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

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

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 35 3 5 1 6 4 3 5 1 4 H 1 V1 V 6 6 F 35 5 F 5 F 4 4 V 6 6 1083 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.

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 3 110 F FE (3 m) 1 110 F FE (3 m) 10 kn (3 m) 5 kn (6 m) E y (3 m) 0 Solving F FE 5 110 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.

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.

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 1 3.6 Ma C t C Stress in C: s C 1 C must equal s Solve for C 1 18.00 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 400 10 6. 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.

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 16.48 kn cos(u) 35 35 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.

1088 ENDIX FE Exam Review roblems g 77 kn m 3 llowable stress: Y 80 Ma FS Y s allow s Y FS Y 140.0 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 0.006 xial stress: E a 436.4 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.

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 0.001 xial strain: a v 3.333 10 3 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 0.003 d 1 d olyethylene bar xial strain: a v 6.50 10 3 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.

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 0.0036 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 1.061 10 3 xial stress in C: C E C 116.7 Ma (well below yield stress of 550 Ma so Hooke s aw applies) ateral strain of C: t C 0.0036 mm t C t C 3.600 10 4 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.

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 1.6875000 10 7 mm 3 Normal strain in bar: ateral strain in bar: Final length of bar: 1.60000 10 3 E b v 5.8000 10 4 f 3004.800 mm Final lateral dimension of bar: b f b b 74.96040 mm Final volume of bar: V final b f f 1.6884156 10 7 mm 3 Increase in volume of bar: V V final V init 9156 mm 3 V V init 0.000543 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.

109 ENDIX FE Exam Review roblems ve. shear stress: b t V 0.474 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 3.600 10 3 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.

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.

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.

ENDIX FE Exam Review roblems 1095 C 45 45 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 1 3 17.00 kn N C 3.00 kn N CD 3 8.00 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.

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.384 10 4 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 1 54.469 mm d 1 d / / p 4 d 113.097 mm Volume of nonprismatic bar: Vol nonprismatic ( 1 ) 459458 mm3 Diameter of prismatic bar of same volume: prismatic p 4 d 184 mm d H Vol nonprismatic 15.30 mm p 4 V prismatic prismatic 459458 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.

ENDIX FE Exam Review roblems 1097 Elongation of nonprismatic bar shown in fig. above: E a 1 1 1 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 3 4 3 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.

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 113.097 mm 4 kn g 11 m 3 W 5.598 N d W E so or d g E d (g ) E 0.053 mm Check max. normal stress at top of bar s max W 0.050 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.

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 1 8.74 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 13.40 kn E b b R-.8: steel rod (E s 10 Ga, d r 1 mm, a s 1 10 6 > 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.

1100 ENDIX FE Exam Review roblems E s 10 Ga d r 1 mm d p 15 mm r p 4 d r 113.097 mm p p 4 d p 176.715 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 1 10 6 > 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 176.7 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.

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 1 1 10.59 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.

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 19.5 10 6 > 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.

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 78.54 mm 11.5 kn d Normal stress in bar: s p 146.4 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 150.000 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.

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 400.0 kn (compression) Max. normal stress in : s 100.8 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.

ENDIX FE Exam Review roblems 1105 Equate above formulas and solve for s x tan(u) 1 so u atana 1 1 b 35.64 s x s u 117.0 Ma cos(u) also u s x sin(u)cos(u) 55.154 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. 3.0.5 D = D 1 1.5 D R D 1 K.0 1.1 1. K = smax snom s nom = p D 1 /4 D D R = 1 1.5 0 0.05 0.10 0.15 0.0 0.5 0.30 R D 1 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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 0.100 d 5 mm 1.50 so K 1.75 d 1 0 mm 3.0.5 D = D 1 1.5 D R D 1 K.0 1.1 1. K = smax snom s nom = p D 1 /4 K = 1.75 D D R = 1 1.5 0 0.05 0.10 0.15 0.0 0.5 0.30 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 0.0005 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 0.0005 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.

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 0.014 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 0.014 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 d4.047 10 5 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.

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 ) 7.608 10 5 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 d4 7.178 10 5 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.

ENDIX FE Exam Review roblems 1109 d 86 mm d 1 5 mm I pipe p 3 (d 4 d 1 4 ) 4.65 10 6 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 1 3.5 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 1 3.5 kn m T 1.5 kn m T 1 d T 1 d C 1 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1110 ENDIX FE Exam Review roblems olar moments of inertia: I p1 p 3 d 1 4 9.655 10 5 mm 4 I p p 3 d 4 7.178 10 5 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 1 0.9 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 1 1 0.9 m 0.75 m f a 1.8 olar moments of inertia: I p1 p 3 d 1 4 1.649 10 5 mm 4 I p p 3 d 4 1.09 10 5 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.

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 1 185 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 975.0 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.

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 1.508 10 3 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 8.0 10 4. 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 1.600 10 3 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.

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 5700000 N mm d 1 5 mm t a 43. Ma d 4 d 1 4 16 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 4 4.396 10 5 mm 4 ower in terms of torque T: pf T Solve for torque T: T 757.9 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.

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 557.0 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 0.060 m d 1 0.040 m I p p 3 (d 4 d 1 4 ) 1.01 10 6 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 3.119 10 5 W T max t a I p d 1191. 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.

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. 3-59 Stress-concentration factor K for a stepped shaft in torsion. (The dashed line is for a full quarter-circular fillet.) K.00 1.1 1. 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 = 1.00 0 0.10 R D 1 0.0 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1116 ENDIX FE Exam Review roblems rismatic shaft T 1max t allow I d 0 Stepped shaft t allow a p d 3 0 16 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 0.100 K 1.35 so from graph (see Fig. 3-59) T max t allow K ap d 3 1 16 b 4 Ma c p 1.35 16 (0 mm)3 d 48.9 N m T 1 max T max 56.6 48.9 1.16 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.

ENDIX FE Exam Review roblems 1117 7.5 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.

