See the solution to Prob Ans. Since. (2E t + 2E c )ch - a. (s max ) t. (s max ) c = 2E c. 2E c. (s max ) c = 3M bh 2E t + 2E c. 2E t. h c.

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1 * The beam has a ectangula coss section and is subjected to a bending moment. f the mateial fom which it is made has a diffeent modulus of elasticity fo tension and compession as shown, detemine the location c of the neutal axis and the maximum compessive stess in the beam. s h c E t b P E c See the solution to Pob h2e c c 2E t + 2E c Since (s max ) c c h - c (s h2e c max) t (2E t + 2E c )ch - a h 1E c bd 1E t + 1E c (s max ) t (s max ) c 2E c 2E t (s max ) t (s max ) c 2E c 3 2E t bh 2 2E t + 2E c 2E c (s max ) c 3 bh 2E t + 2E c 2 2E t 399

2 The beam is subjected to a bending moment of 20 kip # ft diected as shown. Detemine the maximum bending stess in the beam and the oientation of the neutal axis. B y 8 in. C The y and z components of ae negative, Fig. a. Thus, y -20 sin kip # ft z -20 cos kip # ft. z 14 in in. The moments of inetia of the coss-section about the pincipal centoidal y and z axes ae 10 in. D y 1 12 (16)103 B (14)83 B 736 in 4 z 1 12 (10)163 B (8)143 B 1584 in 4 By inspection, the bending stess occus at cones and C ae s - z y z + y z y s max s C (12)(8) (12)( - 5) ksi (T) s max s (12)(-8) (12)(5) ksi 2.01 ksi (C) Hee, u tan a z y tan u tan a tan 225 a 65.1 The oientation of neutal axis is shown in Fig. b. 400

3 Detemine the maximum magnitude of the bending moment that can be applied to the beam so that the bending stess in the membe does not exceed 12 ksi. y 8 in. The y and z components of ae negative, Fig. a. Thus, B C y - sin z - cos z 14 in in. The moments of inetia of the coss-section about pincipal centoidal y and z axes ae 10 in. D y 1 12 (16)103 B (14)83 B 736 in 4 z 1 12 (10)163 B (8)143 B 1584 in 4 By inspection, the maximum bending stess occus at cones and C. Hee, we will conside cone C. s C s allow - z y c z + yz c y (12)(8) (12)(-5) kip # ft 119 kip # ft 401

4 f the esultant intenal moment acting on the coss section of the aluminum stut has a magnitude of 520 N # m and is diected as shown, detemine the bending stess at points and B. The location y of the centoid C of the stut s coss-sectional aea must be detemined. lso, specify the oientation of the neutal axis. z y B 200 mm 20 mm y 20 mm C mm 200 mm 520 N m 20 mm ntenal oment Components: z (520) -480 N # m y 5 13 (520) 200 N # m Section Popeties: y y 0.01(0.4)(0.02) + 2[(0.110)(0.18)(0.02)] 0.4(0.02) + 2(0.18)(0.02) m 57.4 mm z 1 12 (0.4)0.023 B + (0.4)(0.02)( ) (0.04)0.183 B (0.18)( ) B m 4 y 1 12 (0.2)0.43 B (0.18)0.363 B B m 4 aximum Bending Stess: pplying the flexue fomula fo biaxial at points and B s - zy z + yz y s ( ) (10-6 ) + 200(-0.2) (10-3 ) Pa 1.30 Pa (C) s B ( ) (10-6 ) + 200(0.2) (10-3 ) Pa (T) Oientation of Neutal xis: tan a z y tan u tan a (10-6 ) (10-3 tan ( ) ) a

5 * The esultant intenal moment acting on the coss section of the aluminum stut has a magnitude of 520 N # m and is diected as shown. Detemine maximum bending stess in the stut. The location y of the centoid C of the stut s coss-sectional aea must be detemined. lso, specify the oientation of the neutal axis. z y B 200 mm 20 mm y 20 mm C mm 200 mm 520 N m 20 mm ntenal oment Components: z (520) -480 N # m y 5 13 (520) 200 N # m Section Popeties: y y 0.01(0.4)(0.02) + 2[(0.110)(0.18)(0.02)] 0.4(0.02) + 2(0.18)(0.02) m 57.4 mm z 1 12 (0.4)0.023 B + (0.4)(0.02)( ) (0.04)0.183 B (0.18)( ) B m 4 y 1 12 (0.2)0.43 B (0.18)0.363 B B m 4 aximum Bending Stess: By inspection, the maximum bending stess can occu at eithe point o B. pplying the flexue fomula fo biaxial bending at points and B s - z y z + y z y s ( ) (10-6 ) + 200(-0.2) (10-3 ) Pa 1.30 Pa (C) (ax) s B ( ) (10-6 ) Pa (T) Oientation of Neutal xis: + 200(0.2) (10-3 ) tan a z y tan u tan a (10-6 ) (10-3 tan ( ) ) a

6 Conside the geneal case of a pismatic beam subjected to bending-moment components y and z, as shown, when the x, y, z axes pass though the centoid of the coss section. f the mateial is linea-elastic, the nomal stess in the beam is a linea function of position such that s a + by + cz. Using the equilibium conditions 0 1 s d, y 1 zs d, z 1 - ys d, detemine the constants a, b, and c, and show that the nomal stess can be detemined fom the equation s [-1 z y + y yz 2y + 1 y z + z yz 2z]>1 y z - 2 yz 2, whee the moments and poducts of inetia ae defined in ppendix. z z d y z y y s C x Equilibium Condition: s x a + by + cz 0 L s x d 0 L (a + by + cz) d 0 a L d + b L y d + c L z d [1] y L z s x d L z(a + by + cz) d a L z d + b L yz d + c L z 2 d [2] z L -y s x d L -y(a + by + cz) d -a L yd - b L y 2 d - c L yz d [3] Section Popeties: The integals ae defined in ppendix. Note that y d z d 0.Thus, L L Fom Eq. [1] Fom Eq. [2] Fom Eq. [3] a 0 y b yz + c y z -b z - c yz Solving fo a, b, c: a 0 (Since Z 0) b - z y + y yz 2 c y z + z yz 2 y z - yz y z - yz Thus, s x - z y + y yz 2 y + y y + z yz 2 z y z - yz y z - yz (Q.E.D.) 404

7 The cantileveed beam is made fom the Z-section having the coss-section shown. f it suppots the two loadings, detemine the bending stess at the wall in the beam at point. Use the esult of Pob ft 50 lb 50 lb ( y ) max 50(3) + 50(5) 400 lb # ft 4.80(10 3 )lb # in. y 1 12 (3.25)(0.25)3 + 2c 1 12 (0.25)(2)3 + (0.25)(2)(1.125) 2 d in 4 z 1 12 (0.25)(3.25)3 + 2c 1 12 (2)(0.25)3 + (0.25)(2)(1.5) 2 d in 4 2 in. B 2.25 in in. 3 in in. 2 ft yz 2[1.5(1.125)(2)(0.25)] in in. Using the equation developed in Pob s -a z y + y yz by + a y z + z yz bz 2 2 y z - yz y z - yz s {-[0 + (4.80)(103 )(1.6875)](1.625) + [(4.80)(10 3 )( ) + 0](2.125)} [ ( ) - (1.6875) 2 ] 8.95 ksi The cantileveed beam is made fom the Z-section having the coss-section shown. f it suppots the two loadings, detemine the bending stess at the wall in the beam at point B. Use the esult of Pob ft 50 lb 50 lb ( y ) max 50(3) + 50(5) 400 lb # ft 4.80(10 3 )lb # in. y 1 12 (3.25)(0.25)3 + 2c 1 12 (0.25)(2)3 + (0.25)(2)(1.125) 2 d in 4 z 1 12 (0.25)(3.25)3 + 2c 1 12 (2)(0.25)3 + (0.25)(2)(1.5) 2 d in in. 2 in. B 0.25 in. 3 in in. 2 ft yz 2[1.5(1.125)(2)(0.25)] in in. Using the equation developed in Pob s -a z y + y yz by + a y z + z yz bz 2 2 y z - yz y z - yz s B -[0 + (4.80)(103 )(1.6875)](-1.625) + [(4.80)(10 3 )( ) + 0](0.125) [( )( ) - (1.6875) 2 ] 7.81 ksi 405

