REVIEW PROBLEMS & SOLUTIONS JAMES M. GERE BARRY J. GOODNO

Transcription

1 REVIEW ROEMS & SOUTIONS JMES M. GERE RRY J. GOODNO

2 ppendix FE Exam Review roblems R-1.1: plane truss has downward applied load at joint and another load applied leftward at joint 5. The force in member 3 5 is: () 0 () / (C) (D) 1.5 M 1 0 V 6 (3 ) 0 so Method of sections Cut through members 3-5, -5 and -4; use right hand FD M 0 V 6 0 F 35 0 F H 1 V1 V 6 6 F 35 5 F 5 F 4 4 V Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

3 1084 ENDIX FE Exam Review roblems R-1.: The force in member FE of the plane truss below is approximately: () () 1.5 kn. kn (C) 3.9 kn (D) 4.7 kn 15 kn 10 kn 5 kn 3 m 3 m C 3 m D 4.5 m G F E 3 m 1 m Statics M 0 E y (6 m) 15 kn (3 m) 10 kn (6 m) 5 kn (9 m) 0 E y 5 kn x 15 kn 10 kn 5 kn 3 m 3 m C 3 m D y 4.5 m G F E 3 m 1 m E y Method of sections: cut through C, E and FE; use right-hand FD; sum moments about F FE (3 m) F FE (3 m) 10 kn (3 m) 5 kn (6 m) E y (3 m) 0 Solving F FE kn 4 C 10 kn 3 m D 5 kn F FE 3.95 kn F FE 3 10 F FE E 3 m 1 m 1 10 F FE E y 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4 ENDIX FE Exam Review roblems 1085 R-1.3: The moment reaction at in the plane frame below is approximately: () () (C) (D) 1400 N m 80 N m 3600 N m 6400 N m 900 N 900 N 100 N/m 1. m C 1. m C 3 m x 3 m 4 m in connection y C y Statics: use FD of member C to find reaction C y y M 0 C y (3 m) 900 N (1. m) 0 C y 900 N (1. m) 3 m 360 N Sum moments about for entire structure M 0 M C y (3 m) 900 N (1. m) 1 a100 N m b 4 m a 3 4 mb 0 Solving for M M 6400 N m 900 N 100 N/m 1. m C 3 m 4 m C y M x y 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5 1086 ENDIX FE Exam Review roblems R-1.4: hollow circular post C (see figure) supports a load 1 16 kn acting at the top. second load is uniformly distributed around the cap plate at. The diameters and thicknesses of the upper and lower parts of the post are d 30 mm, t 1 mm, d C 60 mm, and t C 9 mm, respectively. The lower part of the post must have the same compressive stress as the upper part. The required magnitude of the load is approximately: () 18 kn () kn (C) 8 kn (D) 46 kn 1 16 kn d 30 mm t 1 mm d C 60 mm t C 9 mm 1 t d p 4 [d (d t ) ] 679 mm C p 4 [d C (d C t C ) ] 144 mm d C Stress in : s Ma C t C Stress in C: s C 1 C must equal s Solve for C kn Check: s C 1 C 3.6 Ma same as in R-1.5: circular aluminum tube of length 650 mm is loaded in compression by forces. The outside and inside diameters are 80 mm and 68 mm, respectively. strain gage on the outside of the bar records a normal strain in the longitudinal direction of The shortening of the bar is approximately: () 0.1 mm () 0.6 mm (C) 0.36 mm (D) 0.5 mm 400 (10 6 ) 650 mm d 0.60 mm 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

6 ENDIX FE Exam Review roblems 1087 Strain gage R-1.6: steel plate weighing 7 kn is hoisted by a cable sling that has a clevis at each end. The pins through the clevises are mm in diameter. Each half of the cable is at an angle of 35 to the vertical. The average shear stress in each pin is approximately: () Ma () 8 Ma (C) 40 Ma (D) 48 Ma W 7 kn d p mm 35 Cross sectional area of each pin: p p 4 d p 380 mm Tensile force in cable: Cable sling a W b T kn cos(u) Clevis Shear stress in each clevis pin (double shear): t T 1.7 Ma Steel plate R-1.7: steel wire hangs from a high-altitude balloon. The steel has unit weight 77kN/m 3 and yield stress of 80 Ma. The required factor of safety against yield is.0. The maximum permissible length of the wire is approximately: () 1800 m () 00 m (C) 600 m (D) 3000 m 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7 1088 ENDIX FE Exam Review roblems g 77 kn m 3 llowable stress: Y 80 Ma FS Y s allow s Y FS Y Ma Weight of wire of length : W Max. axial stress in wire of length : s max W max Max. length of wire: max s allow g 1818 m R-1.8: n aluminum bar (E 7 Ga, 0.33) of diameter 50 mm cannot exceed a diameter of 50.1 mm when compressed by axial force. The maximum acceptable compressive load is approximately: () 190 kn () 00 kn (C) 470 kn (D) 860 kn E 7 Ga d init 50 mm d final 50.1 mm 0.33 ateral strain: d final d init 0.00 d init xial strain: a n xial stress: E a Ma below yield stress of 480 Ma so Hooke s aw applies Max. acceptable compressive load:r max s a p 4 d init b 857 kn R-1.9: n aluminum bar (E 70 Ga, 0.33) of diameter 0 mm is stretched by axial forces, causing its diameter to decrease by 0.0 mm. The load is approximately: () 73 kn () 100 kn (C) 140 kn (D) 339 kn 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8 ENDIX FE Exam Review roblems 1089 E 70 Ga d init 0 mm d 0.0 mm 0.33 ateral strain: d d init d xial strain: a v xial stress: E a 33.3 Ma below yield stress of 70 Ma so Hooke s aw applies Max. acceptable load: max s a p 4 d init b 73.3 kn R-1.10: polyethylene bar (E 1.4 Ga, 0.4) of diameter 80 mm is inserted in a steel tube of inside diameter 80. mm and then compressed by axial force. The gap between steel tube and polyethylene bar will close when compressive load is approximately: () 18 kn () 5 kn (C) 44 kn (D) 60 kn E 1.4 Ga d 1 80 mm d 1 0. mm 0.4 ateral strain: d 1 d 1 Steel tube d 1 d olyethylene bar xial strain: a v xial stress: E a 8.8 Ma well below ultimate stress of 8 Ma so Hooke s aw applies Max. acceptable compressive load: max s a p 4 d 1 b 44.0 kn 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9 1090 ENDIX FE Exam Review roblems R-1.11: pipe (E 110 Ga) carries a load 1 10 kn at and a uniformly distributed load 100 kn on the cap plate at. Initial pipe diameters and thicknesses are: d 38 mm, t 1 mm, d C 70 mm, t C 10 mm. Under loads 1 and, wall thickness t C increases by mm. oisson s ratio v for the pipe material is approximately: () 0.7 () 0.30 (C) 0.31 (D) 0.34 E 110 Ga d 38 mm t 1 mm d C 70 mm t C 10 mm 1 10 kn 100 kn C p 4 [d C (d C t C ) ] 1885 mm C t C d C Cap plate t d 1 xial strain of C: C ( 1 ) E C xial stress in C: C E C Ma (well below yield stress of 550 Ma so Hooke s aw applies) ateral strain of C: t C mm t C t C oisson s ratio: v C 0.34 confirms value for brass given in properties table 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