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) 16 5 7 35 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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 5 1 3 6 Substitute numbers for and : 9m M 5 1 5 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.

110 ENDIX FE Exam Review roblems Sum moments about to get reaction at : R 1 1.6 c15 s1.6) a1.6 b 4.5d 19.4065 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 11.85 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: () 1.5 10 3 () 1.55 10 3 (C) 1.76 10 3 (D) 1.9 10 3 d d max R d ar d b d 1.5 mm R 0.6 m d max ar d 1.48 10 3 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.

ENDIX FE Exam Review roblems 111 Section modulus: q S b h 6.613 106 m 3 h Max. moment at midspan: M max q 8 9.7 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 ) 3.356 10 6 mm 4 Uniformly distributed weight of pipe, q: q g CI 0.47 kn m Vertical force at each lift point: F q 1.484 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.

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 1.979 kn m 4 s M F q a b 1.484 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 3.36 10 6 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 3.375 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) 66.696 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.

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) 6.000 kn m s c M neg (66.4 mm) 118.6 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.

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 1 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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 h3 1 7.031 105 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 0.5 4.0 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.

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 h3 1 1.400 105 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.

ENDIX FE Exam Review roblems 117 0.65 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 h3 1.047 107 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 6.408 105 mm 3 M q 1 q 3 So, since t a 1 Ma q max1 8 t a 9 139 kn m s max M S q max 1 5 s a S s a 5 1 q S 130.0 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.

118 ENDIX FE Exam Review roblems H 7.5 m 1. m W = 700 N W 1 4300 N W 700 N 1 1500 N d 35 mm t 0 mm d 1 d t 195 mm ole cross sectional properties at base: p 4 (d d 1 ) 13509 mm I p 64 (d 4 d 4 1 ) 7.873 10 7 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 10.410 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.

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 ` 9 5 9 5 1.8 R-5.1: rectangular beam with semicircular notches has dimensions h 160 mm and h 1 140 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 3.0.5.0 h h 1 1.1 = 1. h = h 1 + R M h R h 1 K = s max s s = 6M nom nom bh 1 b = thickness M 1.05 1.5 0 0.05 0.10 0.15 0.0 0.5 0.30 FIG. 5-50 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.

1130 ENDIX FE Exam Review roblems R 10 h 1 140 0.071 h 160 h 1 140 1.143 From Fig 5-50: s allow K R 1 (h h 1) 1 (160 mm 140 mm) 10.000 mm K.5 6 M b h so b min 6 M K 1 s allow h 6 (185 N m) (.5) 1 6.5 Ma C(140 mm) D 19.6 mm 3.0 K.5 K =.5.0 h h 1 1.1 = 1. h = h 1 + R M h R h 1 K = s max s s = 6M nom nom bh 1 b = thickness M 1.05 1.5 0 0.05 0.071 0.10 0.15 0.0 0.5 0.30 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.

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 h3 1 4.500 108 mm 4 I cy h b3 1.000 108 mm 4 I ez t h3 1 5.400 107 mm 4 I ey h t3 1 (t h) ab t b 8.099 10 7 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.

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 14400 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 49.07 mm E s s E w w h 1 h t h 118.93 mm Moments of inertia of wood and steel about N: I s b t3 1 s ah t b 1.467 10 6 mm 4 I w b h3 1 w ah 1 h b 5.54 10 7 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 10.11 kn m h E s R-6.3: steel pipe (d 3 104 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.

ENDIX FE Exam Review roblems 1133 d 3 104 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 ) 156.6 mm z C d 1 d d 3 p p 4 (d d 1 ) 1957. mm I s p 64 (d 3 4 d 4 ) 1.573 10 6 mm 4 I p p 64 (d 4 d 1 4 ) 1.950 10 6 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) 1.5 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1134 ENDIX FE Exam Review roblems b 5 mm t 1.5 mm a bt 37.5 mm c a 37.5 mm M 1.75 N m E a 70 Ga E c 110 Ga Equate 1st moments of E about bottom of beam to locate N (distance h above base); then find h 1 dist. from N to top of beam: z b y O C t t E c t c E a a at t b h 1.333 mm E c c E a a h 1 t h 1.667 mm h 1 h 3.000 mm t 3.000 mm Moments of inertia of aluminum and copper strips about N: I c b t 3 1 c ah t b 19.79 mm 4 I a b t 3 1 a ah 1 t b 38.54 mm 4 ending stresses in aluminum and copper: s a Mh 1 E a E a I a E c I c 41.9 Ma s c Mh E c E a I a E c I c 5.6 Ma Ratio of the stress in the aluminum to that of the copper: s a s c 0.795 R-6.5: composite beam of aluminum (E a 7 Ga) and steel (E s 190 Ga) has width b 5 mm and heights h a 4 mm, h s 68 mm. bending moment is applied about the z axis resulting in a maximum stress in the aluminum of 55 Ma. The maximum stress in the steel is approximately: () 86 Ma () 90 Ma (C) 94 Ma (D) 98 Ma y luminum b 5 mm h a 4 mm h s 68 mm Steel h a E a 7 Ga E s 190 Ga s a 55 Ma z O h s a bh a 1050.0 mm s bh s 1700.0 mm b 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.

