Mechanics of materials an integrated learning system 4th edition philpot solutions manual

Mechanics of Materials An Integrated Learning System 4th Edition Philpot Solutions Manual Visit to download the full and correct content document: https://testbankdeal.com/dow nload/mechanics-of-materials-an-integrated-learning-system-4th-edition-philpot-soluti ons-manual/

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.1 During fabrication of a laminated timber arch, one of the 10 in. wide by 1 in. thick Douglas fir [E = 1,900 ksi] planks is bent to a radius of curvature of 40 ft. Determine the maximum bending stress developed in the plank.

Solution From Eq. (8.3): E x = − y = −



1,900 ksi ( 0.5 in.) = 1.979 ksi = 1.979 ksi ( 40 ft )(12 in./ft )

Ans.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.2 A copper wire of diameter d = 2 mm is coiled around a spool of radius r. The elastic modulus of the copper is E = 117 GPa and its yield strength is 310 MPa. Determine the minimum spool radius r that may be used if the bending stress in the wire is not to exceed the yield strength.

Solution From Eq. (8.3): E 117,000 MPa  =− y=− ( 2 mm / 2 ) = 377.419 mm x 310 MPa

 =r+

d 2

r =  −

d 2 mm = 377.419 mm − = 376.419MPa = 376 mm 2 2

Ans.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.3 The boards for a concrete form are to be bent into a circular shape having an inside radius of 10 m.

What maximum thickness can be used for the boards if the normal stress is not to exceed 7 MPa? Assume that the modulus of elasticity for the wood is 12 GPa.

Solution The radius of curvature of the concrete form is dependent on the board thickness: t  = 10,000 mm + 2 From Eq. (8.3): E 12,000 MPa  t  x = − y = −  7 MPa t   2    10,000 mm +  2 Solve for t: t t  12,000 MPa    7 MPa 10,000 mm +   2  2 6,000t  70,000 + 3.5t 5,996.5t  70,000  t  11.67 mm

Ans.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.4 A beam is subjected to equal bending moments of Mz = 3,200 N·m, as shown in Figure P8.4a. The cross-sectional dimensions (Figure P8.4b) are b = 150 mm, c = 30 mm, d = 70 mm, and t = 6 mm.

Determine: (a) the centroid location, the moment of inertia about the z axis, and the controlling section modulus about the z axis. (b) the bending stress at point H. State whether the normal stress at H is tension or compression. (c) the bending stress at point K. State whether the normal stress at K is tension or compression. (d) the maximum bending stress produced in the cross section. State whether the stress is tension or compression.

FIGURE P8.4a

FIGURE P8.4b

Solution

(a) Centroid location in y direction: (reference axis at bottom of shape) yi Shape Area Ai (from bottom) yi Ai 2 (mm ) (mm) (mm3) (1) 828 3 2,484 (2) 840 35 29,400 (3) 288 67 19,296 1,956 51,180 3 yi Ai 51,180 mm y= = = 26.166 mm = 26.2 mm Ai 1,956 mm 2

Ans.

(measured upward from bottom edge of stem) Moment of inertia about the z axis: d = y − yi Shape IC (mm4) (mm) (1) 2,484 23.166 (2) 343,000 –8.834

d²A (mm4) 444,343.774 65,558.508

IC + d²A (mm4) 446,827.774 408,558.508

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed. 864 –40.834 480,224.049 Moment of inertia about the z axis (mm4) =

(3)

Timothy A. Philpot 481,088.049 1,336,474.331

I z = 1,336,000 mm4

Ans.

Section moduli: I 1,336, 474.331 mm 4 S top = z = = 30, 489.197 mm3 ctop 70 mm − 26.166 mm

I z 1,336, 474.331 mm 4 Sbot = = = 51, 077.448 mm3 cbot 26.166 mm  S = 30,500 mm3

Ans.

(b) Bending stress at point H: yH = d − t − y = 70 mm − 6 mm − 26.166 mm = 37.834 mm

x = − =−

My Iz

( 3, 200 N  m )( 37.834 mm )(1,000 mm/m ) = −90.589 MPa = 90.6 MPa (C) 1,336,474.331 mm 4

Ans.

(c) Bending stress at point K: yK = t − y = 6 mm − 26.166 mm = −20.166 mm

x = − =−

My Iz

( 3, 200 N  m )( −20.166 mm )(1,000 mm/m ) = 48.284 MPa = 48.3 MPa (T) 1,336,474.331 mm 4

Ans.