10 ENDIX FE Exam Review roblems 1091 R-1.1: titanium bar (E 100 Ga, v 0.33) with square cross section (b 75 mm) and length 3.0 m is subjected to tensile load 900 kn. The increase in volume of the bar is approximately: () 1400 mm 3 () 3500 mm 3 (C) 4800 mm 3 (D) 900 mm 3 E 100 Ga b 75 mm 3.0 m 900 kn v 0.33 b b Initial volume of bar: V init b mm 3 Normal strain in bar: ateral strain in bar: Final length of bar: E b v f mm Final lateral dimension of bar: b f b b mm Final volume of bar: V final b f f mm 3 Increase in volume of bar: V V final V init 9156 mm 3 V V init R-1.13: n elastomeric bearing pad is subjected to a shear force V during a static loading test. The pad has dimensions a 150 mm and b 5 mm, and thickness t 55 mm. The lateral displacement of the top plate with respect to the bottom plate is 14 mm under a load V 16 kn. The shear modulus of elasticity G of the elastomer is approximately: () 1.0 Ma () 1.5 Ma (C) 1.7 Ma (D) 1.9 Ma V 16 kn a 150 mm b 5 mm d 14 mm t 55 mm 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

11 109 ENDIX FE Exam Review roblems ve. shear stress: b t V Ma a b ve. shear strain: a V g arctana d t b 0.49 t Shear modulus of elastomer: G t 1.90 Ma g R-1.14: bar of diameter d 18 mm and length 0.75 m is loaded in tension by forces. The bar has modulus E 45 Ga and allowable normal stress of 180 Ma. The elongation of the bar must not exceed.7 mm. The allowable value of forces is approximately: () 41 kn () 46 kn (C) 56 kn (D) 63 kn d 18 mm 0.75 m E 45 Ga s a 180 Ma d a.7 mm d (1) allowable value of based on elongation a d a a1 s max a p 4 d b 41. kn s max E a 16.0 Ma elongation governs () allowable load based on tensile stress a s a a p 4 d b 45.8 kn R-1.15: Two flanged shafts are connected by eight 18 mm bolts. The diameter of the bolt circle is 40 mm. The allowable shear stress in the bolts is 90 Ma. Ignore friction between the flange plates. The maximum value of torque T 0 is approximately: () 19 kn m () kn m (C) 9 kn m (D) 37 kn m 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

12 ENDIX FE Exam Review roblems 1093 d b 18 mm d 40 mm t a 90 Ma n 8 olt shear area: s p d b 54.5 mm 4 Max. torque: T max n (t a s ) d.0 kn m T 0 T 0 R-1.16: copper tube with wall thickness of 8 mm must carry an axial tensile force of 175 kn. The allowable tensile stress is 90 Ma. The minimum required outer diameter is approximately: () 60 mm () 7 mm (C) 85 mm (D) 93 mm t 8mm 175 kn s a 90 Ma d Required area based on allowable stress: reqd s a 1944 mm rea of tube of thickness t but unknown outer diameter d: p 4 [d (d t) ] t(d t) Solving for d min : d min s a p t t 85.4 mm so d inner d min t 69.4 mm 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

13 1094 ENDIX FE Exam Review roblems R-.1: Two wires, one copper and the other steel, of equal length stretch the same amount under an applied load. The moduli of elasticity for each is: E s 10 Ga, E c 10 Ga. The ratio of the diameter of the copper wire to that of the steel wire is approximately: () 1.00 () 1.08 (C) 1.19 (D) 1.3 E s 10 Ga E c 10 Ga Copper wire Displacements are equal: d s d c or E s s E c c so E s s E c c Steel wire and Express areas in terms of wire diameters then find ratio: p d c 4 a p d s 4 b E s E c c s E s E c so d c d s E s E c 1.33 R-.: plane truss with span length 4.5 m is constructed using cast iron pipes (E 170 Ga) with cross sectional area of 4500 mm. The displacement of joint cannot exceed.7 mm. The maximum value of loads is approximately: () 340 kn () 460 kn (C) 510 kn (D) 600 kn 4.5 m E 170 Ga 4500 mm d max.7 mm Statics: sum moments about to find reaction at R R 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

14 ENDIX FE Exam Review roblems 1095 C Method of Joints at : F (tension) Force-displ. relation: max E d max 459 kn Check normal stress in bar : s max 10.0 Ma well below yield stress of 90 Ma in tension R-.3: brass rod (E 110 Ga) with cross sectional area of 50 mm is loaded by forces 1 15 kn, 10 kn, and 3 8 kn. Segment lengths of the bar are a.0 m, b 0.75 m, and c 1. m. The change in length of the bar is approximately: () 0.9 mm () 1.6 mm (C).1 mm (D) 3.4 mm E 110 Ga 50 mm a m b 0.75 m 1 3 C D c 1. m a b c 1 15 kn 10 kn 3 8kN Segment forces (tension is positive): N kn N C 3.00 kn N CD kn 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

15 1096 ENDIX FE Exam Review roblems Change in length: d D 1 E (N a N C b N CD c) 0.94 mm d D a b c Check max. stress: positive so elongation N 68.0 Ma well below yield stress for brass so OK R-.4: brass bar (E 110 Ma) of length.5 m has diameter d 1 18 mm over one-half of its length and diameter d 1 mm over the other half. Compare this nonprismatic bar to a prismatic bar of the same volume of material with constant diameter d and length. The elongation of the prismatic bar under the same load 5 kn is approximately: () 3 mm () 4 mm (C) 5 mm (D) 6 mm.5 m 5 kn d 1 18 mm d 1 mm E 110 Ga 1 p 4 d mm d 1 d / / p 4 d mm Volume of nonprismatic bar: Vol nonprismatic ( 1 ) mm3 Diameter of prismatic bar of same volume: prismatic p 4 d 184 mm d H Vol nonprismatic mm p 4 V prismatic prismatic mm 3 Elongation of prismatic bar: d E prismatic 3.09 mm less than d for nonprismatic bar 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