ENDIX FE Exam Review roblems 1135 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: h E s h s s E a a ah s h a b 44.43 mm E a a E s s h 1 h a h s h 65.57 mm h 1 h 110.00 mm Moments of inertia of aluminum and steel parts about N: I s b h s 3 1 s ah h s b 8.401 10 5 mm 4 I a b h a 3 1 aah 1 h a b.40 10 6 mm 4 Set max. bending stress in aluminum to given value then solve for moment M: M s a (E a I a E s I s ) h 1 E a 3.738 kn m Use M to find max. bending stress in steel: s s M h E s E a I a E s I s 98.4 Ma R-7.1: rectangular plate (a 10 mm, b 160 mm) is subjected to compressive stress s x 4.5 Ma and tensile stress s y 15 Ma. The ratio of the normal stress acting perpendicular to the weld to the shear stress acting along the weld is approximately: () 0.7 () 0.54 (C) 0.85 (D) 1. a 10 mm b 160 mm σ y u arctana a b b 36.87 s x 4.5 Ma s y 15 Ma Weld b a σ x t xy 0 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1136 ENDIX FE Exam Review roblems lane stress transformation: normal and shear stresses on y-face of element rotated through angle u (perpendicular to & along weld seam): s u s x s y s x s y cosc au p bd t xy sinc au p bd 7.98 Ma t u s x s y sinc au p bd t xy cosc au p bd 9.36 Ma ` s u t u ` 0.85 R-7.: rectangular plate in plane stress is subjected to normal stresses s x and s y and shear stress t xy. Stress s x is known to be 15 Ma but s y and t xy are unknown. However, the normal stress is known to be 33 Ma at counterclockwise angles of 35 and 75 from the x axis. ased on this, the normal stress s y on the element below is approximately: () 14 Ma () 1 Ma (C) 6 Ma (D) 43 Ma s x 15 Ma s 35 33 Ma s 75 s 35 lane stress transformations for 35 & 75 : s u s x s y s x s y cos( u) t xy sin( u) y σ y τ xy O σ x x For u 35 : u 35 35 s x s y s x s y cos[ (u 35 )] t xy sin[ (u 35 )] s 35 Or s y.8563t xy 69.713 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

ENDIX FE Exam Review roblems 1137 nd for u 75 : u 75 75 s x s y s x s y cos[ (u 75 )] t xy sin[ (u 75 )] s 75 Or s y 0.5359 t xy 34.9 Solving above two equations for s y and t xy gives: a s y b c 1.8563 1 t xy 1 0.5359 d a 69.713 34.9 b a6.1 15.3 b Ma so y 6.1 Ma R-7.3: rectangular plate in plane stress is subjected to normal stresses s x 35 Ma, s y 6 Ma, and shear stress t xy 14 Ma. The ratio of the magnitudes of the principal stresses (s 1 /s ) is approximately: () 0.8 () 1.5 (C).1 (D).9 s x 35 Ma s y 6 Ma t xy 14 Ma y rincipal angles: σ y t xy u 1 1 arctana b 36.091 s x s y u u 1 p 16.091 O τ xy σ x x lane stress transformations: s 1 s x s y s s x s y Ratio of principal stresses: s 1 s.86 s x s y cos(u 1 ) t xy sin(u 1 ) 45.1 Ma s x s y cos(u ) t xy sin(u ) 15.79 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.

1138 ENDIX FE Exam Review roblems R-7.4: drive shaft resists torsional shear stress of 45 Ma and axial compressive stress of 100 Ma. The ratio of the magnitudes of the principal stresses (s 1 /s ) is approximately: () 0.15 () 0.55 (C) 1. (D) 1.9 s x 100 Ma s y 0 t xy 45 Ma rincipal angles: t xy u 1 1 arctana b 0.994 s x s y u u 1 p 110.994 100 Ma lane stress transformations: 45 Ma s u1 s x s y actually s s u s x s y this is s 1 s x s y cos(u 1 ) t xy sin(u 1 ) 117.7 Ma s x s y cos(u ) t xy sin(u ) 17.7 Ma So s 1 max(s u1, s u ) 17.68 Ma s min(s u1, s u ) 117.68 Ma Ratio of principal stresses: ` s 1 s ` 0.15 R-7.5: drive shaft resists torsional shear stress of 45 Ma and axial compressive stress of 100 Ma. The maximum shear stress is approximately: () 4 Ma () 67 Ma (C) 71 Ma (D) 93 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.

ENDIX FE Exam Review roblems 1139 s x 100 Ma s y 0 t xy 45 Ma Max. shear stress: t max a s x s y b t xy 67.3 Ma 100 Ma 45 Ma R-7.6: drive shaft resists torsional shear stress of t xy 40 Ma and axial compressive stress s x 70 Ma. One principal normal stress is known to be 38 Ma (tensile). The stress s y is approximately: () 3 Ma () 35 Ma (C) 6 Ma (D) 75 Ma s x 70 Ma s y is unknown t xy 40 Ma s prin 38 Ma y σ y Stresses s x and s y must be smaller than the given principal stress so: O τ xy σ x x s 1 s prin Substitute into stress transformation equation and solve for s y : s x s y a s x s y b t 66 Ma xy s 1 solve, s y 7 3. Ma R-7.7: cantilever beam with rectangular cross section (b 95 mm, h 300 mm) supports load 160 kn at its free end. The ratio of the magnitudes of the principal stresses (s 1 /s ) at point (at distance c 0.8 m from the free end and distance d 00 mm up from the bottom) is approximately: () 5 () 1 (C) 18 (D) 5 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1140 ENDIX FE Exam Review roblems 160 kn b 95 mm h 300 mm c 0.8 m d 00 mm d h 0.667 Cross section properties: bh 8500 mm h I b h3 1.138 108 mm 4 b c d Q [b (h d)] c h (h d) d 9.500 10 5 mm 3 Moment, shear force and normal and shear stresses at : M c 1.80 10 5 kn mm V t V Q I b 7.485 Ma M ad h b s 9.94 Ma I lane stress state at : s x s t xy t s y 0 rincipal stresses: u 1 arctana t xy b 13.83 s x s y s 1 s x s y s s x s y a s x s y b t xy 31.709 Ma a s x s y b t xy 1.767 Ma Ratio of principal stresses (s 1 / s ): ` s 1 s ` 17.9 R-7.8: simply supported beam ( 4.5 m) with rectangular cross section (b 95 mm, h 80 mm) supports uniform load q 5 kn/m. The ratio of the magnitudes of the principal stresses (s 1 /s ) at a point a 1.0 m from the left support and distance d 100 mm up from the bottom of the beam is approximately: () 9 () 17 (C) 31 (D) 41 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

