δd 0.5 mm Change in ength δl ε l ε ν E E 5 50 ε δl o.5 mm

2 Mechanics of Materials MENG 70 Fall 00 Eam Time allowed: 90min Name Computer No. Problem No (b) [5 Points] A spherical pressure vessel of 900 mm outer diameter is fabricated from steel having ultimate strength of u 400 MPa, knowing that a factor of safety of 4 is desired and the gauge pressure can reach.5 MPa, determine the smallest wall thickness that should be used. Given: d o 900 mm r o 450 mm u 400 MPa p.5 MPa F.S 4 t??? Solution: For spherical shell pr t r r t o p ( r 0 t ) t p ro t + p u 400 F. S 4 all 00 MPa t t 7.74 mm

3 Mechanics of Materials MENG 70 Fall 00 Eam Time allowed: 90min Name Computer No. Problem No [0 Points] The state of plane stress occurs in an aluminum member made of an alloy with tensile yield strength of 0 MPa. Determine factor of safety with respect to yield using (a) maimum shearing stress criterion (b) maimum distortion energy criterion 4 MPa 0 MPa 9 MPa Given: 9 MPa y -4 MPa τ y 0 MPa Y 0 MPa F.S??? Considering; (a) Maimum shearing stress criterion (b) Maimum distortion energy criterion Solution: equation; For plane stress condition the principal stresses are determined by the + y y, ± + τ , ± MPa.MPa y (a) Maimum shear stress theory According to maimum shear stress criterion for plane stress condition,

4 And (b) have; τ ma τ 0.. ma 7.MPa Y 0 τ all 05 MPa 05 F. S Maimum Distortion Energy Theory According to maimum distortion energy criterion for plane stress condition we + Y F. S 0 0 (.) ( 0.)(.) + (.) F. S F. S.7

5 Mechanics of Materials MENG 70 Fall 00 Eam Time allowed: 90min Name Computer No. Problem No [0 Points] Determine equation of elastic curve for the beam and loading shown in figure. Knowing that a.5 m, 5 m and P 50 kn, determine (a) the slope at support A and (b) deflection at point C. Take E 00 GPa. y P C A B a W Given: (a) Elastic curve equation between AB??? Knowing that a.5 m 5 m P 50 kn E 00 GPa find; (b) Deflection at point C (c) Slope at support A Solution: P Free body diagram is shown in Figure -a a M B 0 R A Pb ( let a b) M A 0 FIG: -a Pa R B V Free Body diagram for the section at - where <a M is shown in Figure -b R A Pb M for 0 < < a FIG: -b The differential equation of the deflection curve of bent beam is d y EI M d R A R B

6 d y Pb EI d () The integration of equation () gives dy Pb EI EIθ + C d () The integration of equation () yields P Pb EI y + C + C () a V In the region to the right of the force P at the section - M where a<<, the F.B.D is shown in Figure -c ; The bending moment equation is R A Pb/ Pb M P ( a) FIG: -c d y Pb EI P( a) d (4) The integration of equation (4) yields dy Pb P( a) EI EI θ + C d (5) The integration of equation (5) gives Pb P( a ) EIy + C + C4 () Apply boundary conditions: At 0, y 0 From equation () C 0 (7) At a, θ θ Comparing equations () & (5) and substituting values of constants of integration we get; C C (8) At a y y Therefore comparing equations () & () and substituting values of constants of integration we get, Pba Pba + Ca + Ca + C4 Since C C, we have C 4 0 (9) At, y 0 Equation () gives Pb C ( b ) (0) The values of constants of integration may now be substituted in equations ()&() to give;

7 [ ( b ) ] for a Pb EIy 0 () Pb EIy ( b ) ( a) for a < < b () Equations () & () are valid for in the region indicated Similarly from equations (),the slope at support A is, Pb( b ) θ A () EI Deflection at point C At point C, a, equation () gives Pa b y EI (4) Substituting the given values in equation (4) we get y (I.0 - m 4 from Appendi C ) y -.08 m -.8 mm Slope at Support A Substituting the given values in equation () we get ( 5.5 ) θ A 9. 0 rad BY SINGUARITY METHOD Referring to figure -c Pb M P a d y Pb EI P a d Integrating we have dy Pb P EI EIθ a + C () d Integrating equation () we get Pb P EIy a + C + C () Apply boundary conditions At 0 y0 Equation () gives C 0

8 At y0 From equation () we get 0 Pb P a C P Pb C a Substituting values of constants C and C in equations () and () we get Pb P P Pb EIθ a + a () Pb P P Pb EIy a + a (4) Slope at support A At 0 θ θ A Substituting the given values in equation () we have θ A (.5) θ A rad Deflection at point C From equation (4) Pb y a + a EI b At C, y c (.5) y c.08 m y c.8 mm

9 Mechanics of Materials MENG 70 Fall 00 Eam Time allowed: 90min Name Computer No. Problem No. 4 [0 Points] The steel pipe AB has 0 mm outer diameter and mm wall thickness. Knowing that arm CD is rigidly attached to the pipe, determine the principal stresses and maimum shearing stress at point K. Given: D o 0 mm t mm P 0kN Principal stresses at point K??? Solution: Replace the P by an equivalent force-couple system at the center C of the transverse section containing point K: P 0 kn T (0)(00) 000 kn-mm M 0 50 kn-mm y Stresses, y, τ y at point K 0 T I y π 4 4 ( ) 4 D o D i Mc I M K P Substituting the given values in above equations z

10 y. 55 π 4 4 ( 0 90 ) 4 MPa T r τ y J 4 4 J π ( D o D i ) τ y 4. MPa π 4 4 ( 0 90 ) We note that shearing force P does not cause any shearing stress at point K Principal Stresses + y y ma,min ± + τ y ma, min ± + 4. ma, min 8. ± ma MPa min 48.7 MPa.55 ma + 4. τ τ ma 0.45 MPa

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