8 * The cantileveed wide-flange steel beam is subjected to the concentated foce P at its end. Detemine the lagest magnitude of this foce so that the bending stess developed at does not exceed s allow 180 Pa. ntenal oment Components: Using method of section z 0; z + P cos 30 (2) 0 z P y 0; y + P sin 30 (2) 0 y -1.00P z 200 mm 10 mm 150 mm 10 mm 10 mm y 2 m Section Popeties: z 1 12 (0.2)0.173 B (0.19)0.153 B (10-6 ) m 4 x 30 P y 2c 1 12 (0.01)0.23 B d (0.15)0.013 B (10-6 ) m 4 llowable Bending Stess: By inspection, maximum bending stess occus at points and B. pplying the flexue fomula fo biaxial bending at point. s s allow - zy z + yz y B - (-1.732P)(0.085) (10-6 ) P(-0.1) (10-6 ) P N 14.2 kn The cantileveed wide-flange steel beam is subjected to the concentated foce of P 600 N at its end. Detemine the maximum bending stess developed in the beam at section. ntenal oment Components: Using method of sections z 0; z cos 30 (2) 0 z N # m y 0; y sin 30 (2) 0; y N # m z 200 mm 10 mm 150 mm 10 mm 10 mm y 2 m Section Popeties: z 1 12 (0.2)0.173 B (0.19)0.153 B (10-6 ) m 4 x 30 P y 2c 1 12 (0.01)0.23 B d (0.15)0.013 B (10-6 ) m 4 aximum Bending Stess: By inspection, maximum bending stess occus at and B. pplying the flexue fomula fo biaxial bending at point s - zy z + yz y s (0.085) (10-6 ) (-0.1) (10-6 ) 7.60 Pa (T) (ax) 406

9 f the beam is subjected to the intenal moment of 1200 kn # m, detemine the maximum bending stess acting on the beam and the oientation of the neutal axis. ntenal oment Components: The y component of is positive since it is diected towads the positive sense of the y axis, wheeas the z component of, which is diected towads the negative sense of the z axis, is negative, Fig. a. Thus, y 1200 sin kn # m z cos kn # m Section Popeties: The location of the centoid of the coss-section is given by z 150 mm y 150 mm 300 mm mm x 150 mm y y 0.3(0.6)(0.3) (0.15)(0.15) 0.6(0.3) (0.15) m The moments of inetia of the coss section about the pincipal centoidal y and z axes ae 150 mm y 1 12 (0.6)0.33 B (0.15)0.153 B B m 4 z 1 12 (0.3)0.63 B + 0.3(0.6)( ) 2 - c 1 12 (0.15)0.153 B (0.15)( ) 2 d Bm 4 Bending Stess: By inspection, the maximum bending stess occus at eithe cone o B. s - zy z + yz y s - c B d(0.2893) B B(0.15) B 126 Pa (T) s B - c B d( ) B B(-0.15) B -131 Pa 131 Pa (C)(ax.) Oientation of Neutal xis: Hee, u -30. tan a z y tan u tan a B B tan(-30 ) a The oientation of the neutal axis is shown in Fig. b. 407

10 f the beam is made fom a mateial having an allowable tensile and compessive stess of (s allow ) t 125 Pa and (s allow ) c 150 Pa, espectively, detemine the maximum allowable intenal moment that can be applied to the beam. ntenal oment Components: The y component of is positive since it is diected towads the positive sense of the y axis, wheeas the z component of, which is diected towads the negative sense of the z axis, is negative, Fig. a. Thus, y sin z - cos z 150 mm y 150 mm 300 mm mm x 150 mm Section Popeties: The location of the centoid of the coss section is y y 0.3(0.6)(0.3) (0.15)(0.15) 0.6(0.3) (0.15) m 150 mm The moments of inetia of the coss section about the pincipal centoidal y and z axes ae y 1 12 (0.6)0.33 B (0.15)0.153 B B m 4 z 1 12 (0.3)0.63 B + 0.3(0.6)( ) 2 - c 1 12 (0.15)0.153 B (0.15)( ) 2 d Bm 4 Bending Stess: By inspection, the maximum bending stess can occu at eithe cone o B. Fo cone which is in tension, s (s allow ) t - z y z + y z y B - ( )(0.2893) B + 0.5(0.15) B N # m 1186 kn # m (contols) Fo cone B which is in compession, s B (s allow ) c - z y B z + y z B y B - ( )( ) B + 0.5(-0.15) B N # m 1377 kn # m 408

11 * The shaft is suppoted on two jounal beaings at and B which offe no esistance to axial loading. Detemine the equied diamete d of the shaft if the allowable bending stess fo the mateial is s allow 150 Pa. The FBD of the shaft is shown in Fig. a. The shaft is subjected to two bending moment components z and y, Figs. b and c, espectively. Since all the axes though the centoid of the cicula coss-section of the shaft ae pincipal axes, then the esultant moment 2 2 y + 2 z can be used fo design. The maximum moment occus at D (x 1m). Then, z y 0.5 m 0.5 m C 0.5 m 200 N 0.5 m 200 N 300 N D 300 N B E x 150 N 150 N Then, max N # m s allow max C ; 150(10 6 ) (d>2) p 4 (d>2)4 d m 25 mm 409

12 The 30-mm-diamete shaft is subjected to the vetical and hoizontal loadings of two pulleys as shown. t is suppoted on two jounal beaings at and B which offe no esistance to axial loading. Futhemoe, the coupling to the moto at C can be assumed not to offe any suppot to the shaft. Detemine the maximum bending stess developed in the shaft. x D z 1 m 1 m C 1 m B 1 m E 400 N 100 mm 400 N 60 mm y 150 N 150 N Suppot Reactions: s shown on FBD. ntenal oment Components: The shaft is subjected to two bending moment components y and z. The moment diagam fo each component is dawn. aximum Bending Stess: Since all the axes though the cicle s cente fo cicula shaft ae pincipal axis, then the esultant moment 2 2 y + 2 z can be used to detemine the maximum bending stess. The maximum esultant moment occus at E max N # m. pplying the flexue fomula s max max c 427.2(0.015) p B 161 Pa 410

13 Using the techniques outlined in ppendix, Example.5 o.6, the Z section has pincipal moments of inetia of and z m 4 y m 4, computed about the pincipal axes of inetia y and z, espectively. f the section is subjected to an intenal moment of 250 N # m diected hoizontally as shown, detemine the stess poduced at point. Solve the poblem using Eq y 250 cos N # m z 250 sin N # m y 0.15 cos sin m z -(0.175 cos sin 32.9 ) m 50 mm y 200 mm 32.9 y 250 N m 50 mm z z 300 mm B 200 mm 50 mm s - zy z + yz (0.2210) y 0.471( ( ) ) 60.0(10-6 ) -293 kpa 293 kpa (C) Solve Pob using the equation developed in Pob ntenal oment Components: y 250 N # m z 0 Section Popeties: y 1 12 (0.3)0.053 B + 2c 1 12 (0.05)0.153 B (0.15)0.1 2 B d B m 4 z 1 12 (0.05)0.33 B + 2c 1 12 (0.15)0.053 B (0.05) B d 0.350(10-3 ) m 4 yz 0.15(0.05)(0.125)(-0.1) (0.05)(-0.125)(0.1) 50 mm y 200 mm 32.9 y 250 N m 50 mm z z 300 mm B 200 mm 50 mm B m 4 Bending Stess: Using fomula developed in Pob s -( z y + y yz )y + ( y z + z yz )z 2 y z - yz s -[ ( )(10-3 )](0.15) + [250(0.350)(10-3 ) + 0](-0.175) (10-3 )(0.350)(10-3 ) - [0.1875(10-3 )] kpa 293 kpa (C) 411

14 * Using the techniques outlined in ppendix, Example.5 o.6, the Z section has pincipal moments of inetia of and z m 4 y m 4, computed about the pincipal axes of inetia y and z, espectively. f the section is subjected to an intenal moment of 250 N # m diected hoizontally as shown, detemine the stess poduced at point B. Solve the poblem using Eq mm y 200 mm 32.9 y 250 N m 50 mm z z 300 mm B 200 mm 50 mm ntenal oment Components: y 250 cos N # m z 250 sin N # m Section Popety: y 0.15 cos sin m z 0.15 sin cos m Bending Stess: pplying the flexue fomula fo biaxial bending s z y z + y z y s B 135.8(0.2210) 0.471(10-3 ) ( ) 0.060(10-3 ) 293 kpa 293 kpa (T) 412

15 Detemine the bending stess at point of the beam, and the oientation of the neutal axis. Using the method in ppendix, the pincipal moments of inetia of the coss section ae and y in 4 z in 4, whee z and y ae the pincipal axes. Solve the poblem using Eq z in. 0.5 in. z 4 in. 3 kip ft C 45 4 in. y y in. 0.5 in. ntenal oment Components: Refeing to Fig. a, the y and z components of ae negative since they ae diected towads the negative sense of thei espective axes. Thus, Section Popeties: Refeing to the geomety shown in Fig. b, z œ y œ cos sin in. -(2.817 sin cos 45 ) in. Bending Stess: s - z y œ z + œ y z y - (-2.121)(12)(-2.828) (-2.121)(12)(1.155) ksi 21.0 ksi (C) 413

