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PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators ...

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PROBLEM 4.127 The couple M is applied to a beam of the cross section shown in a plane forming an angle β with the vertical. Determine the stress at (a) point A, (b) point B, (c) point D.

SOLUTION 1 (80)(100)3 = 6.6667 × 106 mm4 = 6.6667 × 10−6 m4 12 1 I y = (100)(80)3 = 4.2667 × 106 mm4 = 4.2667 × 10−6 m4 12 y A = − yB = − yD = 50 mm z A = z B = −z D = 40 mm Iz =

M y = −250 sin 30° = −125 N ⋅ m M z = 250 cos 30° = 216.51 N ⋅ m

(a)

σA = −

M z yA M y z A (216.51)(0.050) (−125)(0.040) + =− + Iz Iy 6.6667 × 10− 6 4.2667 × 10− 6

= −2.80 ×106 Pa (b)

σB = −

M z yB M y z B (216.51)(−0.050) (−125)(0.040) + =− + 6 6 Iz Iy 6.6667 ×10 − 4.2667 ×10 −

= 0.452 ×103 Pa (c)

σD = −

σ A = −2.80 MPa 

σ B = 0.452 MPa 

M z yD M y z D (216.51)(−0.050) (−125)(−0.040) + =− + 6 6 Iz Iy 6.6667 × 10− 4.2667 × 10−

= 2.80 ×10 6Pa

σ D = 2.80 MPa 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 4.128 The couple M is applied to a beam of the cross section shown in a plane forming an angle β with the vertical. Determine the stress at (a) point A, (b) point B, (c) point D.

SOLUTION 1 (80)(32)3 = 218.45 × 103 mm 4 = 218.45 × 10 −9 m4 12 1 I y = (32)(80)3 = 1.36533 × 106 mm4 = 1.36533 × 10 −6 m4 12 y A = yB = − yD = 16 mm z A = −z B = −z D = 40 mm Iz =

M y = 300cos30° = 259.81 N ⋅ m (a)

σA = −

M z = 300sin 30 ° = 150 N ⋅ m

M z y A M y zA (150)(16 × 10−3 ) (259.81)(40 × 10−3 ) + =− + Iz Iy 218.45 × 10−9 1.36533 × 10−6

= −3.37 ×10 6 Pa (b)

σB = −

M z yB Iz

+

M y zB Iy

=−

(150)(16 × 10 −3 ) 218.45× 10−9

σ A = −3.37 MPa  +

(259.81)(−40 × 10 −3 ) 1.36533× 10−6

= −18.60 ×10 6 Pa (c)

σD = −

σ B = −18.60 MPa 

M zy D M y z D (150)( −16 ×10 −3 ) (259.81)( −40 ×10 −3 ) + =− + Iz Iy 218.45 × 10− 9 1.36533 × 10− 6

= 3.37 ×10 6 Pa

σ D = 3.37 MPa 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 4.130 The couple M is applied to a beam of the cross section shown in a plane forming an angle β with the vertical. Determine the stress at (a) point A, (b) point B, (c) point D.

SOLUTION Locate centroid A, mm2 10000 5000 15000

  Σ

z , mm −25 50

A z , mm 3

−250000 250000 0

The centroid lies at point C 1 1 (50)(200)3 + (100)(50)3 = 34.375 × 106 mm4 12 12 1 3 1 I y = (200)(50) + (50)(100) 3 = 25 × 10 6 mm 4 3 3 y A = − y B = 25 mm, y D = − 100 mm z A = z B = −100 mm, z D = 0 Iz =

M z = 1 cos 20° = 0.94 kNm M y = 1 sin 20° = 0.342 kNm

(a)

σA = −

Mz yA M y z A (940)(0.025) (342)( −0.1) + =− + = 2.05 MPa Iz Iy 34.375 × 10− 6 25 × 10− 6



(b)

σB = −

Mz yB M y zB (940)( −0.025) (342)( −0.1) + =− + = −0.684 MPa Iz Iy 34.375× 10−6 25× 10−6



(c)

σD = −

Mz yD M y zD (940)(− 0.1) (342)(0) + =− + = 2.73 MPa −6 Iz Iy 34.375× 10 25× 10−6



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 4.131 The couple M is applied to a beam of the cross section shown in a plane forming an angle β with the vertical. Determine the stress at (a) point A, (b) point B, (c) point D.