ENDIX FE Exam Review roblems 1141 q 5 kn m b 95 mm a 1.0 m 4.5 m h 80 mm d 100 mm q h Cross section properties: a bh 6600 mm b I b h3 1 1.738 108 mm 4 Q [b (h d)] c h (h d) d 8.550 10 5 mm 3 Moment, shear force and normal and shear stresses at distance a from left support: V a q q a 31.50 kn M a q a q a 4.375 104 kn mm t V a Q I b 1.618 Ma M a ad h b s 10.070 Ma I lane stress state: s x s t xy t s y 0 rincipal stresses: u 1 arctana t xy b 8.909 s x s y s 1 s x s y s s x s y a s x s y b t xy 10.34 Ma a s x s y b t xy 0.54 Ma Ratio of principal stresses (s 1 / s ): ` s 1 s ` 40.7 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

114 ENDIX FE Exam Review roblems R-8.1: thin wall spherical tank of diameter 1.5 m and wall thickness 65 mm has internal pressure of 0 Ma. The maximum shear stress in the wall of the tank is approximately: () 58 Ma () 67 Ma (C) 115 Ma (D) 17 Ma d 1.5 m t 65 mm p 0 Ma t Thin wall tank since: a d 0.087 b Weld iaxial stress: p a d b s 115.4 Ma t Max. shear stress at 45 deg. rotation is 1/ of s t max s 57.7 Ma R-8.: thin wall spherical tank of diameter 0.75 m has internal pressure of 0 Ma. The yield stress in tension is 90 Ma, the yield stress in shear is 475 Ma, and the factor of safety is.5. The modulus of elasticity is 10 Ga, oisson s ratio is 0.8, and maximum normal strain is 10 10 6. The minimum permissible thickness of the tank is approximately: () 8.6 mm () 9.9 mm (C) 10.5 mm (D) 11.1 mm d 0.75 m p 0 Ma E 10 Ga s Y 90 Ma t Y 475 Ma FS Y.5 0.8 a 10(10 6 ) Thickness based on tensile stress: Weld p a d b t 1 a s 10.190 mm Y b FS 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.

ENDIX FE Exam Review roblems 1143 Thickness based on shear stress: p a d b t 4 a t 9.868 mm Y b FS Y Thickness based on normal strain: p a d b t 3 (1 n) a E t 3 10.54 mm largest value controls R-8.3: thin wall cylindrical tank of diameter 00 mm has internal pressure of 11 Ma. The yield stress in tension is 50 Ma, the yield stress in shear is 140 Ma, and the factor of safety is.5. The minimum permissible thickness of the tank is approximately: () 8. mm () 9.1 mm (C) 9.8 mm (D) 11.0 mm d 00 mm p 11 Ma s Y 50 Ma t Y 140 Ma FS Y.5 Wall thickness based on tensile stress: p a d b t 1 11.00 mm s Y FS Y Wall thickness based on shear stress: p a d b t a t 9.81 mm Y b FS Y larger value governs t 1 a d b 0.110 t a d b 0.098 R-8.4: thin wall cylindrical tank of diameter.0 m and wall thickness 18 mm is open at the top. The height h of water (weight density 9.81 kn/m 3 ) in the tank at which the circumferential stress reaches 10 Ma in the tank wall is approximately: () 14 m () 18 m (C) 0 m (D) 4 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.

1144 ENDIX FE Exam Review roblems d m t 18 mm s a 10 Ma ressure at height h: p h g w h kn g w 9.81 m 3 d p h a d b Circumferential stress: s c t Set s c equal to s a and solve for h: h s a t (g w ) a d 18.3 m b (g w h) a d b s c t h R-8.5: The pressure relief valve is opened on a thin wall cylindrical tank, with radius to wall thickness ratio of 18, thereby decreasing the longitudinal strain by 150 10 6. ssume E 73 Ga and v 0.33. The original internal pressure in the tank was approximately: () 370 ka () 450 ka (C) 500 ka (D) 590 ka r t r t r t 18 148 (10 6 ) E 73 Ga n 0.33 ongitudinal strain: p E ar b (1 ) t Set to and solve for pressure p: strain gage p E 497 ka r t (1 n) R-8.6: cylindrical tank is assembled by welding steel sections circumferentially. Tank diameter is 1.5 m, thickness is 0 mm, and internal pressure is.0 Ma. The maximum stress in the heads of the tank is approximately: () 38 Ma () 45 Ma (C) 50 Ma (D) 59 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.

ENDIX FE Exam Review roblems 1145 d 1.5 m t 0 mm p.0 Ma p a d b s h 37.5 Ma t Welded seams R-8.7: cylindrical tank is assembled by welding steel sections circumferentially. Tank diameter is 1.5 m, thickness is 0 mm, and internal pressure is.0 Ma. The maximum tensile stress in the cylindrical part of the tank is approximately: () 45 Ma () 57 Ma (C) 6 Ma (D) 75 Ma d 1.5 m t 0 mm p.0 Ma p a d b s c 75.0 Ma t Welded seams R-8.8: cylindrical tank is assembled by welding steel sections circumferentially. Tank diameter is 1.5 m, thickness is 0 mm, and internal pressure is.0 Ma. The maximum tensile stress perpendicular to the welds is approximately: () Ma () 9 Ma (C) 33 Ma (D) 37 Ma d 1.5 m t 0 mm p.0 Ma p a d b s w 37.5 Ma t Welded seams R-8.9: cylindrical tank is assembled by welding steel sections circumferentially. Tank diameter is 1.5 m, thickness is 0 mm, and internal pressure is.0 Ma. The maximum shear stress in the heads is approximately: () 19 Ma () 3 Ma (C) 33 Ma (D) 35 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.