(d) Maximum bending stress: The maximum bending stress occurs at either the top or the bottom surface of the beam. The top of the cross section is at y = 43.834 mm, and the bottom of the cross section is at y = −26.166 mm. The larger bending stress magnitude occurs at the larger magnitude of these two values; in this case, at the top of the cross section.

My x = − Iz =−

( 3, 200 N  m )( 43.834 mm )(1,000 mm/m ) = −104.955 MPa = 105.0 MPa (C) 1,336,474.331 mm4

Ans.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.5 A beam is subjected to equal bending moments of Mz = 45 kip·ft, as shown in Figure P8.5a. The cross-sectional dimensions (Figure P8.5b) are b1 = 7.5 in., d1 = 1.5 in., b2 = 0.75 in., d2 = 6.0 in., b3 = 3.0 in., and d3 = 2.0 in. Determine: (a) the centroid location, the moment of inertia about the z axis, and the controlling section modulus about the z axis. (b) the bending stress at point H. State whether the normal stress at H is tension or compression. (c) the bending stress at point K. State whether the normal stress at K is tension or compression. (d) the maximum bending stress produced in the cross section. State whether the stress is tension or compression.

FIGURE P8.5a

FIGURE P8.5b

Solution

(a) Centroid location in y direction: (reference axis at bottom of shape) yi Shape Area Ai (from bottom) yi Ai 2 (in. ) (in.) (in.3) (1) 11.25 8.75 98.437 (2) 4.5 5.0 22.5 (3) 6.0 1.0 6.0 21.75 126.937 yi Ai 126.937 in.3 y= = = 5.836 in. = 5.84 in. Ai 21.75 in.2

Ans.

(measured upward from bottom edge of section) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed. Moment of inertia about the z axis: d = y − yi Shape IC d²A 4 (in. ) (in.) (in.4) (1) 2.109 –2.914 95.515 (2) 13.500 0.836 3.147 (3) 2.000 4.836 140.333 Moment of inertia about the z axis (in.4) =

Timothy A. Philpot

IC + d²A (in.4) 97.624 16.647 142.333 256.604 I z = 256.6 in.4

Ans.

Section moduli: I 256.604 in.4 S top = z = = 70.038 in.3 ctop 9.5 in. − 5.836 in.

Sbot =

Iz 256.604 in.4 = = 43.968 in.3 cbot 5.836 in.

 S = 43.97 in.3

Ans.

(b) Bending stress at point H: yH = d3 − y = 2.0 in. − 5.836 in. = −3.836 in.

x = − =−

My Iz

( 45 kip  ft )( −3.836 in.)(12 in./ft ) = 8.073 ksi = 8.07 ksi (T) 256.604 in.4

Ans.

(c) Bending stress at point K: yK = d 2 + d3 − y = 6.0 in. + 2.0 in. − 5.836 in. = 2.164 in.

x = − =−

My Iz

( 45 kip  ft )( 2.164 in.)(12 in./ft ) = −4.554 ksi = 4.55 ksi (C) 256.604 in.4

Ans.

(d) Maximum bending stress: The maximum bending stress occurs at either the top or the bottom surface of the beam. The top of the cross section is at y = 3.664 in., and the bottom of the cross section is at y = −5.836 in. The larger bending stress magnitude occurs at the larger magnitude of these two values; in this case, at the bottom of the cross section.

My x = − Iz

=−

( 45 kip  ft )( −5.836 in.)(12 in./ft ) = 12.282 ksi = 12.28 ksi (T) 256.604 in.4

Ans.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.6 A beam is subjected to equal bending moments of Mz = 240 N·m, as shown in Figure P8.6a. The cross-sectional dimensions (Figure P8.6b) are a = 20 mm, b = 40 mm, d = 80 mm, and r = 12 mm.

Determine: (a) the centroid location, the moment of inertia about the z axis, and the controlling section modulus about the z axis. (b) the bending stress at point H. State whether the normal stress at H is tension or compression. (c) the bending stress at point K. State whether the normal stress at K is tension or compression. (d) the maximum bending stress produced in the cross section. State whether the stress is tension or compression.

FIGURE P8.6a

FIGURE P8.6b

Solution (a) Centroid location in y direction: (reference axis at bottom of shape) yi Shape Area Ai (from bottom) yi Ai 2 (mm ) (mm) (mm3) rectangle 3,200 40.0 128,000.000 circle cutout –452.389 20.0 –9,047.787 2,747.611 118,952.213 3 yi Ai 118,952.213 mm y= = = 43.293 mm = 43.3 mm Ai 2,747.611 mm 2

Ans.