16 ENDIX FE Exam Review roblems 1097 Elongation of nonprismatic bar shown in fig. above: E a b 3.63 mm R-.5: nonprismatic cantilever bar has an internal cylindrical hole of diameter d/ from 0 to x, so the net area of the cross section for Segment 1 is (3/4). oad is applied at x, and load -/ is applied at x. ssume that E is constant. The length of the hollow segment, x, required to obtain axial displacement d /E at the free end is: () x /5 () x /4 (C) x /3 (D) x 3/5 Forces in Segments 1 & : N 1 3 N Displacement at free end: d Segment 1 Segment d x x d 3 N 1 x E a 3 4 b N ( x) E 3 x d 3 E a 3 4 b ( x) ( 5 x) E E Set d 3 equal to /E and solve for x ( 5 x) E E ( 5 x) E So x 3/5 or (3 5 x) 0 simplify S 0 E E 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

17 1098 ENDIX FE Exam Review roblems R-.6: nylon bar (E.1 Ga) with diameter 1 mm, length 4.5 m, and weight 5.6 N hangs vertically under its own weight. The elongation of the bar at its free end is approximately: () 0.05 mm () 0.07 mm (C) 0.11 mm (D) 0.17 mm E.1 Ga 4.5 m d 1 mm p d mm 4 kn g 11 m 3 W N d W E so or d g E d (g ) E mm Check max. normal stress at top of bar s max W Ma ok - well below ult. stress for nylon R-.7: monel shell (E m 170 Ga, d 3 1 mm, d 8 mm) encloses a brass core (E b 96 Ga, d 1 6 mm). Initially, both shell and core are of length 100 mm. load is applied to both shell and core through a cap plate. The load required to compress both shell and core by 0.10 mm is approximately: () 10. kn () 13.4 kn (C) 18.5 kn (D) 1.0 kn E m 170 Ga d 1 6mm d 3 1 mm E b 96 Ga d 8mm 100 mm 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

18 ENDIX FE Exam Review roblems 1099 Monel shell rass core d 1 d d 3 m p 4 (d 3 d ) 6.83 mm b p 4 d mm Compatibility: m b E m m E b b d m d b m E m m E b b b Statics: m b so b a1 E m m E b b b Set d equal to 0.10 mm and solve for load : d b b E b b so b E b b with d b 0.10 mm d b and then E b b d b a1 E m m b kn E b b R-.8: steel rod (E s 10 Ga, d r 1 mm, a s > C ) is held stress free between rigid walls by a clevis and pin (d p 15 mm) assembly at each end. If the allowable shear stress in the pin is 45 Ma and the allowable normal stress in the rod is 70 Ma, the maximum permissible temperature drop T is approximately: () 14 C () 0 C (C) 8 C (D) 40 C 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

19 1100 ENDIX FE Exam Review roblems E s 10 Ga d r 1 mm d p 15 mm r p 4 d r mm p p 4 d p mm s 1(10 6 ) > C t a 45 Ma s a 70 Ma Force in rod due to temperature drop T: pin, d p ΔT rod, d r Clevis and normal stress in rod: F r E s r ( s ) T So T max associated with normal stress in rod s r F r r T maxrod s a E s a s 7.8 degrees Celsius (decrease) Controls Now check T based on shear stress in pin (in double shear): T maxpin t a ( p ) E s r a s 55.8 t pin F r p R-.9: threaded steel rod (E s 10 Ga, d r 15 mm, s > C ) is held stress free between rigid walls by a nut and washer (d w mm) assembly at each end. If the allowable bearing stress between the washer and wall is 55 Ma and the allowable normal stress in the rod is 90 Ma, the maximum permissible temperature drop T is approximately: () 5 C () 30 C (C) 38 C (D) 46 C E s 10 Ga d r 15 mm d w mm r p 4 d r mm w p 4 (d w d r ) 03.4 mm s 1(10 6 ) > C s ba 55 Ma s a 90 Ma 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

20 ENDIX FE Exam Review roblems 1101 ΔT rod, d r washer, dw Force in rod due to temperature drop T: and normal stress in rod: F r E s r ( s ) T So T max associated with normal stress in rod s r F r r T maxrod s a E s a s 35.7 degrees Celsius (decrease) Now check T based on bearing stress beneath washer: T maxwasher s ba ( w ) E s r a s 5.1 degrees Celsius (decrease) Controls s b F r w R-.10: steel bolt (area 130 mm, E s 10 Ga) is enclosed by a copper tube (length 0.5 m, area 400 mm, E c 110 Ga) and the end nut is turned until it is just snug. The pitch of the bolt threads is 1.5 mm. The bolt is now tightened by a quarter turn of the nut. The resulting stress in the bolt is approximately: () 56 Ma () 6 Ma (C) 74 Ma (D) 81 Ma E s 10 Ga E c 110 Ga 0.5 m c 400 mm s 130 mm n 0.5 p 1.5 mm Copper tube Compatibility: shortening of tube and elongation of bolt applied displacement of n p c E c c Statics: s E s s n p c s Steel bolt Solve for s s E c c s E s s n p or n p s a kn b E c c E s s 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

21 110 ENDIX FE Exam Review roblems Stress in steel bolt: s s s s 81.0 Ma Stress in copper tube: s c s c 6.3 Ma tension compression R-.11: steel bar of rectangular cross section (a 38 mm, b 50 mm) carries a tensile load. The allowable stresses in tension and shear are 100 Ma and 48 Ma respectively. The maximum permissible load max is approximately: () 56 kn () 6 kn (C) 74 kn (D) 91 kn a 38 mm b 50 mm ab 1900 mm s a 100 Ma t a 48 Ma b ar is in uniaxial tension so T max s max /; since t a s a, shear stress governs max t a 91. kn a R-.1: brass wire (d.0 mm, E 110 Ga) is pretensioned to T 85 N. The coefficient of thermal expansion for the wire is > C. The temperature change at which the wire goes slack is approximately: () 5.7 C () 1.6 C (C) 1.6 C (D) 18. C E 110 Ga d.0 mm b 19.5 (10 6 ) > C T 85 N T d T p 4 d 3.14 mm 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