1146 ENDIX FE Exam Review roblems d 1.5 m t 0 mm p.0 Ma p a d b t h 18.8 Ma 4 t Welded seams R-8.10: cylindrical tank is assembled by welding steel sections circumferentially. Tank diameter is 1.5 m, thickness is 0 mm, and internal pressure is.0 Ma. The maximum shear stress in the cylindrical part of the tank is approximately: () 17 Ma () 6 Ma (C) 34 Ma (D) 38 Ma d 1.5 m t 0 mm p.0 Ma p a d b t max 37.5 Ma t Welded seams R-8.11: cylindrical tank is assembled by welding steel sections in a helical pattern with angle a 50 degrees. Tank diameter is 1.6 m, thickness is 0 mm, and internal pressure is.75 Ma. Modulus E 10 Ga and oisson s ratio n 0.8. The circumferential strain in the wall of the tank is approximately: () 1.9 10 4 () 3. 10 4 (C) 3.9 10 4 (D) 4.5 10 4 d 1.6 m t 0 mm p.75 Ma E 10 Ga n 0.8 a 50 Circumferential stress: p a d b s c 110.0 Ma t Helical weld α Circumferential strain: c s c ( n) 4.50 10 4 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.

ENDIX FE Exam Review roblems 1147 R-8.1: cylindrical tank is assembled by welding steel sections in a helical pattern with angle a 50 degrees. Tank diameter is 1.6 m, thickness is 0 mm, and internal pressure is.75 Ma. Modulus E 10 Ga and oisson s ratio n 0.8. The longitudinal strain in the the wall of the tank is approximately: () 1. 10 4 ().4 10 4 (C) 3.1 10 4 (D) 4.3 10 4 d 1.6 m t 0 mm p.75 Ma E 10 Ga n 0.8 a 50 ongitudinal stress: p a d b s 55.0 Ma t Helical weld α ongitudinal strain: s (1 n) 1.15 10 4 E R-8.13: cylindrical tank is assembled by welding steel sections in a helical pattern with angle a 50 degrees. Tank diameter is 1.6 m, thickness is 0 mm, and internal pressure is.75 Ma. Modulus E 10 Ga and oisson s ratio n 0.8. The normal stress acting perpendicular to the weld is approximately: () 39 Ma () 48 Ma (C) 78 Ma (D) 84 Ma d 1.6 m t 0 mm p.75 Ma E 10 Ga 0.8 a 50 Helical weld α ongitudinal stress: p a d b s 55.0 Ma So s x s 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.

1148 ENDIX FE Exam Review roblems Circumferential stress: p a d b s c 110.0 Ma so s y s c t ngle perpendicular to the weld: u 90 a 40.000 Normal stress perpendicular to the weld: s 40 s x s y s x s y cos ( u) 77.7 Ma R-8.14: segment of a drive shaft (d 00 mm, d 1 160 mm) is subjected to a torque T 30 kn?m. The allowable shear stress in the shaft is 45 Ma. The maximum permissible compressive load is approximately: () 00 kn () 86 kn (C) 38 kn (D) 44 kn d 00 mm d 1 160 mm t a 45 Ma T 30 kn m Cross section properties: T p 4 (d d 1 ) 11310 mm I p p 3 (d 4 d 1 4 ) 9.74 10 7 mm 4 Normal and in-plane shear stresses: T s x 0 s y t xy T a d b I 3.349 Ma Maximum in-plane shear stress: set max allow then solve for s y t max a s x s y b t xy so s y #4 (t a t xy ) 5.303 Ma Finally solve for s y : max y 86 kn R-8.15: thin walled cylindrical tank, under internal pressure p, is compressed by a force F 75 kn. Cylinder diameter is d 90 mm and wall thickness t 5.5 mm. llowable normal stress is 110 Ma and allowable shear stress is 60 Ma. The maximum allowable internal pressure p max is approximately: () 5 Ma () 10 Ma (C) 13 Ma (D) 17 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.

ENDIX FE Exam Review roblems 1149 d 90 mm t 5.5 mm s a 110 Ma F 75 kn p d t 1555 mm Circumferential normal stress: F F p max a d b s c t and setting s c s a and solving for p max : p maxc s a a t b 13.4 Ma d controls ongitudinal normal stress: p max a d b s F t or s p max d 4 t F So set s s a and solve for p max : p max as a F b 4 t d 38.7 Ma Check also in-plane & out-of-plane shear stresses: all are below allowable shear stress so circumferential normal stress controls as noted above. R-9.1: n aluminum beam (E 7 Ga) with a square cross section and span length.5 m is subjected to uniform load q 1.5 kn/m. The allowable bending stress is 60 Ma. The maximum deflection of the beam is approximately: () 10 mm () 16 mm (C) mm (D) 6 mm E 7 (10 3 )Ma q 1.5 N mm 500 mm s a 60 Ma q = 1.5 kn/m Max. moment and deflection at /: =.5 m M max q 8 d max 5 q 4 384 E I 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.

1150 ENDIX FE Exam Review roblems Moment of inertia and section modulus for square cross section (height width b) I b4 1 S I a b b b 3 6 Flexure formula s max M max S Max. deflection formula q 8 s max a b3 6 b so b 3 3 4 q s max 5q 4 d max so solve for d 384 E a b4 max if s max s a 1 b d max 5q 4 ca 3 q b 1 4 d 3 384 E 4 s max 1. mm R-9.: n aluminum cantilever beam (E 7 Ga) with a square cross section and span length.5 m is subjected to uniform load q 1.5 kn/m. The allowable bending stress is 55 Ma. The maximum deflection of the beam is approximately: () 10 mm () 0 mm (C) 30 mm (D) 40 mm E 7(10 3 ) Ma q 1.5 N mm 500 mm s a 55 Ma Max. moment at support & max. deflection at : q M max q Moment of inertia and section modulus for square cross section (height width b) I b4 1 d max q 4 8 E I S I a b b b 3 6 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