16 Detemine the bending stess at point of the beam using the esult obtained in Pob The moments of inetia of the coss sectional aea about the z and y axes ae z y in 4 and the poduct of inetia of the coss sectional aea with espect to the z and y axes is yz in 4. (See ppendix ) z in. 0.5 in. 4 in. 45 C 3 kip ft 4 in. y z y in. 0.5 in. ntenal oment Components: Since is diected towads the negative sense of the y axis, its y component is negative and it has no z component. Thus, y -3 kip # ft z 0 Bending Stess: s - z y + y yz By + y z + z yz Bz y z - yz 2 - C0(5.561) + (-3)(12)(-3.267)D(-1.183) + C -3(12)(5.561) + 0(-3.267)D(2.817) 5.561(5.561) - (-3.267) ksi 21.0 ksi 414

17 The composite beam is made of 6061-T6 aluminum () and C83400 ed bass (B). Detemine the dimension h of the bass stip so that the neutal axis of the beam is located at the seam of the two metals. What maximum moment will this beam suppot if the allowable bending stess fo the aluminum is 1s allow 2 al 128 Pa and fo the bass 1s allow 2 b 35 Pa? Section Popeties: n E al E b 68.9(109 ) 101(10 9 ) b b nb al (0.15) m y y B 150 mm h 50 mm ( )(0.05) + ( h)(0.15)h (0.05) + (0.15)h h m 41.3 mm N 1 12 ( )0.053 B (0.05)( ) (0.15) B ( )( ) B m 4 llowable Bending Stess: pplying the flexue fomula ssume failue of ed bass (s allow ) b c N B ( ) (10-6 ) 6598 N # m 6.60 kn # m (contols!) ssume failue of aluminium (s allow ) al n c N (0.05) B c (10-6 ) d N # m 29.2 kn # m 415

18 * The composite beam is made of 6061-T6 aluminum () and C83400 ed bass (B). f the height h 40 mm, detemine the maximum moment that can be applied to the beam if the allowable bending stess fo the aluminum is 1s allow 2 al 128 Pa and fo the bass 1s allow 2 b 35 Pa. Section Popeties: Fo tansfomed section. n E al E b 68.9(109 ) 101.0(10 9 ) b b nb al (0.15) m y y B 150 mm h 50 mm 0.025( )(0.05) + (0.07)(0.15)(0.04) (0.05) (0.04) m N 1 12 ( )0.053 B (0.05)( ) (0.15)0.043 B (0.04)( ) B m 4 llowable Bending Stess: pplying the flexue fomula ssume failue of ed bass (s allow ) b c N B ( ) (10-6 ) 6412 N # m 6.41 kn # m (contols!) ssume failue of aluminium (s allow ) al n c N B c ( ) (10-6 ) d N # m 28.4 kn # m 416

19 Segment of the composite beam is made fom 2014-T6 aluminum alloy and segment B is -36 steel. f w 0.9 kip>ft, detemine the absolute maximum bending stess developed in the aluminum and steel. Sketch the stess distibution on the coss section. w 15 ft B 3 in. 3 in. 3 in. aximum oment: Fo the simply-suppoted beam subjected to the unifom distibuted load, the maximum moment in the beam is max wl B kip # ft. Section Popeties: The coss section will be tansfomed into that of steel as shown in Fig. a. Hee, n E al E st Then b st nb al (3) in. The location of the centoid of the tansfomed section is y y 1.5(3)(3) + 4.5(3)(1.0965) 3(3) + 3(1.0965) in. The moment of inetia of the tansfomed section about the neutal axis is + d (3)33 B + 3(3)( ) in 4 aximum Bending Stess: Fo the steel, (1.0965)33 B (3)( ) 2 (s max ) st maxc st (12)(2.3030) ksi t the seam, s st y in. maxy Fo the aluminium, (12)(0.6970) ksi (s max ) al n maxc al t the seam, (12)( ) c d 13.3 ksi s al y in. n maxy c (12)(0.6970) d 2.50 ksi The bending stess acoss the coss section of the composite beam is shown in Fig. b. 417

20 Segment of the composite beam is made fom 2014-T6 aluminum alloy and segment B is -36 steel. f the allowable bending stess fo the aluminum and steel ae (s allow ) al 15 ksi and (s allow ) st 22 ksi, detemine the maximum allowable intensity w of the unifom distibuted load. w 15 ft B 3 in. 3 in. 3 in. aximum oment: Fo the simply-suppoted beam subjected to the unifom distibuted load, the maximum moment in the beam is max wl2 w152 B w. 8 8 Section Popeties: The coss section will be tansfomed into that of steel as shown in Fig. a. Hee, n E al E st Then b st nb al (3) in. The location of the centoid of the tansfomed section is y y 1.5(3)(3) + 4.5(3)(1.0965) 3(3) + 3(1.0965) in. The moment of inetia of the tansfomed section about the neutal axis is + d (3)33 B + 3(3)( ) (1.0965)33 B in 4 Bending Stess: ssuming failue of steel, B (3)( ) 2 (s allow ) st max c st ; 22 (28.125w)(12)(2.3030) w kip>ft (contols) ssuming failue of aluminium alloy, (s allow ) al n max c al (28.125w)(12)( ) ; c d w 1.02 kip>ft 418

21 The Douglas fi beam is einfoced with -36 staps at its cente and sides. Detemine the maximum stess developed in the wood and steel if the beam is subjected to a bending moment of z 7.50 kip # ft. Sketch the stess distibution acting ove the coss section. y 0.5 in. 0.5 in. 0.5 in. 6 in. z 2 in. 2 in. Section Popeties: Fo the tansfomed section. n E w E st 1.90(103 ) 29.0(10 3 ) b st nb w (4) in. N 1 12 ( )63 B in 4 aximum Bending Stess: pplying the flexue fomula (s max ) st c 7.5(12)(3) 8.51 ksi (s max ) w n c 7.5(12)(3) c d ksi

22 * The top plate is made of 2014-T6 aluminum and is used to einfoce a Kevla 49 plastic beam. Detemine the maximum stess in the aluminum and in the Kevla if the beam is subjected to a moment of 900 lb # ft. 0.5 in. 6 in. 0.5 in. 0.5 in. 12 in. 0.5 in. Section Popeties: n E al E k 10.6(103 ) 19.0(10 3 ) b k n b al (12) in. y y 0.25(13)(0.5) + 2[(3.25)(5.5)(0.5)] (6.6947)(0.5) 13(0.5) + 2(5.5)(0.5) (0.5) in. N 1 12 (13)0.53 B + 13(0.5)( ) (1)5.53 B + 1(5.5)( ) in (6.6947)0.53 B (0.5)( ) 2 aximum Bending Stess: pplying the flexue fomula (s max ) al n c 900(12)( ) c d 245 psi (s max ) k c 900(12)( ) psi 420

23 The top plate made of 2014-T6 aluminum is used to einfoce a Kevla 49 plastic beam. f the allowable bending stess fo the aluminum is (s allow ) al 40 ksi and fo the Kevla (s allow ) k 8 ksi, detemine the maximum moment that can be applied to the beam. Section Popeties: 0.5 in. 6 in. 0.5 in. n E al E k 10.6(103 ) 19.0(10 3 ) in. 12 in. b k n b al (12) in. 0.5 in. y y 0.25(13)(0.5) + 2[(3.25)(5.5)(0.5)] (6.6947(0.5) 13(0.5) + 2(5.5)(0.5) (0.5) N 1 12 (13)0.53 B + 13(0.5)( ) in 4 aximum Bending Stess: pplying the flexue fomula ssume failue of aluminium (s allow ) al n c ( ) c d kip # in kip # ft ssume failue of Kevla 49 (s allow ) k c in (1)5.53 B + 1(5.5)( ) (6.6947)0.53 B (0.5)( ) 2 8 ( ) kip # in 16.4 kip # ft (Contols!) 421