SOLUTION M y = 25sin 15 ° = 6.4705 kN ⋅ m M z = 25cos15 ° = 24.148 kN ⋅ m 1 1 (80)(90)3 + (80)(30) 3 = 5.04 × 10 6 mm 4 12 12 4 −6 5.04 10 m = × Iy Iy =

1 1 1 I z = (90)(60) 3 + (60)(20) 3 + (30)(100) 3 =16.64 ×10 6 mm 4 = 16.64 ×10−6 m 4 3 3 3

Stress:

(a)

σA =

σ =

M yz Iy

−

M zy Iz

(6.4705 kN ⋅ m)(0.045 m) (24.148 kN ⋅ m)(0.060 m) − 5.04 ×10 −6 m 4 16.64 ×10 −6 m 4

σ A = − 29.3 MPa 

= 57.772 MPa − 87.072 MPa

(b)

σB =

(6.4705 kN ⋅ m)(− 0.045 m) −6

5.04 × 10

m

4

−

(24.148 kN ⋅ m)(0.060 m) −6

16.64× 10

m4

= −57.772 MPa − 87.072 MPa

(c)

σD =

σ B = − 144.8 MPa 

(6.4705 kN ⋅ m)(− 0.015 m) (24.148 kN⋅ m)(− 0.100 m) − 5.04 ×10 −6 m 4 16.64 ×10 −6 m 4

= −19.257 MPa +145.12 MPa

σ D = − 125.9 MPa 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 4.132 The couple M is applied to a beam of the cross section shown in a plane forming an angle β with the vertical. Determine the stress at (a) point A, (b) point B, (c) point D.

SOLUTION 1 (200)(12)3 + (200)(12)(119)2 12 = 34015200 mm 4

Flange:

Iz =

Web:

1 (12)(200)3 = 8× 106 mm4 12 1 I z = (8)(226) 3 = 7695451 mm 4 12 1 3 4 I y = (226)(8) = 9643 mm 12 Iy =

I z = (2)(34015200) + 7695451 = 75725851 mm 4

Total:

I y = (2)(8 × 106 ) + 9643 = 16009643 mm4 y A = yB = − yD = 125 mm; z A = − z B = − zC = 100 mm M z = 28 cos 30° = 24.25 kNm M y = − 28 sin 30° = − 14 kNm

(a)

σA = −

Mz yA M y z A (24.25 ×10 3)(0.125) ( −14 ×10 3)(0.1) + =− + = − 127.5 MPa −12 −12 Iz Iy 75725851× 10 16009643× 10



(b)

σB = −

Mz yB M yz B (24250)(0.125) ( −14000)( −0.1) + =− + = 47.4 MPa −12 Iz Iy 75725851 × 10 16009643 ×10 −12



(c)

σD = −

Mz yD M y zD (24250)( −0.125) ( −14000)( −0.1) + =− + = 127.5 MPa Iz Iy 75725851× 10− 12 16009643× 10− 12



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 4.133 The couple M is applied to a beam of the cross section shown in a plane forming an angle β with the vertical. Determine the stress at (a) point A, (b) point B, (c) point D.

SOLUTION 1 1 (140)(70)3 − (120)(50)3 = 2.7517 × 106 mm4 12 12 − = 2.7517 × 10 6 m4 1 1 I y = (70)(140)3 − (50)(120)3 = 8.8067 × 106 mm4 12 12 = 8.8067 × 10−6 m4 Iz =

yA = yB = − yD = 35 mm = 0.035 m zA = − zB = − zD = 70 mm = 0.070 m 3 3 M z = M sin15 ° = (15 ×10 ) sin 15° = 3.8823 ×10 N ⋅ m 3 3 My = M cos15 ° = (15 ×10 ) cos 15° =14.4889 ×10 N ⋅m

(a)

σA = −

M z yA M y z A (3.8823× 10 3)(0.035) (14.4889× 10 3)(0.070) + =− + − − Iz Iy 2.7517 × 10 6 8.8067 × 10 6

= 65.8 ×10 6 Pa (b)

σB = −

σ A = 65.8 MPa 

M z yB M y z B (3.8823× 10 3)(0.035) (14.4889× 10 3)(− 0.070) + =− + − − Iz Iy 2.7517 × 10 6 8.8067 × 10 6

= −164.5 ×10 6 Pa

σB = −164.5 MPa 

M z yD M y z D (3.8823 ×10 )( −0.035) (14.4889 ×10 )( −0.070) + =− + Iz Iy 2.7517 ×10 −6 8.8067 ×10 −6 3

(c)

σD = −

6 = − 65.8× 10 Pa

3

σ D = −65.8 MPa 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 4.135 The couple M acts in a vertical plane and is applied to a beam oriented as shown. Determine (a) the angle that the neutral axis forms with the horizontal, (b) the maximum tensile stress in the beam.