ENDIX FE Exam Review roblems 1151 Flexure formula s max M max S Max. deflection formula q s max a b3 6 b so b 3 q 3 s max d max q 4 d max 8 E a b4 1 b 8 E q 4 q ca3 s max so solve for d max if s max s a 1 1 4 b3d 9.9 mm R-9.3: steel beam (E 10 Ga) with I 119 10 6 mm 4 and span length 3.5 m is subjected to uniform load q 9.5 kn/m. The maximum deflection of the beam is approximately: () 10 mm () 13 mm (C) 17 mm (D) 19 mm E 10(10 3 ) Ma I 119(10 6 ) mm 4 strong axis I for W310 5 q 9.5 N mm 3500 mm M y q x k = 48EI/ 3 R = kδ Max. deflection at by superposition of SS beam mid-span deflection & R /k: d max 5 q ( )4 384 E I (q ) a 48 E I 13.07 mm b 3 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

115 ENDIX FE Exam Review roblems R-9.4: steel bracket C (EI 4. 10 6 N?m ) with span length 4.5 m and height H m is subjected to load 15 kn at C. The maximum rotation of joint is approximately: () 0.1 () 0.3 (C) 0.6 (D) 0.9 E 10 Ga I 0 10 6 mm 4 strong axis I for W00.5 EI 4.0 10 6 N m 15 kn C 4.5 m H m H Max. rotation at : apply statically-equivalent moment H at on SS beam u max ( H) 3 E I 0.614 u max 0.011 rad R-9.5: steel bracket C (EI 4. 10 6 N?m ) with span length 4.5 m and height H m is subjected to load 15 kn at C. The maximum horizontal displacement of joint C is approximately: () mm () 31 mm (C) 38 mm (D) 40 mm E 10 Ga I 0 10 6 mm 4 strong axis I for W00.5 EI 4.0 10 6 N m 15 kn C 4.5 m H m H 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.

ENDIX FE Exam Review roblems 1153 Max. rotation at : apply statically-equivalent moment H at on SS beam u max ( H) 3 E I 0.614 u max 0.011 rad Horizontal deflection of vertical cantilever C: d C H3 9.54 mm 3 E I Finally, superpose u H and d C d C u max H d C 31.0 mm R-9.6: nonprismatic cantilever beam of one material is subjected to load at its free end. Moment of inertia I I 1. The ratio r of the deflection d to the deflection d 1 at the free end of a prismatic cantilever with moment of inertia I 1 carrying the same load is approximately: () 0.5 () 0.40 (C) 0.56 (D) 0.78 Max. deflection of prismatic cantilever (constant I 1 ) d 1 3 3 E I 1 Rotation at C due to both load & moment / at C for nonprismatic beam: a b u C E I a b 3 E I 8 E I I C I 1 Deflection at C due to both load & moment / at C for nonprismatic beam: a 3 b d Cl 3 E I a b a b 5 3 E I 48 E I Total deflection at : a 3 d d Cl u C b 3 E I 1 d 5 3 a 3 a 3 b 48 E I 8 E I b 3 (7 I 1 I ) 3 E I 1 4 E I 1 I 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.

1154 ENDIX FE Exam Review roblems Ratio d /d 1 : r 3 (7 I 1 I ) 4 E I 1 I 3 3 E I 1 7 I 1 8 I 1 8 so r 7 8 a1 b 1 8 0.563 R-9.7: steel bracket CD (EI 4. 10 6 N?m ), with span length 4.5 m and dimension a m, is subjected to load 10 kn at D. The maximum deflection at is approximately: () 10 mm () 14 mm (C) 19 mm (D) 4 mm E 10 Ga I 0 10 6 mm 4 strong axis I for W00.5 EI 4.0 10 6 N m 10 kn 4.5 m h 06 mm a m D a C Statically-equivalent loads at end of cantilever : downward load CCW moment a Downward deflection at by superposition: d 3 ( a) 4.1 mm 3 E I E I d 0.005 R-10.1: ropped cantilever beam has moment M 1 applied at joint. Framework C has moment M applied at C. oth structures have constant flexural rigidity EI. If the ratio of the applied moments M 1 /M 3/, the ratio of the reactive moments M 1 /M at clamped support is: () 1 () 3/ (C) (D) 5/ 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.

ENDIX FE Exam Review roblems 1155 y y M C M 1 / x x M 1 M From rob. 10.3-1: M 1 M 1 Statically-equivalent moment at is M so M M Ratio of reactive moments is M 1 M M 1 M so M 1 M 3 R-10.: ropped cantilever beam has moment M 1 applied at joint. Framework C has moment M applied at C. oth structures have constant flexural rigidity EI. If the ratio of the applied moments M 1 /M 3/, the ratio of the joint rotations at, u 1 /u, is: () 1 () 3/ (C) (D) 5/ y y M C M 1 / x x M 1 θ 1 M 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.

1156 ENDIX FE Exam Review roblems From rob 10.3-1: u 1 M 1 4 EI Statically-equivalent moment at is M so u M 4 EI Ratio of joint rotations is u 1 M 1 u M so u 1 u 3 R-10.3: Structure 1 with member C of length / has force 1 applied at joint C. Structure with member C of length has force applied at C. oth structures have constant flexural rigidity EI. If the ratio of the applied forces 1 / 5/, the ratio of the joint rotations u 1 /u is: () 1 () 5/4 (C) 3/ (D) C 1 C y / y x x M1 θ 1 M θ Statically-equivalent moment at is M 1 1 / a 1 b From rob. 10.3-1: u 1 1 4 EI 8 EI Statically-equivalent moment at is M so u 4 EI 4 EI Ratio of joint rotations is u 1 u 1 so u 1 u 5 4 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

ENDIX FE Exam Review roblems 1157 R-10.4: Structure 1 with member C of length / has force 1 applied at joint C. Structure with member C of length has force applied at C. oth structures have constant flexural rigidity EI. The required ratio of the applied forces 1 / so that joint rotations u 1 and u are equal is approximately: () 1 () 5/4 (C) 3/ (D) C 1 C y / y x x M 1 θ 1 M θ Statically-equivalent moment at is M 1 1 / a 1 b From rob. 10.3-1: u 1 1 4 EI 8 EI Statically-equivalent moment at is M so u ( ) 4 EI 4 EI Ratio of joint rotations is u 1 u 1 and u 1 u 1 so 1 R-10.5: Structure 1 with member C of length / has force 1 applied at joint C. Structure with member C of length has force applied at C. oth structures have constant flexural rigidity EI. If the ratio of the applied forces 1 / 5/, the ratio of the joint reactions R 1 /R is: () 1 () 5/4 (C) 3/ (D) 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.