24 The membe has a bass coe bonded to a steel casing. f a couple moment of 8 kn # m is applied at its end, detemine the maximum bending stess in the membe. E b 100 GPa, E st 200 GPa. n E b E st (0.14)(0.14) (0.05)(0.1) (10-6 )m 4 aximum stess in steel: 8 kn m 3 m 20 mm 100 mm 20 mm 20 mm 20 mm 100 mm (s st ) max c 1 8(103 )(0.07) 20.1 Pa (max) (10-6 ) aximum stess in bass: (s b ) max nc 2 0.5(8)(103 )(0.05) (10-6 ) 7.18 Pa The steel channel is used to einfoce the wood beam. Detemine the maximum stess in the steel and in the wood if the beam is subjected to a moment of 850 lb # ft. E st 29(10 3 ) ksi, E w 1600 ksi. 4 in. (0.5)(16)(0.25) + 2(3.5)(0.5)(2.25) + (0.8276)(3.5)(2.25) y in. 0.5(16) + 2(3.5)(0.5) + (0.8276)(3.5) 1 12 (16)(0.53 ) + (16)(0.5)( ) + 2a 1 12 b(0.5)(3.53 ) + 2(0.5)(3.5)( ) 0.5 in. 0.5 in. 15 in. 850 lb ft 0.5 in (0.8276)(3.53 ) + (0.8276)(3.5)( ) in 4 aximum stess in steel: (s st ) c 850(12)( ) psi 1.40 ksi aximum stess in wood: (s w ) n(s st ) max (1395) 77.0 psi 422

25 * white spuce beam is einfoced with -36 steel staps at its top and bottom as shown. Detemine the bending moment it can suppot if (s allow ) st 22 ksi and (s allow ) w 2.0 ksi. y 0.5 in. 4 in. 0.5 in. z 3 in. x Section Popeties: Fo the tansfomed section. n E w E st 1.40(103 ) 29.0(10 3 ) b st nb w (3) in. N 1 12 (3)53 B ( )43 B in 4 llowable Bending Stess: pplying the flexue fomula ssume failue of steel (s allow ) st c 22 (2.5) kip # in 11.7 kip # ft (Contols!) ssume failue of wood (s allow ) w n y (2) c d kip # in 27.7 kip # ft 423

26 f the beam is subjected to an intenal moment of 45 kn # m, detemine the maximum bending stess developed in the -36 steel section and the 2014-T6 aluminum alloy section B. 50 mm 15 mm 150 mm B Section Popeties: The coss section will be tansfomed into that of steel as shown in Fig. a. Hee, n E al B Thus, b st nb al (0.015) m. The E st B location of the tansfomed section is y y 0.075(0.15)( ) + 0.2cp B d 0.15( ) + p B m The moment of inetia of the tansfomed section about the neutal axis is + d ( )0.153 B (0.15)( ) p0.054 B + p B( ) B m 4 aximum Bending Stess: Fo the steel, (s max ) st c st B( ) B 154 Pa Fo the aluminum alloy, (s max ) al n c al C B(0.1882) S 171 Pa B 424

27 The concete beam is einfoced with thee 20-mm diamete steel ods. ssume that the concete cannot suppot tensile stess. f the allowable compessive stess fo concete is (s allow ) con 12.5 Pa and the allowable tensile stess fo steel is (s allow ) st 220 Pa, detemine the equied dimension d so that both the concete and steel achieve thei allowable stess simultaneously. This condition is said to be balanced. lso, compute the coesponding maximum allowable intenal moment that can be applied to the beam. The moduli of elasticity fo concete and steel ae E con 25 GPa and E st 200 GPa, espectively. 200 mm d Bending Stess: The coss section will be tansfomed into that of concete as shown in Fig. a. Hee, n E st 200. t is equied that both concete and steel E con 25 8 achieve thei allowable stess simultaneously. Thus, (s allow ) con c con ; B c con B c con (1) (s allow ) st n c st ; B 8B (d - c con) R B d - c con (2) Equating Eqs. (1) and (2), B B c con d - c con c con d (3) (3) Section Popeties: The aea of the steel bas is st 3c p B d Bp m 2 - Thus, the tansfomed aea of concete fom steel is ( con) t n s 8C BpD Bp m 2. Equating the fist moment of the aea of concete above and below the neutal axis about the neutal axis, 0.2(c con )(c con >2) Bp (d - c con ) 0.1c con Bpd Bpc con c con pd pc con (4) Solving Eqs. (3) and (4), d m 531 mm c con m Thus, the moment of inetia of the tansfomed section is 1 3 (0.2) B Bp( ) 2 425

28 Continued B m 4 Substituting this esult into Eq. (1), B C B S N # m 98.6 kn # m The beam is made fom thee types of plastic that ae identified and have the moduli of elasticity shown in the figue. Detemine the maximum bending stess in the PVC. (b bk ) 1 n 1 b Es 160 (3) 0.6 in lb 500 lb PVC E PVC 450 ksi Escon E E 160 ksi Bakelite E B 800 ksi (b bk ) 2 n 2 b pvc 450 (3) in ft 4 ft 3 ft y y (1)(3)(2) + 3(0.6)(2) + 4.5(1.6875)(1) 3(2) + 0.6(2) (1) in. 1 in. 2 in. 2 in (3)(23 ) + 3(2)( ) (0.6)(23 ) + 0.6(2)( ) 3 in (1.6875)(13 ) (1)( ) in 4 (s max ) pvc n c 2 a b 1500(12)(3.0654) ksi 426

29 * The low stength concete floo slab is integated with a wide-flange -36 steel beam using shea studs (not shown) to fom the composite beam. f the allowable bending stess fo the concete is (s allow ) con 10 Pa, and allowable bending stess fo steel is (s allow ) st 165 Pa, detemine the maximum allowable intenal moment that can be applied to the beam. 1 m 100 mm 15 mm 400 mm 15 mm 15 mm Section Popeties: The beam coss section will be tansfomed into that of steel. Hee, n E con Thus, E st b st nb con (1) m. The location of the tansfomed section is 200 mm y y (0.015)(0.2) + 0.2(0.37)(0.015) (0.015)(0.2) (0.1)(0.1105) 0.015(0.2) (0.015) (0.2) + 0.1(0.1105) m The moment of inetia of the tansfomed section about the neutal axis is + d (0.2) B Bending Stess: ssuming failue of steel, (s allow ) st c st B m 4 ssuming failue of concete, + 0.2(0.015)( ) (0.015)0.373 B (0.37)( ) (0.2) B + 0.2(0.015)( ) (0.1105)0.13 B (0.1)( ) 2 ; B (0.3222) B N # m 332 kn # m (s allow ) con n c con ; ( ) B C S B N # m 330 kn # m (contols) 427

30 The einfoced concete beam is used to suppot the loading shown. Detemine the absolute maximum nomal stess in each of the -36 steel einfocing ods and the absolute maximum compessive stess in the concete. ssume the concete has a high stength in compession and yet neglect its stength in suppoting tension. 4 ft 10 kip 8 ft 10 kip 4 ft 8 in. 15 in. 2 in. 1 in. diamete ods max (10 kip)(4 ft) 40 kip # ft st 3(p)(0.5) in 2 E st 29.0(10 3 ) ksi E con 4.20(10 3 ) ksi n st 29.0(103 ) 4.20(10 3 (2.3562) in2 ) y 0; 8(h )a h b (13 - h ) 0 2 h h Solving fo the positive oot: h in. c 1 12 (8)(5.517)3 + 8(5.517)(5.517>2) 2 d ( ) in 4 (s con ) max y 40(12)(5.517) ksi (s st ) max na y b a 29.0(103 ) ) 4.20(10 3 ba40(12)(13 b 18.3 ksi )

31 The einfoced concete beam is made using two steel einfocing ods. f the allowable tensile stess fo the steel is (s st ) allow 40 ksi and the allowable compessive stess fo the concete is (s conc ) allow 3 ksi, detemine the maximum moment that can be applied to the section. ssume the concete cannot suppot a tensile stess. E st 29(10 3 ) ksi, E conc 3.8(10 3 ) ksi. 4 in. 8 in. 6 in. 8 in. 18 in. 2 in. 1-in. diamete ods st 2(p)(0.5) in 2 n st 29(103 ) 3.8(10 3 (1.5708) in2 ) y 0; 22(4)(h +2) + h (6)(h >2) (16 - h ) 0 Solving fo the positive oot: c 1 12 (22)(4)3 + 22(4)( ) 2 d + c 1 12 (6)( )3 + 6( )( >2) 2 d ssume concete fails: (s con ) allow y ssume steel fails: 3h h h in ( ) in 4 ; 3 ( ) kip # in. (s st ) allow na y b; 40 29(103 ) ) 3.8(10 3 (16 ) kip # in kip # ft (contols) 429