22 ENDIX FE Exam Review roblems 1103 Normal tensile stress in wire due to pretension T and temperature increase T: s T E a b T Wire goes slack when normal stress goes to zero; solve for T T T E a b 1.61 degrees Celsius (increase in temperature) R-.13: copper bar (d 10 mm, E 110 Ga) is loaded by tensile load 11.5 kn. The maximum shear stress in the bar is approximately: () 73 Ma () 87 Ma (C) 145 Ma (D) 150 Ma E 110 Ga d 10 mm p 4 d mm 11.5 kn d Normal stress in bar: s p Ma For bar in uniaxial stress, max. shear stress is on a plane at 45 deg. to axis of bar and equals 1/ of normal stress: t max s 73. Ma R-.14: steel plane truss is loaded at and C by forces 00 kn. The cross sectional area of each member is 3970 mm. Truss dimensions are H 3 m and 4 m. The maximum shear stress in bar is approximately: () 7 Ma () 33 Ma (C) 50 Ma (D) 69 Ma 00 kn 3970 mm H 3m 4m Statics: sum moments about to find vertical reaction at vert H kn (downward) 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

23 1104 ENDIX FE Exam Review roblems C H Method of Joints at : C vert vert C horiz H C vert 00.0 kn So bar force in is: C horiz kn (compression) Max. normal stress in : s Ma Max. shear stress is 1/ of max. normal stress for bar in uniaxial stress and is on plane at 45 deg. to axis of bar: t max s 50.4 Ma R-.15: plane stress element on a bar in uniaxial stress has tensile stress of s 78 Ma (see fig.). The maximum shear stress in the bar is approximately: () 9 Ma () 37 Ma (C) 50 Ma (D) 59 Ma u 78 Ma lane stress transformation formulas for uniaxial stress: σ θ / τ θ τ θ σ θ θ s x s u cos(u) on element face at angle and s x s u sin(u) on element face at angle 90 τ θ τ θ 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

24 ENDIX FE Exam Review roblems 1105 Equate above formulas and solve for s x tan(u) 1 so u atana 1 1 b s x s u Ma cos(u) also u s x sin(u)cos(u) Ma Max. shear stress is 1/ of max. normal stress for bar in uniaxial stress and is on plane at 45 deg. to axis of bar: t max s x 58.5 Ma R-.16: prismatic bar (diameter d 0 18 mm) is loaded by force 1. stepped bar (diameters d 1 0 mm, d 5 mm, with radius R of fillets mm) is loaded by force. The allowable axial stress in the material is 75 Ma. The ratio 1 / of the maximum permissible loads that can be applied to the bars, considering stress concentration effects in the stepped bar, is: () 0.9 () 1. (C) 1.4 (D).1 1 d 1 d d 1 d 0 1 FIG. -66 Stress-concentration factor K for round bars with shoulder fillets. The dashed line is for a full quater-circular fillet D = D D R D 1 K K = smax snom s nom = p D 1 /4 D D R = R D Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

25 1106 ENDIX FE Exam Review roblems rismatic bar 1 max s allow a p d 0 (18 mm) b (75 Ma) cp d 19.1 kn 4 4 Stepped bar from stress conc. Fig. -66 R mm d 1 0 mm d 5 mm 1.50 so K 1.75 d 1 0 mm D = D D R D 1 K K = smax snom s nom = p D 1 /4 K = 1.75 D D R = R D 1 max s allow K ap d 1 Ma mm) b a75 b cp(0 d 13.5 kn 4 K 4 1 max 19.1 kn max 13.5 kn 1.41 R-3.1: brass rod of length 0.75 m is twisted by torques T until the angle of rotation between the ends of the rod is 3.5. The allowable shear strain in the copper is rad. The maximum permissible diameter of the rod is approximately: () 6.5 mm () 8.6 mm (C) 9.7 mm (D) 1.3 mm 0.75 m f 3.5 g a Max. shear strain: T d T a d g fb max so d max g a f 1.8 mm 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

26 ENDIX FE Exam Review roblems 1107 R-3.: The angle of rotation between the ends of a nylon bar is 3.5. The bar diameter is 70 mm and the allowable shear strain is rad. The minimum permissible length of the bar is approximately: () 0.15 m () 0.7 m (C) 0.40 m (D) 0.55 m d 70 mm f 3.5 T d T g a Max. shear strain: g r f so min d f 0.15 m g a R-3.3: brass bar twisted by torques T acting at the ends has the following properties:.1 m, d 38 mm, and G 41 Ga. The torsional stiffness of the bar is approximately: () 100 N m () 600 N m (C) 4000 N m (D) 4800 N m G 41 Ga.1 m T d T d 38 mm olar moment of inertia, I p : I p p 3 d mm 4 Torsional stiffness, k T : k T G I p 3997 N m R-3.4: brass pipe is twisted by torques T 800 N m acting at the ends causing an angle of twist of 3.5 degrees. The pipe has the following properties:.1 m, d 1 38 mm, and d 56 mm. The shear modulus of elasticity G of the pipe is approximately: () 36.1 Ga () 37.3 Ga (C) 38.7 Ga (D) 40.6 Ga 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

27 1108 ENDIX FE Exam Review roblems T T d 1 d.1 m d 1 38 mm d 56 mm f 3.5 T 800 N m olar moment of inertia: I p p 3 (d 4 d 1 4 ) mm 4 Solving torque-displacement relation for shear modulus G: G T f I p 36.1 Ga R-3.5: n aluminum bar of diameter d 5 mm is twisted by torques T 1 at the ends. The allowable shear stress is 65 Ma. The maximum permissible torque T 1 is approximately: () 1450 N m () 1675 N m (C) 1710 N m (D) 1800 N m d 5 mm T 1 d T 1 t a 65 Ma I p p 3 d mm 4 From shear formula: T t a I p 1 max a d 1795 N m b R-3.6: steel tube with diameters d 86 mm and d 1 5 mm is twisted by torques at the ends. The diameter of a solid steel shaft that resists the same torque at the same maximum shear stress is approximately: () 56 mm () 6 mm (C) 75 mm (D) 8 mm 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