1158 ENDIX FE Exam Review roblems C 1 C y / y x x R 1 R Statically-equivalent moment at is M 1 1 / From rob. 10.3-1: R 1 3 M 1 or R 1 3 Statically-equivalent moment at is M a 1 b 3 1 4 so R 3 3 Ratio of reactions at is R 1 R 1 so R 1 R 1 a5 b 5 4 R-10.6: Structure 1 with member C of length / has force 1 applied at joint C. Structure with member C of length has force applied at C. oth structures have constant flexural rigidity EI. The required ratio of the applied forces 1 / so that joint reactions R 1 and R are equal is: () 1 () 5/4 (C) 3/ (D) 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.

ENDIX FE Exam Review roblems 1159 C 1 C y / y x x R 1 R Statically-equivalent moment at is M 1 1 / From rob. 10.3-1: R 1 3 M 1 or R 1 3 Statically-equivalent moment at is M so R 3 3 a 1 b 3 1 4 Ratio of reactions at is R 1 R 1 and R 1 R 1 so 1 R-10.7: Structure 1 with member C of length / has force 1 applied at joint C. Structure with member C of length has force applied at C. oth structures have constant flexural rigidity EI. If the ratio of the applied forces 1 / 5/, the ratio of the joint C lateral deflections d C1 /d C is approximately: () 1/ () 4/5 (C) 3/ (D) 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.

1160 ENDIX FE Exam Review roblems 1 δ C1 C y / x 1 a 3 b d C1 3 EI d C1 d C 5 1 8 R-11.1: eam C has a sliding support at and is supported at C by a pinned end steel column with square cross section (E 00 Ga, b 40 mm) and height 3.75 m. The column must resist a load Q at with a factor of safety.0 with respect to the critical load. The maximum permissible value of Q is approximately: () 10.5 kn () 11.8 kn (C) 13. kn (D) 15.0 kn 5 8 4 EI 1 5 5 56 5 3 1 48 EI d C 3 3 EI 4 EI 7 3 1 EI 5 56 0.446 E 00 Ga n.0 b 40 mm 3.75 m d C d I b4 1.133 105 mm 4 Q Statics: sum vertical forces to find reaction at D: R D Q So force in pin-pin column is Q cr Q cr Q cr p E I 9.9 kn D llowable value of Q: Q allow Q cr n 15.0 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.

ENDIX FE Exam Review roblems 1161 R-11.: eam C has a pin support at and is supported at C by a steel column with square cross section (E 190 Ga, b 4 mm) and height 5.5 m. The column is pinned at C and fixed at D. The column must resist a load Q at with a factor of safety.0 with respect to the critical load. The maximum permissible value of Q is approximately: () 3.0 kn () 6.0 kn (C) 9.4 kn (D) 10.1 kn E 190 Ga n.0 b 4 mm 5.5 m Effective length of pinned-fixed column: d C d e 0.699 3.670 m Q I b4 1.593 105 mm 4 Statics: use FD of C and sum moments about to find force in column as a multiple of Q: D F CD Q (3 d) d 3Q So force in pin-fixed column is 3Q cr 3 Q cr cr p E I e llowable value of Q: 36.1 kn So Q cr cr 3 1.0 kn Q allow Q cr n 6.0 kn R-11.3: steel pipe column (E 190 Ga, a 14 10 6 per degree Celsius, d 8 mm, d 1 70 mm) of length 4.5 m is subjected to a temperature increase T. The column is pinned at the top and fixed at the bottom. The temperature increase at which the column will buckle is approximately: () 36 C () 4 C (C) 54 C (D) 58 C E 190 Ga 4.5 m a [14 (10 6 )]/ C d 8 mm d 1 70 mm p 4 (d d 1 ) 143.57 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.

116 ENDIX FE Exam Review roblems Effective length of pinned-fixed column: e 0.699 3.0 m I p 64 (d 4 d 1 4 ) 1.04076 10 6 mm 4 xial compressive load in bar: E a( T) Equate to Euler buckling load and solve for T: T p E I e E a Or T p I a e 58.0 C ΔT R-11.4: steel pipe (E 190 Ga, a 14 10 6 / C, d 8 mm, d 1 70 mm) of length 4.5 m hangs from a rigid surface and is subjected to a temperature increase T 50 C. The column is fixed at the top and has a small gap at the bottom. To avoid buckling, the minimum clearance at the bottom should be approximately: ().55 mm () 3.4 mm (C) 4.17 mm (D) 5.3 mm E 190 Ga 450 mm a [14(10 6 )] / C T 50 C d 8 mm d 1 70 mm p 4 (d d 1 ) 1433 mm I p 64 (d 4 d 1 4 ) 1.041 10 6 mm 4 ΔT Effective length of fixed-roller support column: e.0 8500.0 mm Column elongation due to temperature increase: d 1 a T.975 mm Euler buckling load for fixed-roller column: cr p E I 7.013 kn e Column shortening under load of cr : d cr 0.4 mm E Mimimum required gap size to avoid buckling: gap d 1 d.55 mm gap frictionless surface 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.