32 Fo the cuved beam in Fig. 6 40a, show that when the adius of cuvatue appoaches infinity, the cuved-beam fomula, Eq. 6 24, educes to the flexue fomula, Eq Nomal Stess: Cuved-beam fomula s (R - ) d whee ( - R) L and R 1 d s ( - ) ( -) [1] + y [2] L d a L + y bd L a - - y + y + 1b d - L y + y d [3] Denominato of Eq. [1] becomes, y ( -) - L + y d - - y L + y d Using Eq. [2], ( -) - y L + y + y - y d - y y L + y d y 2 L + y d - y 1 y d - y L + y d L y2 1 + y d - 1 y d - y L y 1 + y d But, 1 y d 0, as y : 0 Then, Eq. [1] becomes Using Eq. [2], ( -) : s ( - ) s ( - -y ) Using Eq. [3], s C - - L y + y d - y L d + y S C L y + y d - y d L + y S 430

33 Continued C L y 1 + y d - y L d 1 + y S s y : 0 L y 1 + y d 0 and y L d 1 + y y 1 d y Theefoe, s a - y b -y (Q.E.D.) * The membe has an elliptical coss section. f it is subjected to a moment of 50 N # m, detemine the stess at points and B. s the stess at point, which is located on the membe nea the wall, the same as that at? Explain. L d 2p b a 2p(0.0375) ( a 2 ) ( ) m 75 mm 150 mm 250 mm 100 mm p ab p(0.075)(0.0375) (10-3 )p R 1 d (10-3 )p B - R s (R - ) ( - R) 50( ) ( k Pa (T) )p (0.1)( ) s B (R - B) B ( - R) 50( ) ( kpa (C) )p (0.25)( ) No, because of localized stess concentation at the wall. 431

34 The membe has an elliptical coss section. f the allowable bending stess is s allow 125 Pa detemine the maximum moment that can be applied to the membe. 150 mm 75 mm 250 mm 100 mm B a m; b m p(0.075)(0.0375) p L d 2pb a ( a 2 ) 2p(0.0375) ( ) m R 1 d p m - R (10-3 ) m s (R - ) ( - R) ssume tension failue. 125(10 6 ) ( ) p(0.1)( )(10-3 ) 14.0 kn # m (contols) ssume compession failue: -125(10 6 ) ( ) p(0.25)( )(10-3 ) 27.9 kn # m 432

35 Detemine the geatest magnitude of the applied foces P if the allowable bending stess is (s allow ) c 50 Pa in compession and (s allow ) t 120 Pa in tension. 75 mm P 10 mm 160 mm P 150 mm 10 mm 10 mm 250 mm 150 mm ntenal oment: 0.160P adius of cuvatue. Section Popeties: is positive since it tends to incease the beam s y 0.255(0.15)(0.01) (0.15)(0.01) (0.075)(0.01) 0.15(0.01) (0.01) (0.01) m 0.15(0.01) (0.01) (0.01) m 2 L d 0.15 ln ln ln m R 1 d m - R m llowable Nomal Stess: pplying the cuved-beam fomula ssume tension failue (s allow ) t B (R - ) ( - R) 0.16P( ) (0.25)( ) ssume compession failue P N kn (s allow ) t B (R - ) ( - R) 0.16P( ) (0.42)( ) P N 55.2 kn (Contols!) 433

36 f P 6 kn, detemine the maximum tensile and compessive bending stesses in the beam. 75 mm P 10 mm 160 mm P 150 mm 10 mm 10 mm 250 mm 150 mm ntenal oment: 0.160(6) kn # m is positive since it tends to incease the beam s adius of cuvatue. Section Popeties: y 0.255(0.15)(0.01) (0.15)(0.01) (0.075)(0.01) 0.15(0.01) (0.01) (0.01) m 0.15(0.01) (0.01) (0.01) m 2 L d 0.15 ln ln ln m R 1 d m - R m Nomal Stess: pplying the cuved-beam fomula (s max ) t (R - ) ( - R) 0.960(103 )( ) (0.25)( ) 4.51 Pa (s max ) c (R - ) ( - R) 0.960(103 )( ) (0.42)( ) Pa 434

37 * The cuved beam is subjected to a bending moment of 900 N # m as shown. Detemine the stess at points and B, and show the stess on a volume element located at each of these points. C 30 B 400 mm 100 mm C 20 mm 15 mm 150 mm B ntenal oment: -900 N # m is negative since it tends to decease the beam s adius cuvatue. Section Popeties: 0.15(0.015) + 0.1(0.02) m (0.15)(0.015) (0.1)(0.02) (10-3 ) m (10-3 ) m L d ln ln (10-3 ) m R 1 d (10-3 ) m - R (10-3 ) m Nomal Stess: pplying the cuved-beam fomula s (R - ) ( - R) -900( ) (0.57)( )(10-3 ) 3.82 Pa (T) s B (R - B) B ( - R) -900( ) (0.4)( )(10-3 ) Pa 9.73 Pa (C) 435

38 The cuved beam is subjected to a bending moment of 900 N # m. Detemine the stess at point C. C 30 B 400 mm 100 mm C 20 mm 15 mm 150 mm B ntenal oment: -900 N # m is negative since it tends to decease the beam s adius of cuvatue. Section Popeties: 0.15(0.015) + 0.1(0.02) m (0.15)(0.015) (0.1)(0.02) (10-3 ) m (10-3 ) m L d ln ln (10-3 ) m R 1 d (10-3 ) m - R (10-3 ) m Nomal Stess: pplying the cuved-beam fomula s C (R - C) C ( - R) -900( ) (0.55)( )(10-3 ) 2.66 Pa (T) 436

39 The elbow of the pipe has an oute adius of 0.75 in. and an inne adius of 0.63 in. f the assembly is subjected to the moments of 25 lb # in., detemine the maximum stess developed at section a-a. 30 a 1 in. 25 lb in. L d 2p ( c 2 ) a 2p( ) - 2p ( ) in in in. p( ) - p( ) p R 1 d p in lb in. - R in. (s max ) t (R - ) ( - R) 25( ) 204 psi (T) p(1)( ) (s max ) c (R - B) 25( ) 120 psi (C) B ( - R) p(2.5)( ) The cuved membe is symmetic and is subjected to a moment of 600 lb # ft. Detemine the bending stess in the membe at points and B. Show the stess acting on volume elements located at these points. B 0.5 in. 2 in. 0.5(2) + 1 (1)(2) 2 in2 2 8 in. 1.5 in. 9(0.5)(2) B(1)(2) in. 2 L d 0.5 ln c 1(10) 10 cln d - 1 d in. (10-8) 8 R 1 d in R in. s (R - ) ( - R) s 600(12)( ) 2(8)( ) 10.6 ksi (T) s B 600(12)( ) 2(10)( ) ksi 12.7 ksi (C) 437

40 * The cuved ba used on a machine has a ectangula coss section. f the ba is subjected to a couple as shown, detemine the maximum tensile and compessive stess acting at section a-a. Sketch the stess distibution on the section in thee dimensions. a a mm 50 mm 75 mm 250 N N 150 mm 75 mm a+ O 0; cos 60 (0.075) sin 60 (0.15) N # m L d b ln ln m (0.075)(0.05) 3.75(10-3 ) m 2 R 1 d 3.75(10-3 ) m - R s (R - ) ( ) ( - R) 3.75( kpa )(0.2375)( ) 792 kpa (C) s B (R - B) ( ) B ( - R) 3.75( Pa (T) )(0.1625)( ) 438

41 The ceiling-suspended C-am is used to suppot the X-ay camea used in medical diagnoses. f the camea has a mass of 150 kg, with cente of mass at G, detemine the maximum bending stess at section. G 1.2 m 200 mm 20 mm 40 mm 100 mm Section Popeties: 1.22(0.1)(0.04) (0.2)(0.02) 0.1(0.04) + 0.2(0.02) m L d 0.1 ln ln Bm 0.1(0.04) + 0.2(0.02) m 2 R 1 d (10-3 ) m - R B m ntenal oment: The intenal moment must be computed about the neutal axis as shown on FBD N # m is negative since it tends to decease the beam s adius of cuvatue. aximum Nomal Stess: pplying the cuved-beam fomula s (R - ) ( - R) ( ) 0.008(1.26)( )(10-3 ) 18.1 Pa (T) s B (R - B) B ( - R) ( ) 0.008(1.20)( )(10-3 ) Pa 26.2 Pa (C) (ax) 439

42 The cicula sping clamp poduces a compessive foce of 3 N on the plates. Detemine the maximum bending stess poduced in the sping at. The sping has a ectangula coss section as shown. 10 mm 20 mm ntenal oment: s shown on FBD, N # m is positive since it tends to incease the beam s adius of cuvatue. 210 mm 200 mm Section Popeties: m 220 mm L d b ln ln B m (0.01)(0.02) B m 2 R 1 d 0.200(10-3 ) (10-3 ) m - R B m aximum Nomal Stess: pplying the cuved-beam fomula s C (R - 2) 2 ( - R) 0.660( ) 0.200(10-3 )(0.21)( )(10-3 ) -1.95Pa 1.95 Pa (C) s t (R - 1) 1 ( - R) 0.660( ) 0.200(10-3 )(0.2)( )(10-3 ) 2.01 Pa (T) (ax) 440