28 ENDIX FE Exam Review roblems 1109 d 86 mm d 1 5 mm I pipe p 3 (d 4 d 1 4 ) mm 4 Shear formula for hollow pipe: t max T a d b I pipe Shear formula for solid shaft: T a d b 16 T t max p p d 3 3 d4 Equate and solve for d of solid shaft: d D 16 T p T a d b d 1 d T 1 3 d I pipe d a 3 1 I pipe b3 8.0 mm p d R-3.7: stepped steel shaft with diameters d 1 56 mm and d 5 mm is twisted by torques T kn m and T 1.5 kn m acting in opposite directions. The maximum shear stress is approximately: () 54 Ma () 58 Ma (C) 6 Ma (D) 79 Ma d 1 56 mm d 5 mm T kn m T 1.5 kn m T 1 d T 1 d C Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

29 1110 ENDIX FE Exam Review roblems olar moments of inertia: I p1 p 3 d mm 4 I p p 3 d mm 4 Shear formula - max. shear stresses in segments 1 & : t max1 d 1 (T 1 T ) 58.0 Ma I p1 t max T a d b I p 54.3 Ma R-3.8: stepped steel shaft (G 75 Ga) with diameters d 1 36 mm and d 3 mm is twisted by torques T at each end. Segment lengths are m and 0.75 m. If the allowable shear stress is 8 Ma and maximum allowable twist is 1.8 degrees, the maximum permissible torque is approximately: () 14 N m () 180 N m (C) 185 N m (D) 57 N m d 1 36 mm d 3 mm T d 1 d T G 75 Ga C t a 8 Ma m 0.75 m f a 1.8 olar moments of inertia: I p1 p 3 d mm 4 I p p 3 d mm 4 Max torque based on allowable shear stress - use shear formula: t max1 d 1 T I p1 t max T a d b I p T max1 t a a I p1 d 1 b 57 N m T max t a a I p d b 180 N m controls 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

30 ENDIX FE Exam Review roblems 1111 Max. torque based on max. rotation & torque-displacement relation: f T G a 1 b I p1 I p G f a T max a N m b I p1 I p R-3.9: gear shaft transmits torques T 975 N m, T 1500 N m, T C 650 N m and T D 85 N m. If the allowable shear stress is 50 Ma, the required shaft diameter is approximately: () 38 mm () 44 mm (C) 46 mm (D) 48 mm t a 50 Ma T T 975 N m T 1500 N m T C 650 N m T D 85 N m Find torque in each segment of shaft: T T N m T CD T D 85.0 N m T T C C T C T T 55.0 N m D T D Shear formula: T a d b 16 T t p p d 3 3 d4 Set t to t allowable and T to torque in each segment; solve for required diameter d (largest controls) 1 Segment : d a 16 T 3 b 46.3 mm p t a Segment C: d a 16 T C p t a 1 3 b 37.7 mm Segment CD: d a 16 T CD p t a 1 3 b 43.8 mm 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

31 111 ENDIX FE Exam Review roblems R-3.10: hollow aluminum shaft (G 7 Ga, d 96 mm, d 1 5 mm) has an angle of twist per unit length of 1.8 /m due to torques T. The resulting maximum tensile stress in the shaft is approximately: () 38 Ma () 41 Ma (C) 49 Ma (D) 58 Ma G 7 Ga T d T d 96 mm d 1 5 mm u 1.8 >m Max. shear strain due to twist per unit length: d 1 d g max a d b u radians Max. shear stress: t max Gg max 40.7 Ma Max. tensile stress on plane at 45 degrees & equal to max. shear stress: s max t max 40.7 Ma R-3.11: Torques T 5.7 kn m are applied to a hollow aluminum shaft (G 7 Ga, d 1 5 mm). The allowable shear stress is 45 Ma and the allowable normal strain is The required outside diameter d of the shaft is approximately: () 38 mm () 56 mm (C) 87 mm (D) 91 mm T 5.7 kn m G 7 Ga d 1 5 mm t a1 45 Ma a 8.0(10 4 ) d 1 d llowable shear strain based on allowable normal strain for pure shear g a a so resulting allow. shear stress is: t a Gg a 43. Ma 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

32 ENDIX FE Exam Review roblems 1113 So allowable shear stress based on normal strain governs t a t a Use torsion formula to relate required d to allowable shear stress: t max T a d b p 3 (d 4 d 1 4 ) and rearrange equation to get Solve resulting 4th order equation numerically, or use a calculator and trial & error T N mm d 1 5 mm t a 43. Ma d 4 d p T d t a f(d ) d 4 a 16 p T 4 b d t d 1 a gives d 91 mm R-3.1: motor drives a shaft with diameter d 46 mm at f 5.5 Hz and delivers 5 kw of power. The maximum shear stress in the shaft is approximately: () 3 Ma () 40 Ma (C) 83 Ma (D) 91 Ma f 5.5 Hz 5 kw d 46 mm I p p 3 d mm 4 ower in terms of torque T: pf T Solve for torque T: T N m p f Max. shear stress using torsion formula: T a d b t max 39.7 Ma I p d f R-3.13: motor drives a shaft at f 10 Hz and delivers 35 kw of power. The allowable shear stress in the shaft is 45 Ma. The minimum diameter of the shaft is approximately: () 35 mm () 40 mm (C) 47 mm (D) 61 mm 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

33 1114 ENDIX FE Exam Review roblems f 10 Hz t a 45 Ma 35 kw d f ower in terms of torque T: pf T Solve for torque T: T N m p f Shear formula: T a d b t p 3 d 4 or t 16 T p d 3 Solve for diameter d: 1 d a 16 T 3 b 39.8 mm p t a R-3.14: drive shaft running at 500 rpm has outer diameter 60 mm and inner diameter 40 mm. The allowable shear stress in the shaft is 35 Ma. The maximum power that can be transmitted is approximately: () 0 kw () 40 kw (C) 88 kw (D) 31 kw n 500 rpm t a 35 (10 6 ) d m d m I p p 3 (d 4 d 1 4 ) m 4 Shear formula: N m d n d 1 d t T a d b I p or ower in terms of torque T: pf T p(n/60) T max p n 60 T max W T max t a I p d N m max 31 kw 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

34 ENDIX FE Exam Review roblems 1115 R-3.15: prismatic shaft (diameter d 0 19 mm ) is loaded by torque T 1. stepped shaft (diameters d 1 0 mm, d 5 mm, radius R of fillets mm) is loaded by torque T. The allowable shear stress in the material is 4 Ma. The ratio T 1 /T of the maximum permissible torques that can be applied to the shafts, considering stress concentration effects in the stepped shaft is: () 0.9 () 1. (C) 1.4 (D).1 T 1 d 0 T 1 D R D 1 T T FIG Stress-concentration factor K for a stepped shaft in torsion. (The dashed line is for a full quarter-circular fillet.) K T R D D 1 T 1.50 = + R D 1 D 1.5 t max = Kt nom t nom = 16T pd3 1 D D 1 = R D Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