ENDIX FE Exam Review roblems 1163 R-11.5: pinned-end copper strut (E 110 Ga) with length 1.6 m is constructed of circular tubing with outside diameter d 38 mm. The strut must resist an axial load 14 kn with a factor of safety.0 with respect to the critical load. The required thickness t of the tube is: ().75 mm () 3.15 mm (C) 3.89 mm (D) 4.33 mm E 110 Ga 1.6 m d 38 mm n.0 14 kn cr n cr 8.0 kn Solve for required moment of inertia I in terms of cr then find tube thickness t cr p E I I cr p E d I 6605 mm 4 Moment of inertia Solve numerically for min. thickness t: I p 64 [d 4 (d t) 4 ] t min 4.33 mm d 4 (d t) 4 I 64 p d t min 9.3 mm inner diameter R-11.6: plane truss composed of two steel pipes (E 10 Ga, d 100 mm, wall thickness 6.5 mm) is subjected to vertical load W at joint. Joints and C are 7 m apart. The critical value of load W for buckling in the plane of the truss is nearly: () 138 kn () 146 kn (C) 153 kn (D) 164 kn E 10 Ga 7m d 100 mm t 6.5 mm Moment of inertia I p 64 [d4 (d t) 4 ].097 10 6 mm 4 W d Member lengths: cos(40 ) 5.36 m 40 50 C C cos(50 ) 4.500 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.

1164 ENDIX FE Exam Review roblems Statics at joint to find member forces F and F C : Sum horizontal forces at joint : F F F cos(50 ) cos(40 ) F C cos(50 ) C cos(40 ) where a cos(50 ) cos(40 ) 0.839 Sum vertical forces at joint : W F sin(40 ) F C sin(50 ) W F cos(50 ) C cos(40 ) sin(40 ) F C sin(50 ) 1 F C W b where b a cos(50 ) sin(40 ) sin(50 )b cos(40 ) 0.766 So member forces in terms of W are: F C Wb and F F C a or F W(ab) with ab 0.643 Euler buckling loads in & C: F _cr p E I 151.118 kn so W _cr b a F _cr 138 kn lower value controls F C_cr p E I C 14.630 kn so W C_cr b F C_cr 164 kn R-11.7: beam is pin-connected to the tops of two identical pipe columns, each of height h, in a frame. The frame is restrained against sidesway at the top of column 1. Only buckling of columns 1 and in the plane of the frame is of interest here. The ratio (a/) defining the placement of load Q cr, which causes both columns to buckle simultaneously, is approximately: () 0.5 () 0.33 (C) 0.67 (D) 0.75 Draw FD of beam only; use statics to show that Q cr causes forces 1 and in columns 1 & respectively: 1 a a b Q cr EI h 1 a Q cr -a EI h a Q cr 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.

ENDIX FE Exam Review roblems 1165 uckling loads for columns 1 & : cr1 p EI cr p EI h (0.699 h) a a a a b Q cr b Q cr Solve above expressions for Q cr, then solve for required a/ so that columns buckle at the same time: p EI (0.699 h) a a b p EI h a a b Or Or So p EI (0.699 h) a a b p EI h a a b 0 0.699 ( a) a 0 a 0.699 (1 0.699 ) 0.38 Or a a 0.699 R-11.8: steel pipe column (E 10 Ga) with length 4.5 m is constructed of circular tubing with outside diameter d 90 mm and inner diameter d 1 64 mm. The pipe column is fixed at the base and pinned at the top and may buckle in any direction. The Euler buckling load of the column is most nearly: () 303 kn () 560 kn (C) 690 kn (D) 70 kn E 10 Ga d 90 mm Moment of inertia 4.5 mm d 1 64 mm d 1 d I p 64 (d 4 d 1 4 ) I.397 10 6 mm 4 Effective length of column for fixed-pinned case: e 0.699.971 m cr p E I e 563 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.

1166 ENDIX FE Exam Review roblems R-11.9: n aluminum tube (E 7 Ga) of circular cross section has a pinned support at the base and is pin-connected at the top to a horizontal beam supporting a load Q 600 kn. The outside diameter of the tube is 00 mm and the desired factor of safety with respect to Euler buckling is 3.0. The required thickness t of the tube is most nearly: () 8 mm () 10 mm (C) 1 mm (D) 14 mm E 7 Ga.5 m n 3.0 Q 600 kn d 00 mm SM c 0.5 Q 1.5 1000 kn Find required I based on critical buckling load Critical load C Q = 600 kn cr n cr 3000 kn 1.5 m 1.0 m cr p E I I cr p E I 6.386 10 6 mm 4 d 00 mm.5 m Moment of inertia I p 64 [d4 (d t) 4 ] t min d 4 d 4 I 64 p t min 9.73 mm R-11.10: Two pipe columns are required to have the same Euler buckling load cr. Column 1 has flexural rigidity E I and height 1 ; column has flexural rigidity (4/3)E I and height. The ratio ( / 1 ) at which both columns will buckle under the same load is approximately: () 0.55 () 0.7 (C) 0.81 (D) 1.10 013 Cengage earning. ll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

ENDIX FE Exam Review roblems 1167 Equate Euler buckling load expressions for the two columns considering their different properties, base fixity conditions and lengths: p p a Eb ( I) E I (0.699 1 ) 3 Simplify then solve for / 1 : a b 0.699 1 4 3 4 1 3 0.699 0.807 cr E I 1 cr E/3 I R-11.11: Two pipe columns are required to have the same Euler buckling load cr. Column 1 has flexural rigidity E I 1 and height ; column has flexural rigidity (/3)E I and height. The ratio (I /I 1 ) at which both columns will buckle under the same load is approximately: () 0.8 () 1.0 (C). (D) 3.1 Equate Euler buckling load expressions for the two columns considering their different properties, base fixity conditions and lengths: p p a E I 1 (0.699 ) 3 Eb (I ) Simplify then solve for I /I 1 : I E I 1 (0.699 ) 3 E 3 I I 1 (0.699) 3.07 cr E I 1 cr E/3 I 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.