43 Detemine the maximum compessive foce the sping clamp can exet on the plates if the allowable bending stess fo the clamp is s allow 4 Pa. 10 mm 20 mm 210 mm 200 mm 220 mm Section Popeties: m L d b ln ln B m (0.01)(0.02) B m 2 R 1 d 0.200(10-3 ) (10-3 ) m - R B m ntenal oment: The intenal moment must be computed about the neutal axis as shown on FBD. max P is positive since it tends to incease the beam s adius of cuvatue. llowable Nomal Stess: pplying the cuved-beam fomula ssume compession failue s c s allow (R - 2) 2 ( - R) B P( ) 0.200(10-3 )(0.21)( )(10-3 ) ssume tension failue P N s t s allow (R - 1) 1 ( - R) B P( ) 0.200(10-3 )(0.2)( )(10-3 ) P 3.09 N (Contols!) 441

44 * While in flight, the cuved ib on the jet plane is subjected to an anticipated moment of 16 N # m at the section. Detemine the maximum bending stess in the ib at this section, and sketch a two-dimensional view of the stess distibution. 16 N m 5 mm 20 mm 5 mm 30 mm 0.6 m 5 mm d> (0.03)ln (0.005)ln (0.03)ln L (10-3 ) in. 2(0.005)(0.03) + (0.02)(0.005) 0.4(10-3 ) in 2-3 ) R 1 d> 0.4( (10-3 ) (s c ) max (R - c) ( - R) 16( ) 0.4(10-3 )(0.630)( ) Pa (s s ) max (R - s) ( - R) 16( ) 0.4( Pa )(0.6)( ) f the adius of each notch on the plate is 0.5 in., detemine the lagest moment that can be applied. The allowable bending stess fo the mateial is s allow 18 ksi in. 1 in in. b b h Fom Fig. 6-44: K 2.60 s max K c 1.0 in. 18(10 3 ) 2.60c ()(6.25) 1 d 12 (1)(12.5) lb # in kip # ft 442

45 The symmetic notched plate is subjected to bending. f the adius of each notch is 0.5 in. and the applied moment is 10 kip # ft, detemine the maximum bending stess in the plate. b h in in. 1 in. Fom Fig. 6-44: K 2.60 s max K c 2.60c (10)(12)(6.25) 1 12 (1)(12.5)3 d 12.0 ksi The ba is subjected to a moment of 40 N # m. Detemine the smallest adius of the fillets so that an allowable bending stess of s allow 124 Pa is not exceeded. llowable Bending Stess: 20 mm 80 mm 7 mm s allow K c (0.01) B KB 1 12 (0.007)(0.023 ) R K 1.45 Stess Concentation Facto: Fom the gaph in the text w with and, then. h K 1.45 h mm * The ba is subjected to a moment of 80 mm 17.5 N # m. f 5 mm, detemine the maximum bending 20 mm stess in the mateial. Stess Concentation Facto: Fom the gaph in the text with w and, then K h 5 h aximum Bending Stess: s max K c 17.5(0.01) 1.45B 1 12 (0.007)(0.023 ) R 7 mm 54.4 Pa 443

46 The simply suppoted notched ba is subjected to two foces P. Detemine the lagest magnitude of P that can be applied without causing the mateial to yield. The mateial is -36 steel. Each notch has a adius of in. P 1.25 in. P 0.5 in in. 20 in. 20 in. 20 in. 20 in. b b ; h Fom Fig K 1.92 c s Y K 0.25 ; c 20P(0.625) 1 d 12 (0.5)(1.25)3 P 122 lb The simply suppoted notched ba is subjected to the two loads, each having a magnitude of P 100 lb. Detemine the maximum bending stess developed in the ba, and sketch the bending-stess distibution acting ove the coss section at the cente of the ba. Each notch has a adius of in. P 1.25 in. P 0.5 in in. 20 in. 20 in. 20 in. 20 in. b b ; h Fom Fig. 6-44, K 1.92 c s max K 2000(0.625) 1.92c d 29.5 ksi (0.5)(1.25)

47 Detemine the length L of the cente potion of the ba so that the maximum bending stess at, B, and C is the same. The ba has a thickness of 10 mm. w h h Fom Fig. 6-43, K mm 60 mm 350 N 40 mm 7 mm L L 200 mm 200 mm 2 2 C B (s ) max K c (35)(0.02) 1.5c d Pa 1 12 (0.01)(0.043 ) (s B ) max (s ) max B c (10 6 ) 175(0.2 + L 2 )(0.03) 1 12 (0.01)(0.063 ) L 0.95 m 950 mm * The stepped ba has a thickness of 15 mm. Detemine the maximum moment that can be applied to its ends if it is made of a mateial having an allowable bending stess of s allow 200 Pa. 45 mm 30 mm 3 mm 10 mm 6 mm Stess Concentation Facto: w Fo the smalle section with and, we have K 1.2 h 6 h obtained fom the gaph in the text. w Fo the lage section with and, we have K 1.75 h 3 h obtained fom the gaph in the text. llowable Bending Stess: Fo the smalle section s max s allow K c ; (0.005) B 1.2B 1 12 (0.015)(0.013 ) R 41.7 N # m (Contols!) Fo the lage section s max s allow K c ; (0.015) B 1.75B 1 12 (0.015)(0.033 ) R 257 N # m 445

48 The beam is made of an elastic plastic mateial fo which s Y 250 Pa. Detemine the esidual stess in the beam at its top and bottom afte the plastic moment p is applied and then eleased. x 1 12 (0.2)(0.23) (0.18)(0.2) (10-6 )m 4 C 1 T 1 s Y (0.2)(0.015) 0.003s Y 15 mm 20 mm 200 mm p C 2 T 2 s Y (0.1)(0.02) 0.002s Y p 0.003s Y (0.215) s Y (0.1) s Y (250)(10 6 ) kn # m 200 mm 15 mm s p c (103 )(0.115) (10-6 ) Pa y ; y m 98.0 mm s top s bottom Pa The wide-flange membe is made fom an elasticplastic mateial. Detemine the shape facto. Plastic analysis: T 1 C 1 s Y bt; T 2 C 2 s Y a h - 2t bt 2 t t t h P s Y bt(h - t) + s Y a h - 2t 2 b(t)a h - 2t b 2 b s Y cbt(h - t) + t 4 (h - 2t)2 d Elastic analysis: 1 12 bh3-1 (b - t)(h - 2t) [bh3 - (b - t)(h - 2 t) 3 ] Y s y c s Y 1 12 B[bh 3 - (b - t)(h - 2t) 3 ] h 2 Shape facto: k P Y bh3 - (b - t)(h - 2t) 3 6h [bt(h - t) + t 4 (h - 2t)2 ]s Y bh 3 - (b - t)(h - 2t) 3 6h s Y s Y 3h 2 c 4bt(h - t) + t(h - 2t)2 bh 3 - (b - t)(h - 2t) 3 d 446

49 Detemine the shape facto fo the coss section. aximum Elastic oment: The moment of inetia about neutal axis must be detemined fist. N 1 12 (a)(3a) (2a)a3 B a 4 pplying the flexue fomula with s s Y, we have a a a s Y Y c Y s Y c s Y ( a 4 ) 1.5a a 3 s Y a a a Plastic oment: P s Y (a)(a)(2a) + s Y (0.5a)(3a)(0.5a) Shape Facto: 2.75a 3 s Y k P Y 2.75a3 s Y a 3 s Y 1.71 * The beam is made of elastic pefectly plastic mateial. Detemine the maximum elastic moment and the plastic moment that can be applied to the coss section. Take a 2 in. and s Y 36 ksi. a aximum Elastic oment: The moment of inetia about neutal axis must be detemined fist. N 1 12 (2)63 B (4)23 B in 4 pplying the flexue fomula with s s Y, we have s Y Y c a a a a a Y s Y c Plastic oment: 36(38.667) kip # in 38.7 kip # ft P 36(2)(2)(4) + 36(1)(6)(1) 792 kip # in 66.0 kip # ft 447