35 1116 ENDIX FE Exam Review roblems rismatic shaft T 1max t allow I d 0 Stepped shaft t allow a p d b 4 Ma c p 16 (19 mm)3 d 56.6 N m d 5 mm d 1 0 mm 1.50 R mm d 1 0 mm K 1.35 so from graph (see Fig. 3-59) T max t allow K ap d b 4 Ma c p (0 mm)3 d 48.9 N m T 1 max T max R-4.1: simply supported beam with proportional loading ( 4.1 kn) has span length 5 m. oad is 1. m from support and load is 1.5 m from support. The bending moment just left of load is approximately: () 5.7 kn m () 6. kn m (C) 9.1 kn m (D) 10.1 kn m a 1. m b.3 m c 1.5 m a b c 5.00 m 4.1 kn Statics to find reaction force at : a b c R 1 [ a (a b)] 6.74 kn Moment just left of load : M R c 10.1 kn m compression on top of beam R-4.: simply-supported beam is loaded as shown in the figure. The bending moment at point C is approximately: () 5.7 kn m () 6.1 kn m (C) 6.8 kn m (D) 9.7 kn m 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

36 ENDIX FE Exam Review roblems kn C 1.8 kn/m 0.5 m 1.0 m 1.0 m 3.0 m 5.0 m Statics to find reaction force at : R 1 5 m cc1.8 kn (3 m 0.5 m) d 7.5 kn (3 m 1 m)d 7.15 kn m Moment at point C, m from : M R ( m) 7.5 kn(1.0 m) 6.75 kn m compression on top of beamr R-4.3: cantilever beam is loaded as shown in the figure. The bending moment at 0.5 m from the support is approximately: () 1.7 kn m () 14. kn m (C) 16.1 kn m (D) 18.5 kn m 4.5 kn 1.8 kn/m 1.0 m 1.0 m 3.0 m Cut beam at 0.5 m from support; use statics and right-hand FD to find internal moment at that point M 0.5 m (4.5 kn) a0.5 m 1.0 m 3.0 m b 1.8 kn (3.0 m) m 18.5 kn m (tension on top of beam) R-4.4: n -shaped beam is loaded as shown in the figure. The bending moment at the midpoint of span is approximately: () 6.8 kn m () 10.1 kn m (C) 1.3 kn m (D) 15.5 kn m 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

37 1118 ENDIX FE Exam Review roblems 4.5 kn 9 kn 1.0 m C 5.0 m 1.0 m Use statics to find reaction at ; sum moments about R 1 [9 kn (6 m) 4.5 kn (1. m)] 9.90 kn 5 m Cut beam at midpoint of ; use right hand FD, sum moments M R a 5 m b 9 kn a5 m 1 mb 6.75 kn m tension on top of beam R-4.5: T-shaped simple beam has a cable with force anchored at and passing over a pulley at E as shown in the figure. The bending moment just left of C is 1.5 kn m. The cable force is approximately: ().7 kn () 3.9 kn (C) 4.5 kn (D) 6. kn M C 1.5 kn m Sum moments about D to find vertical reaction at : V 1 [ (4 m)] 7 m V 4 7 (downward) Now cut beam & cable just left of CE & use left FD; show V downward & show vertical cable force component m 3 m m of (4/5) upward at ; sum moments at C to get M C and equate to given numerical value of M C to find : Cable C E D 4 m M C 4 5 (3) V ( 3) M C 4 (3) a 4b ( 3) Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

38 ENDIX FE Exam Review roblems 1119 Solve for : 35 (1.5).73 kn 16 R-4.6: simple beam ( 9 m) with attached bracket DE has force 5 kn applied downward at E. The bending moment just right of is approximately: () 6 kn m () 10 kn m (C) 19 kn m (D) kn m Sum moments about to find reaction at C: C R C 1 c a 6 3 bd D E Cut through beam just right of, then use FD of C to find moment at : M R C a 3 b Substitute numbers for and : 9m M kn 18.8 kn m R-4.7: simple beam with an overhang C is loaded as shown in the figure. The bending moment at the midspan of is approximately: () 8 kn m () 1 kn m (C) 17 kn m (D) 1 kn m 15 kn/m 4.5 kn m C 1.6 m 1.6 m 1.6 m 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

39 110 ENDIX FE Exam Review roblems Sum moments about to get reaction at : R c15 s1.6) a1.6 b 4.5d kn 3. Cut beam at midspan, use left FD & sum moments to find moment at midspan: M mspan R s1.6) 15 s1.6) a 1.6 b kn m R-5.1: copper wire (d 1.5 mm) is bent around a tube of radius R 0.6 m. The maximum normal strain in the wire is approximately: () () (C) (D) d d max R d ar d b d 1.5 mm R 0.6 m d max ar d b R d R-5.: simply supported wood beam ( 5 m) with rectangular cross section (b 00 mm, h 80 mm) carries uniform load q 6.5 kn/m which includes the weight of the beam. The maximum flexural stress is approximately: () 8.7 Ma () 10.1 Ma (C) 11.4 Ma (D) 14.3 Ma 5m b 00 mm h 80 mm q 9.5 kn m 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

40 ENDIX FE Exam Review roblems 111 Section modulus: q S b h m 3 h Max. moment at midspan: M max q kn m Max. flexural stress at midspan: b s max M max S 11.4 Ma R-5.3: cast iron pipe ( 1 m, weight density 7 kn/m 3, d 100 mm, d 1 75 mm) is lifted by a hoist. The lift points are 6 m apart. The maximum bending stress in the pipe is approximately: () 8 Ma () 33 Ma (C) 47 Ma (D) 59 Ma s d 1 d 1 m s 4m d 100 mm d 1 75 mm g CI 7 kn m 3 ipe cross sectional properties: p 4 sd d 1 ) 3436 mm I p 64 sd 4 d 1 4 ) mm 4 Uniformly distributed weight of pipe, q: q g CI 0.47 kn m Vertical force at each lift point: F q kn 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