SOLUTION For C 200 × 17.1 rolled steel shape,

I z = 0.538 ×10 6mm 4 = 0.538 ×10 − 6m 4 6 4 6 4 I y = 13.4 × 10 mm = 13.4 × 10− m

zE = zD = −zA = −zB = yD = yB = −14.4 mm

1 (203) = 101.5 mm 2 yE = y A = 57 − 14.4 = 42.6 mm

M z = (2.8 ×103 ) cos 10 ° = 2.7575 ×10 3 N ⋅ m My = (2.8 ×103 ) sin 10 ° = 486.21 N ⋅ m

(a)

Angle of neutral axis. tanϕ =

0.538 Iz tanθ = tan10° = 0.007079 Iy 13.4

ϕ = 0.4056° α = 9.59 ° 

α = 10° − 0.4056°

(b)

Maximum tensile stress occurs at point D.

σD = −

Mz yD M y zD (2.7575 × 10 3 )( −14.4 × 10 −3 ) (486.21)(0.1015) + =− + Iz Iy 0.538× 10− 6 13.4 × 10− 6 6

6

6

= 73.807 × 10 + 3.682 × 10 = 77.5 ×10 Pa

σ D = 77.5 MPa 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 4.136 The couple M acts in a vertical plane and is applied to a beam oriented as shown. Determine (a) the angle that the neutral axis forms with the horizontal, (b) the maximum tensile stress in the beam.

SOLUTION For W 310 × 38.7 rolled steel shape, I z = 85.1× 106 mm4 = 85.1× 10−6 m4 I y = 7.27× 106 mm4 = 7.27× 10 −6 m4 1 y A = y B = − y D = − y E =   (310) = 155 mm 2  1 zA = zE = − zB = − zD =   (165)= 82.5 mm  2 M z = (16 × 10 3 ) cos 15° = 15.455× 10 3 N ⋅ m M y = (16 × 10 3 ) sin 15° = 4.1411× 10 3 N ⋅ m

(a)

Angle of neutral axis. tan ϕ =

Iz 85.1× 10− 6 tan θ = tan15° = 3.1365 − Iy 7.27 ×10 6

ϕ = 72.3° α = 72.3° − 15° (b)

α = 57.3° 

Maximum tensile stress occurs at point E.

σE = −

Mz yE M yz E + Iz Iy (15.455 ×10 )( −155 ×10− ) (4.1411 ×10 )(82.5 ×10− ) + 85.1× 10−6 7.27 × 10−6 3

=−

= 75.1 × 10 6 Pa

3

3

3

σ E = 75.1 MPa 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 4.138 The couple M acts in a vertical plane and is applied to a beam oriented as shown. Determine (a) the angle that the neutral axis forms with the horizontal, (b) the maximum tensile stress in the beam.

SOLUTION Iz ′ = 176.9 × 103 mm 4 = 176.9 × 10− 9 m 4 Iy ′ = 281 × 103 mm4 = 281 × 10 −9 m4 y′E = −18.57 mm, z E = 25 mm Mz ′ = 400cos 30° = 346.41 N ⋅ m M y′ = 400sin 30° = 200 N ⋅ m

(a)

− Iz ′ 176.9 × 10 9 tanθ = ⋅ tan 30° = 0.36346 Iy ′ 281 × 10 − 9

tanϕ =

ϕ = 19.97° α = 30° − 19.97° (b)

α = 10.03° 

Maximum tensile stress occurs at point E.

σE = −

M z′ y′E My′ zE′ (346.41)(− 18.57 × 10 −3 ) (200)(25 × 10 −3 ) + =− + Iz′ Iy′ 176.9 × 10 −9 281 × 10 −9 = 36.36 × 10 6 + 17.79 × 10 6 = 54.2 × 10 6 Pa

σ E = 54.2 MPa 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 4.139 The couple M acts in a vertical plane and is applied to a beam oriented as shown. Determine (a) the angle that the neutral axis forms with the horizontal plane, (b) the maximum tensile stress in the beam.

SOLUTION For S150 × 18.6 rolled steel shape

I z = − 9.11× 10 6mm 4= 9.11× 10− 6 m 4 I y = 0.782× 10 6mm 4 = 0.782× 10 − 6 m 4

 1 z E = −z A = −z B = z D =   (85) = 42.5 mm  2 1 y A = y B = −y D = −y E =   (152) = 76 mm 2

Mz = 1.5 × 103 sin 20° = 0.51303 ×103 N ⋅ m M y = (1.5 ×10 3) cos 20 ° =1.4095 ×10 3 N ⋅ m

(a)

tan φ =

6 Iz 9.11× 10− tanθ = tan(90° − 20° ) = 32.007 Iy 0.782 × 10−6

φ = 88.21 α = 88.21 − 70° = 18.21° (b)



Maximum tensile stress occurs at point D.