50 The box beam is made of an elastic pefectly plastic mateial fo which s Y 250 Pa. Detemine the esidual stess in the top and bottom of the beam afte the plastic moment is applied and then eleased. Plastic oment: p P B (0.2)(0.025)(0.175) B (0.075)(0.05)(0.075) N # m odulus of Ruptue: The modulus of uptue can be detemined using the flexue fomula with the application of evese, plastic moment P N # m. s 25 mm 150 mm 25 mm 150 mm 25 mm 25 mm 1 12 (0.2)0.23 B (0.15)0.153 B B m 4 s P c (0.1) B Pa Residual Bending Stess: s shown on the diagam. s œ top s œ bot s - s Y Pa Detemine the shape facto fo the wideflange beam. x 1 12 (0.2)(0.23) (0.18)(0.2) B m 4 C 1 T 1 s Y (0.2)(0.015) 0.003s Y C 2 T 2 s Y (0.1)(0.02) 0.002s Y 15 mm 20 mm 200 mm p p 0.003s Y (0.215) s Y (0.1) s Y s Y Y c 200 mm 15 mm Y s Y )10-6 B s Y k p Y s Y s Y

51 Detemine the shape facto of the beam s coss section. Refeing to Fig. a, the location of centoid of the coss-section is 3 in. y y 7.5(3)(6) + 3(6)(3) 3(6) + 6(3) 5.25 in. 6 in. The moment of inetia of the coss-section about the neutal axis is 1 12 (3)63 B + 3(6)(5.25-3) (6)33 B + 6(3)( ) in 4 Hee s max s Y and c y 5.25 in. Thus s max c ; s Y Y (5.25) Y s Y Refeing to the stess block shown in Fig. b, 1.5 in. 3 in. 1.5 in. sd 0; T - C 1 - C 2 0 L d(3)s Y - (6 - d)(3)s Y - 3(6)s Y 0 d 6 in. Since d 6 in., c 1 0, Fig. c. Hee Thus, T C 3(6) s Y 18 s Y Thus, P T(4.5) 18 s Y (4.5) 81 s Y k P Y 81 s Y s Y

52 * The beam is made of elastic-pefectly plastic mateial. Detemine the maximum elastic moment and the plastic moment that can be applied to the coss section. Take s Y 36 ksi. Refeing to Fig. a, the location of centoid of the coss-section is y y 7.5(3)(6) + 3(6)(3) 3(6) + 6(3) 5.25 in. 3 in. 6 in. The moment of inetia of the coss-section about the neutal axis is 1 12 (3)(63 ) + 3(6)(5.25-3) (6)(33 ) + 6(3)( ) in 4 Hee, s max s Y 36 ksi and y 5.25 in. Then s max c ; 36 Y (5.25) in. 3 in. 1.5 in. Y kip # in 143 kip # ft Refeing to the stess block shown in Fig. b, Since d 6 in., c 1 0, Hee, Thus, sd 0; T - C 1 - C 2 0 L d(3) (36) - (6 - d)(3)(36) - 3(6) (36) 0 T C 3(6)(36) 648 kip d 6 in. P T(4.5) 648(4.5) 2916 kip # in 243 kip # ft 450

53 Detemine the shape facto fo the coss section of the H-beam. x 1 12 (0.2)(0.023 ) + 2a 1 12 b(0.02)(0.23 ) 26.8(10-6 )m mm C 1 T 1 s Y (2)(0.09)(0.02) s y C 2 T 2 s Y (0.01)(0.24) s y 20 mm p 200 mm 20 mm p s Y (0.11) s Y (0.01) s Y s Y Yc 20 mm Y s Y(26.8)(10-6 ) s Y k p Y s Y s Y The H-beam is made of an elastic-plastic mateial fo which s Y 250 Pa. Detemine the esidual stess in the top and bottom of the beam afte the plastic moment p is applied and then eleased. x 1 12 (0.2)(0.023 ) + 2a 1 12 b(0.02)(0.23 ) 26.8(10-6 )m 4 C 1 T 1 s Y (2)(0.09)(0.02) s y 20 mm p 200 mm 200 mm 20 mm C 2 T 2 s Y (0.01)(0.24) s y 20 mm p s Y (0.11) s Y (0.01) s Y p (250)10 6 B 105 kn # m s p c 105(103 )(0.1) 26.8(10-6 ) 392 Pa y ; y mm 392 s T s B Pa 451

54 Detemine the shape facto of the coss section. The moment of inetia of the coss-section about the neutal axis is 1 12 (3)(93 ) (6) (33 ) in 4 Hee, s max s Y and c 4.5 in. Then 3 in. 3 in. 3 in. s max c ; s Y Y(4.5) Y 43.5 s Y Refeing to the stess block shown in Fig. a, 3 in. 3 in. 3 in. T 1 C 1 3(3)s Y 9 s Y T 2 C 2 1.5(9)s Y 13.5 s Y Thus, P T 1 (6) + T 2 (1.5) 9s Y (6) s Y (1.5) s Y k P Y s Y 43.5 s Y

55 * The beam is made of elastic-pefectly plastic mateial. Detemine the maximum elastic moment and the plastic moment that can be applied to the coss section. Take s Y 36 ksi. The moment of inetia of the coss-section about the neutal axis is 1 12 (3)(93 ) (6)(33 ) in 4 3 in. 3 in. 3 in. Hee, s max s Y 36 ksi and c 4.5 in. Then Refeing to the stess block shown in Fig. a, Thus, s max c ; 36 Y (4.5) T 1 C 1 3(3)(36) 324 kip T 2 C 2 1.5(9)(36) 486 kip P T 1 (6) + T 2 (1.5) 324(6) + 486(1.5) Y 1566 kip # in kip # ft 2673 kip # in kip # ft 223 kip # ft 3 in. 3 in. 3 in. 453

56 Detemine the shape facto of the coss section fo the tube. The moment of inetia of the tube s coss-section about the neutal axis is p 4 o 4 - i4 B p B p in 4 Hee, s max s Y and C o 6 in, 5 in. 6 in. s max c ; s Y Y (6) p Y s Y The plastic oment of the table s coss-section can be detemined by supe posing the moment of the stess block of the solid cicula coss-section with adius o 6 in and i 5 in. as shown in Figue a, Hee, T 1 C p(62 )s Y 18ps Y Thus, T 2 C p(52 )s Y 12.5p s Y P T 1 b 2c 4(6) 3p d - T 2b 2c 4(5) 3p d (18ps Y )a 16 p b ps Ya 40 3p b s Y k P Y s Y s Y

57 The beam is made fom elastic-pefectly plastic mateial. Detemine the shape facto fo the thick-walled tube. aximum Elastic oment. The moment of inetia of the coss-section about the neutal axis is o i p 4 o 4 - i4 B With c o and s max s Y, s max c ; s Y Y Y ( o ) p 4 o 4 - i 4 B p 4 o o 4 - i 4 Bs Y Plastic oment. The plastic moment of the coss section can be detemined by supeimposing the moment of the stess block of the solid beam with adius 0 and i as shown in Fig. a, Refeing to the stess block shown in Fig. a, T 1 c 1 p 2 o 2 s Y T 2 c 2 p 2 i 2 s Y P T 1 c2a 4 o 3p bd - T 2c2a 4 i 3p bd p 2 o 2 s Y a 8 o 3p b - p 2 i 2 s Y a 8 i 3p b 4 3 o 3 - i 3 Bs Y Shape Facto. k P Y 4 3 o 3-3 i Bs Y 16 o 3 o - 3 i B p 4 4 o - 4 i Bs 3p 4 o - 4 i Y B o 455

58 Detemine the shape facto fo the membe. Plastic analysis: T C 1 2 (b)a h 2 bs Y b h 4 s Y P b h 4 s Ya h 3 b b h2 12 s Y h 2 h 2 Elastic analysis: 2c 1 12 (b)a h 3 2 b d b h3 48 b Y s Y c s Y bh3 48 B h 2 b h2 24 s Y Shape facto: k p Y bh 2 12 s Y 2 bh 2 24 s Y * The membe is made fom an elastic-plastic mateial. Detemine the maximum elastic moment and the plastic moment that can be applied to the coss section. Take b 4 in., h 6 in., s Y 36 ksi. h 2 Elastic analysis: 2c 1 12 (4)(3)3 d 18 in 4 h 2 Y s Y c 36(18) kip # in. 18 kip # ft Plastic analysis: b T C 1 (4)(3)(36) 216 kip 2 p 2160a 6 3 b 432 kip # in. 36 kip # ft 456

59 The beam is made of a mateial that can be assumed pefectly plastic in tension and elastic pefectly plastic in compession. Detemine the maximum bending moment that can be suppoted by the beam so that the compessive mateial at the oute edge stats to yield. h s Y sd 0; C - T 0 L 1 2 s Y(d)(a) - s Y (h - d)a 0 a s Y d 2 3 h 1 2 s Ya a h2 hb(a)a hb s Y The box beam is made fom an elastic-plastic mateial fo which s Y 25 ksi. Detemine the intensity of the distibuted load w 0 that will cause the moment to be (a) the lagest elastic moment and (b) the lagest plastic moment. w 0 Elastic analysis: 1 12 (8)(163 ) (6)(123 ) in 4 max s Y c w kip>ft ; 27w 0 (12) 25( ) 8 9 ft 9 ft 12 in. 8 in. 16 in. Plastic analysis: C 1 T 1 25(8)(2) 400 kip 6 in. C 2 T 2 25(6)(2) 300 kip P 400(14) + 300(6) 7400 kip # in. 27w 0 (12) 7400 w kip>ft 457