41 11 ENDIX FE Exam Review roblems Max. moment is either at lift points (M 1 ) or at midspan (M ): M 1 q a s b a s b kn m 4 s M F q a b kn m 4 controls, tension on top tension on top Max. bending stress at lift point: u M 1 u a d b s max 9.5 Ma I R-5.4: beam with an overhang is loaded by a uniform load of 3 kn/m over its entire length. Moment of inertia I z mm 4 and distances to top and bottom of the beam cross section are 0 mm and 66.4 mm, respectively. It is known that reactions at and are 4.5 kn and 13.5 kn, respectively. The maximum bending stress in the beam is approximately: () 36 Ma () 67 Ma (C) 10 Ma (D) 119 Ma 3 kn/m 4 m m C z C y 0 mm 66.4 mm R 4.5 kn I z 3.36 (10 6 )mm 4 q 3 kn m ocation of max. positive moment in (cut beam at location of zero shear & use left FD): x max R q 1.5 m M pos R x max 3 kn m x max kn m compression on top of beam Compressive stress on top of beam at x max : s c1 M pos (0 mm) 0.1 Ma I z Tensile stress at bottom of beam at x max : s t1 M pos (66.4 mm) Ma I z 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

42 ENDIX FE Exam Review roblems 113 Max. negative moment at (use FD of C to find moment; compression on bottom of beam): M neg a3 kn m b ( m) kn m s c M neg (66.4 mm) Ma I z s t M neg (0 mm) 35.7 Ma I z R-5.5: steel hanger with solid cross section has horizontal force 5.5 kn applied at free end D. Dimension variable b 175 mm and allowable normal stress is 150 Ma. Neglect self weight of the hanger. The required diameter of the hanger is approximately: () 5 cm () 7 cm (C) 10 cm (D) 13 cm 5.5 kn b 175 mm a 150 Ma 6b Reactions at support: N (leftward) M (b) 1.9 kn m D C b (tension on bottom) b Max. normal stress at bottom of cross section at : s max a p d 4 b ( b) a d b a p d 4 64 b s max 4 (16 b d) p d 3 Set s max s a and solve for required diameter d: ( s a )d 3 (4)d 64b 0 solve numerically or by trial & error to find d reqd 5.11 cm 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

43 114 ENDIX FE Exam Review roblems R-5.6: cantilever wood pole carries force 300 N applied at its free end, as well as its own weight (weight density 6 kn/m 3 ). The length of the pole is 0.75 m and the allowable bending stress is 14 Ma. The required diameter of the pole is approximately: () 4. cm () 5.5 cm (C) 6.1 cm (D) 8.5 cm 300 N s a 14 Ma 0.75 m g w 6 kn m 3 Uniformly distributed weight of pole: w g w a p d 4 b Max. moment at support: d M max w Section modulus of pole cross section: S I a d b S p d 4 64 p d 3 a d b 3 Set M max equal to s a S and solve for required min. diameter d: cg w a p d 4 bd s aa p d3 3 b 0 Or a p s a 3 b d3 a p g w b d 0 8 solve numerically or by trial & error to find d reqd 5.50 cm Since wood pole is light, try simpler solution which ignores self weight: s a S Or a p s a 3 b d3 d reqd c a 3 3 bd 5.47 cm p s a Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

44 ENDIX FE Exam Review roblems 115 R-5.7: simply supported steel beam of length 1.5 m and rectangular cross section (h 75 mm, b 0 mm) carries a uniform load of q 48 kn/m, which includes its own weight. The maximum transverse shear stress on the cross section at 0.5 m from the left support is approximately: () 0 Ma () 4 Ma (C) 30 Ma (D) 36 Ma 1.5 m q 48 kn m h 75 mm Cross section properties: b 0 mm q bh 1500 mm h Q ab h b h 1406 mm3 4 I b h mm 4 Support reactions: R q 36.0 kn Transverse shear force at 0.5 m from left support: V 0.5 R q (0.5 m) 4.0 kn b Max. shear stress at N at 0.5 m from left support: t max V 0.5 Q I b 4.0 Ma Or more simply... t max 3 V Ma R-5.8: simply supported laminated beam of length 0.5 m and square cross section weighs 4.8 N. Three strips are glued together to form the beam, with the allowable shear stress in the glued joint equal to 0.3 Ma. Considering also the weight of the beam, the maximum load that can be applied at /3 from the left support is approximately: () 40 N () 360 N (C) 434 N (D) 510 N 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

45 116 ENDIX FE Exam Review roblems q at /3 1 mm 1 mm 1 mm 36 mm 36 mm 0.5 m W 4.8 N q W 9.60 N m h 36 mm b 36 mm t a 0.3 Ma Cross section properties: bh 196 mm I b h mm 4 h Q joint ab 3 b ah h b 5184 mm3 6 Max. shear force at left support: V max q a 3 b Shear stress on glued joint at left support; set t t a then solve for max : t V max Q joint I b Or t V max a b h 9 b a b h3 1 b b Or t a 4 V max 3 b h t a 4 3 b h c q a 3 bd so for t a 0.3 Ma max 3 a3 b h t a 4 q b 434 N R-5.9: n aluminum cantilever beam of length 0.65 m carries a distributed load, which includes its own weight, of intensity q/ at and q at. The beam cross section has width 50 mm and height 170 mm. llowable bending stress is 95 Ma and allowable shear stress is 1 Ma. The permissible value of load intensity q is approximately: () 110 kn/m () 1 kn/m (C) 130 kn/m (D) 139 kn/m 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

46 ENDIX FE Exam Review roblems m b 50 mm h 170 mm s a 95 Ma t a 1 Ma q q Cross section properties: bh 8500 mm I b h mm 4 Reaction force and moment at : R 1 aq qb M 5 1 q Compare max. permissible values of q based on shear and moment allowable stresses; smaller value controls t max 3 R R 3 4 q 3 t a 3 ± 4 q S b h mm 3 M q 1 q 3 So, since t a 1 Ma q max1 8 t a kn m s max M S q max 1 5 s a S s a 5 1 q S kn m So, since s a 95 Ma R-5.10: n aluminum light pole weighs 4300 N and supports an arm of weight 700 N, with arm center of gravity at 1. m left of the centroidal axis of the pole. wind force of 1500 N acts to the right at 7.5 m above the base. The pole cross section at the base has outside diameter 35 mm and thickness 0 mm. The maximum compressive stress at the base is approximately: () 16 Ma () 18 Ma (C) 1 Ma (D) 4 Ma 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