σD = −

M yZ D M z yD (0.51303 × 10 3)( −76 × 10 −3) (1.4095 × 10 3)(42.5 × 10 − 3) + =− + Iz Iy 9.11 × 10− 6 0.782 × 10− 6

= 80.9 × 106 Pa = 80.9 MPa



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 4.140 The couple M acts in a vertical plane and is applied to a beam oriented as shown. Determine (a) the angle that the neutral axis forms with the horizontal, (b) the maximum tensile stress in the beam.

SOLUTION I z′ = 53.6× 103 mm4 = 53.6× 10−9 m4 3 4 9 4 I y′ = 14.77 × 10 mm = 14.77 × 10− m

M z′ = 120 sin 70° = 112.763 N ⋅ m M y′ = 120 cos 70° = 41.042 N ⋅ m

(a)

Angle of neutral axis.

θ = 20° . tan ϕ =

I z′ I y′

tan θ =

53.6 × 10−9 tan 20° = 1.32084 14.77 ×10−9

ϕ = 52.871° α = 52.871° − 20°

(b)

α = 32.9 ° 

The maximum tensile stress occurs at point E. yE′ = −16 mm = −0.016 m zE =10 mm = 0.010 m M ′ z′ M y′ σ E = − z′ E + y E I z′ I y′ =−

(112.763)( −0.016) −9

53.6 × 10

= 61.448 ×10 6Pa

+

(41.042)(0.010) 14.77 × 10− 9

σ E = 61.4 MPa 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 4.141 The couple M acts in a vertical plane and is applied to a beam oriented as shown. Determine the stress at point A.

SOLUTION 1 3 6 4 I y = 2  (180)(60)  = 25.92 × 10 mm 6  1  I z = 2  (60)(180) 3 + (60)(180)(30) 2  = 77.76 × 10 6 mm 4 12   I

yz

= 2 {(60)(180)(30)(30)} = 19.44 × 106 mm4

Using Mohr’s circle determine the principal axes and principal moments of inertia. Y = (25.92 × 106 mm4 ,19.44 × 106 mm4 ) Z = (77.76 × 106 mm4 , − 19.44 × 106 mm4 ) 6

4

E = (51.84 × 10 mm , 0)

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 4.141 (Continued)

DY 19.44 = DE 25.92 2θm = 36.87 ° θm =18.435 °

tan 2θm =

2

2

R = DE + DY = 32.4 × 106 mm6 I u = (51.84 − 32.4)10 6 = 19.44 × 10 6 mm 4 I v = (51.84 + 32.4)10 6 = 84.24 × 10 6 mm 4 Mu = 14sin18.435° = 4.43 kNm M v = 14cos18.435° = 13.3 kNm u A = 0.12 cos 18.435° + 0.06 sin 18.435° = 0.1328 m

ν A = − 0.12 sin 18.435° + 0.06 cos 18.435° = 0.01897 m M u Mν σA = − ν A + u A Iν

=−

Iu

(13300)(0.1328) −6

84.24× 10

= − 16.64 MPa

+

(4430)(0.01897) 19.44× 10−6

σ A = −16.6 MPa 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 4.142 The couple M acts in a vertical plane and is applied to a beam oriented as shown. Determine the stress at point A.

SOLUTION Using Mohr’s circle, determine the principal axes and principal moments of inertia

I yz(106 ) mm 4 Y (3.65, 3.45) 10 6 mm 4 6 4 Z (10.1, 3.45) 10 mm

E (6.875, 0) 10 6 mm 4 6 4 EF = 3.225 ×10 mm 6 4 FZ = 3.45 ×10 mm

3.45 FZ = = 1.0698 EF 3.225 Iu = (6.875 − 4.72)10 4 = 2.155 × 10 6 mm 4 , Iv = (6.875 + 4.72)10 6 = 11.6 × 106 mm4

R = 3.225 2 + 3.45 2 = 4.72 × 10 6 mm 4

θ m = 23.5°

tan 2θ m =

Mu = (6.8) sin 23.5° = 2.71 kNm M v = (6.8) cos 23.5° = 6.24 kNm uA = yA cos θm + zA sin θm = −98cos 23.5 ° − 27 sin 23.5 ° = −100.6 mm v A = z A cos θm − yA sin θm = − 27 cos 23.5° + 98 sin 23.5° = 14.3 mm

σA = −

M vu A M u v A (6240)(− 0.1006) (2710)(0.0143) + =− + Iv Iu 11.6 ×10 −6 2.155× 10−6

= 72.1 MPa



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their indiv...