60 The box beam is made fom an elastic-plastic mateial fo which s Y 36 ksi. Detemine the magnitude of each concentated foce P that will cause the moment to be (a) the lagest elastic moment and (b) the lagest plastic moment. P P Fom the moment diagam shown in Fig. a, max 6 P. The moment of inetia of the beam s coss-section about the neutal axis is 1 12 (6)(123 ) (5)(103 ) in 4 Hee, s max s Y 36 ksi and c 6 in. s max c ; 36 Y (6) t is equied that Y 2684 kip # in kip # ft 6 ft 8 ft 6 in. 10 in. 12 in. 5 in. 6 ft max Y 6P P kip 37.3 kip Refeing to the stess block shown in Fig. b, T 1 C 1 6(1)(36) 216 kip T 2 C 2 5(1)(36) 180 kip Thus, P T 1 (11) + T 2 (5) 216(11) + 180(5) t is equied that 3276 kip # in 273 kip # ft max P 6P 273 P 45.5 kip 458

61 * The beam is made of a polyeste that has the stess stain cuve shown. f the cuve can be epesented by the equation s [20 tan P2] ksi, whee tan P2 is in adians, detemine the magnitude of the foce P that can be applied to the beam without causing the maximum stain in its fibes at the citical section to exceed P max in.>in. P 8 ft 8 ft 2 in. 4 in. s(ksi) s 20 tan 1 (15 P) P(in./in.) aximum ntenal oment: The maximum intenal moment 4.00P occus at the mid span as shown on FBD. Stess Stain Relationship: Using the stess stain elationship. the bending stess can be expessed in tems of y using e y. s 20 tan - 1 (15e) 20 tan - 1 [15(0.0015y)] 20 tan - 1 (0.0225y) When e max in.>in., y 2 in. and s max ksi Resultant ntenal oment: The esultant intenal moment can be evaluated fom the integal ysd. L 2 L ysd Equating 2in 2 yc20 tan - 1 (0.0225y)D(2dy) L 0 2in 80 y tan - 1 (0.0225y) dy L 0 80B 1 + (0.0225)2 y 2 2(0.0225) 2 tan - 1 (0.0225y) kip # in 4.00P(12) y 2(0.0225) R 2 2in. 0 P kip 100 lb 459

62 The plexiglass ba has a stess stain cuve that can be appoximated by the staight-line segments shown. Detemine the lagest moment that can be applied to the ba befoe it fails. 20 mm 20 mm compession s (Pa) 60 failue 40 tension P (mm/mm) Ultimate oment: s d 0; C - T 2 - T 1 0 L sc 1 2 ( d)(0.02) d B c 1 2 a d 2 b(0.02) d (60 + d 40)106 B c(0.02) 2 d 0 ssume. s Pa; d m Fom the stain diagam, e e mm>mm Fom the stess stain diagam, s (OK! Close to assumed value) s Pa 0.04 Theefoe, s - 50s d (10 6 )d 0 C B c 1 ( )(0.02) d N 2 T ( ) 106 B c(0.02)a bd N 2 T B c (0.02)a bd N 2 y 1 2 ( ) m 3 y a b m 2 y c1-1 2(40) + 60 a bda b m ( ) ( ) ( ) 94.7 N # m 460

63 The stess stain diagam fo a titanium alloy can be appoximated by the two staight lines. f a stut made of this mateial is subjected to bending, detemine the moment esisted by the stut if the maximum stess eaches a value of (a) and (b). s s B 3 in. s (ksi) s B 180 s in. B P (in./in.) a) aximum Elastic oment : Since the stess is linealy elated to stain up to point, the flexue fomula can be applied. s c s c 140C 1 12 (2)(33 )D kip # in 35.0 kip # ft b) The Ultimate oment : C 1 T 1 1 ( )(1.125)(2) 360 kip 2 C 2 T 2 1 (140)(0.375)(2) 52.5 kip 2 360( ) (0.5) kip # in 59.8 kip # ft Note: The centoid of a tapezodial aea was used in calculation of moment. 461

64 beam is made fom polypopylene plastic and has a stess stain diagam that can be appoximated by the cuve shown. f the beam is subjected to a maximum tensile and compessive stain of P0.02 mm>mm, detemine the maximum moment. s(pa) e max 0.02 s 10(10 6 )P 1/4 100 mm s max B(0.02) 1> Pa 30 mm e y P (mm/mm) e 0.4 y s B(0.4) 1>4 y 1> y s d 2 y(7.9527)10 6 By 1>4 (0.03)dy L L B y 5>4 dy B a 4 L 5 b(0.05)9> N # m * The beam has a ectangula coss section and is made of an elastic-plastic mateial having a stess stain diagam as shown. Detemine the magnitude of the moment that must be applied to the beam in ode to ceate a maximum stain in its oute fibes of P max mm 200 mm s(pa) P (mm/mm) C 1 T B(0.1)(0.2) 4000 kn C 2 T (200)106 B(0.1)(0.2) 2000 kn 4000(0.3) (0.1333) 1467 kn # m 1.47 N # m 462

65 The ba is made of an aluminum alloy having a stess stain diagam that can be appoximated by the staight line segments shown. ssuming that this diagam is the same fo both tension and compession, detemine the moment the ba will suppot if the maximum stain at the top and bottom fibes of the beam is P max s(ksi) in. s ; s 82 ksi C 1 T 1 1 (0.3333)( )(3) 81 kip P(in./in.) in. C 2 T 2 1 (1.2666)( )(3) 266 kip 2 C 3 T 3 1 (0.4)(60)(3) 36 kip 2 81(3.6680) + 266(2.1270) + 36(0.5333) kip # in kip # ft Note: The centoid of a tapezodial aea was used in calculation of moment aeas The beam is made fom thee boads nailed togethe as shown. f the moment acting on the coss section is 650 N # m, detemine the esultant foce the bending stess poduces on the top boad. Section Popeties: 15 mm y (0.29)(0.015) + 2[0.0775(0.125)(0.02)] 0.29(0.015) + 2(0.125)(0.02) 125 mm 650 N m 20 mm m N 1 12 (0.29) B (0.015) ( ) 2 20 mm 250 mm (0.04) B (0.125)( ) B m 4 Bending Stess: pplying the flexue fomula s y s B 650( ) (10-6 ) Pa s 650( ) Pa (10-6 ) Resultant Foce: F R 1 2 ( )106 B (0.015)(0.29) 5883 N 5.88 kn 463

66 The beam is made fom thee boads nailed togethe as shown. Detemine the maximum tensile and compessive stesses in the beam. 15 mm 125 mm 650 N m 20 mm 20 mm 250 mm Section Popeties: y (0.29)(0.015) + 2[0.0775(0.125)(0.02)] 0.29(0.015) + 2(0.125)(0.02) m N 1 12 (0.29) B (0.015)( ) (0.04) B (0.125)( ) B m 4 aximum Bending Stess: pplying the flexue fomula s y (s max ) t 650( ) (10-6 ) 3.43 Pa (T) (s max ) c 650( ) 1.62 Pa (C) (10-6 ) 464

67 * Detemine the bending stess distibution in the beam at section a a. Sketch the distibution in thee dimensions acting ove the coss section. 80 N 80 N a a 400 mm 300 mm 300 mm 400 mm a+ 0; - 80(0.4) 0 32 N # m z 1 12 (0.075)( ) + 2a 1 12 b(0.015)(0.13 ) (10-6 )m 4 s max c 32(0.05) 635 kpa (10-6 ) 80 N 80 N 15 mm 100 mm 15 mm 75 mm The composite beam consists of a wood coe and two plates of steel. f the allowable bending stess fo the wood is (s allow ) w 20 Pa, and fo the steel (s allow ) st 130 Pa, detemine the maximum moment that can be applied to the beam. E w 11 GPa, E st 200 GPa. z 125 mm y n E st E w 200(109 ) 11(10 9 ) ( )( ) (10-3 )m 4 Failue of wood : (s w ) max c x 20 mm 75 mm 20 mm 20(10 6 ) (0.0625) (10-3 ) ; 41.8 kn # m Failue of steel : (s st ) max nc 130(10 6 ) ()(0.0625) (10-3 ) 14.9 kn # m (contols) 465