47 118 ENDIX FE Exam Review roblems H 7.5 m 1. m W = 700 N W N W 700 N N d 35 mm t 0 mm d 1 d t 195 mm ole cross sectional properties at base: p 4 (d d 1 ) mm I p 64 (d 4 d 4 1 ) mm 4 1. m 7.5 m x 1 = 1500 N W 1 = 4300 N 0 mm z y x y 35 mm Compressive (downward) force at base of pole: N W 1 W 5.0 kn ending moment at base of pole: M W 1 H kn m Compressive stress at right side at base of pole: d s c N M a b I 15.9 Ma results in compression at right R-5.11: Two thin cables, each having diameter d t/6 and carrying tensile loads, are bolted to the top of a rectangular steel block with cross section dimensions b t. The ratio of the maximum tensile to compressive stress in the block due to loads is: () 1.5 () 1.8 (C).0 (D).5 Cross section properties of block: bt I b t 3 1 d t 6 t b Tensile stress at top of block: d s t a t b at b 9 I b t 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

48 ENDIX FE Exam Review roblems 119 Compressive stress at bottom of block: d s c a t b at b I 5 b t Ratio of max. tensile to compressive stress in block: ratio ` s t s c ` R-5.1: rectangular beam with semicircular notches has dimensions h 160 mm and h mm. The maximum allowable bending stress in the plastic beam is s max 6.5 Ma, and the bending moment is M 185 N m. The minimum permissible width of the beam is: () 1 mm () 0 mm (C) 8 mm (D) 3 mm R M M h h 1 K h h = 1. h = h 1 + R M h R h 1 K = s max s s = 6M nom nom bh 1 b = thickness M FIG Stress-concentration factor K for a notched beam of rectangular cross section in pure bending ( h height of beam; b thickness of beam, perpendicular to the plane of the figure). The dashed line is for semicircular notches ( h h 1 R) R h1 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

49 1130 ENDIX FE Exam Review roblems R 10 h h 160 h From Fig 5-50: s allow K R 1 (h h 1) 1 (160 mm 140 mm) mm K.5 6 M b h so b min 6 M K 1 s allow h 6 (185 N m) (.5) Ma C(140 mm) D 19.6 mm 3.0 K.5 K =.5.0 h h = 1. h = h 1 + R M h R h 1 K = s max s s = 6M nom nom bh 1 b = thickness M R h1 R-6.1: composite beam is made up of a 00 mm 300 mm core (E c 14 Ga) and an exterior cover sheet (300 mm 1 mm, E e 100 Ga) on each side. llowable stresses in core and exterior sheets are 9.5 Ma and 140 Ma, respectively. The ratio of the maximum permissible bending moment about the z-axis to that about the y-axis is most nearly: () 0.5 () 0.7 (C) 1. (D) 1.5 y b 00 mm h 300 mm E c 14 Ga t 1 mm E e 100 Ga z C 300 mm s ac 9.5 Ma s ae 140 Ma 1 mm 00 mm 1 mm 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

50 ENDIX FE Exam Review roblems 1131 Composite beam is symmetric about both axes so each N is an axis of symmetry Moments of inertia of cross section about z and y axes: I cz b h mm 4 I cy h b mm 4 I ez t h mm 4 I ey h t3 1 (t h) ab t b mm 4 ending about z axis based on allowable stress in each material (lesser value controls) ae c I cz E e I ez b M max_cz s ac 5.9 kn m h E c ae c I cz E e I ez b M max_ez s ae 109. kn m h E e ending about y axis based on allowable stress in each material (lesser value controls) M max_cy s (E c I cy E e I ey ) ac 74.0 kn m b E c M max_ey s (E c I cy E e I ey ) ae a b 136.kN m tbe e ratio z_to_y M max_cz M max_cy 0.7 allowable stress in the core, not exterior cover sheet, controls moments about both axes R-6.: composite beam is made up of a 90 mm 160 mm wood beam (E w 11 Ga) and a steel bottom cover plate (90 mm 8 mm, E s 190 Ga). llowable stresses in wood and steel are 6.5 Ma and 110 Ma, respectively. The allowable bending moment about the z-axis of the composite beam is most nearly: ().9 kn?m () 3.5 kn?m (C) 4.3 kn?m (D) 9.9 kn?m 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

51 113 ENDIX FE Exam Review roblems b 90 mm h 160 mm t 8mm y E w 11 Ga E s 190 Ga s aw 6.5 Ma s as 110 Ma w bh mm z O 160 mm 8 mm s bt 70 mm 90 mm ocate N (distance h above base) by summing 1st moments of E about base of beam; then find h 1 dist. from N to top of beam: t E s s E w w at h b h mm E s s E w w h 1 h t h mm Moments of inertia of wood and steel about N: I s b t3 1 s ah t b mm 4 I w b h3 1 w ah 1 h b mm 4 llowable moment about z axis based on allowable stress in each material (lesser value controls) M max_w s (E w I w E s I s ) aw 4.6 kn m h 1 E w M max_s s (E w I w E s I s ) as kn m h E s R-6.3: steel pipe (d mm, d 96 mm) has a plastic liner with inner diameter d 1 8 mm. The modulus of elasticity of the steel is 75 times that of the modulus of the plastic. llowable stresses in steel and plastic are 40 Ma and 550 ka, respectively. The allowable bending moment for the composite pipe is approximately: () 1100 N?m () 130 N?m (C) 1370 N?m (D) 1460 N?m 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

52 ENDIX FE Exam Review roblems 1133 d mm d 96 mm d 1 8 mm s as 40 Ma s ap 550 ka y Cross section properties: s p 4 (d 3 d ) mm z C d 1 d d 3 p p 4 (d d 1 ) mm I s p 64 (d 3 4 d 4 ) mm 4 I p p 64 (d 4 d 1 4 ) mm 4 Due to symmetry, N of composite beam is the z axis llowable moment about z axis based on allowable stress in each material (lesser value controls) M max_p s (E p I p E s I s ) M max_s s (E p I p E s I s ) as ap a d 3 b E s Modular ratio: n E s n 75 E p Divide through by E p in moment expressions above M max_s s (I p ni s ) as a d 130 N m 3 b n M max_ p s (I p ni s ) ap a d 1374 N m b a d b E p R-6.4: bimetallic beam of aluminum (E a 70 Ga) and copper (E c 110 Ga) strips has width b 5 mm; each strip has thickness t 1.5 mm. bending moment of 1.75 N?m is applied about the z axis. The ratio of the maximum stress in the aluminum to that in the copper is approximately: () 0.6 () 0.8 (C) 1.0 (D) Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.