(measured upward from bottom edge of section) Moment of inertia about the z axis: d = y − yi Shape IC d²A (mm4) (mm) (mm4) rectangle 1,706,666.667 3.293 34,699.588 circle cutout –16,286.016 23.293 –245,449.374 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 1,741,366.254 –261,735.390 1,479,630.864

I z = 1,480,000 mm4

Ans.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Section moduli: I 1,479,630.864 mm4 S top = z = = 40,309.2 mm3 ctop 80 mm − 43.293 mm

I z 1,479,630.864 mm4 Sbot = = = 34,177.2 mm3 cbot 43.293 mm  S = 34, 200 mm3

Ans.

(b) Bending stress at point H: yH = d − a − y = 80 mm − 20 mm − 43.293 mm = 16.707 mm

x = − =−

My Iz

( −240 N  m )(16.707 mm )(1,000 mm/m ) 1,479,630.864 mm 4

= 2.710 MPa = 2.71 MPa (T)

Ans.

(c) Bending stress at point K: yK = y − ( a + r ) = 43.293 mm − ( 20 mm + 12 mm ) = −11.293 mm

x = − =−

My Iz

( −240 N  m )( −11.293 mm )(1,000 mm/m ) 1,479,630.864 mm4

= −1.832 MPa = 1.832 MPa (C)

Ans.

(d) Maximum bending stress: The maximum bending stress occurs at either the top or the bottom surface of the beam. The top of the cross section is at y = 36.707 mm, and the bottom of the cross section is at y = −43.293 mm. The larger bending stress magnitude occurs at the larger magnitude of these two values; in this case, at the bottom of the cross section.

My x = − Iz

=−

( −240 N  m )( −43.293 mm )(1,000 mm/m ) 1,479,630.864 mm 4

= −7.022 MPa = 7.02 MPa (C)

Ans.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.7 The dimensions of the beam cross section shown in Figure P8.7 are a = 7 mm and b = 45 mm.

The internal bending moment about the z centroidal axis is Mz = 325 N·m. What is the magnitude of the maximum bending stress in the beam?

FIGURE P8.7

Solution Moment of inertia about the z axis: Shape Rectangle Two cutouts on sides Moment of inertia about the z axis (mm4) =

IC (mm4) 277,830 –30,171.856 247,658.144

y = 3a = 3 ( 7 mm ) = 21 mm

x = − =−

My Iz

( 325 N  m )( 21 mm )(1,000 mm/m ) 247,658.144 mm 4

= 27.558 MPa = 27.6 MPa

Ans.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.8 The dimensions of the double-box beam cross section shown in Figure P8.8 are b = 150 mm, d = 50 mm, and t = 4 mm. If the maximum allowable bending stress is b = 17 MPa, determine the maximum internal bending moment Mz magnitude that can be applied to the beam.

FIGURE P8.8

Solution Moment of inertia about the z axis: Shape

IC (mm4) 1,562,500 –852,012 710,488

Outer rectangle Two cutout rectangles Moment of inertia about the z axis (mm4) =

x =

M zc Iz

Mz =

 xIz

(17 N/mm )( 710, 488 mm ) = 2

4

c 50 mm / 2 = 483,132 N  mm = 483 N  m

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.9 An aluminum alloy [E = 10,300 ksi] tee-shaped bar is used as a beam. The cross section of the tee shape, shown in Figure P8.9, has a total depth of d = 7.5 in. The centroidal z axis of this cross section is located 2.36 in. above point H. The beam spans in the x direction and it bends about the z centroidal axis. A strain gage is affixed to the side of the tee stem at a distance of c = 1.25 in. below the top surface of the tee shape. After loads are applied to the beam, a normal strain of x = –810  is measured by the strain gage. What is the bending stress at point H? FIGURE P8.9

Solution The y coordinate at the strain gage is: ygage = d − c − y = 7.5 in. − 1.25 in. − 2.36 in. = 3.890 in.

The bending strain at H is thus:

 gage

ygage

=

H yH

 H =

yH −2.36 in.  gage = −810 10−6 in./in.) = 491.414 10−6 in./in. ( ygage 3.890 in.

The bending stress at H is:  x = E x = (10,300 ksi ) ( 491.414 10−6 in./in.) = 5.062 ksi = 5.06 ksi

Ans.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.10 The cross-sectional dimensions of the beam shown in Figure P8.10 are d = 17 in., bf = 10 in., tf = 1.0 in., tw = 0.60 in., and a = 3.5 in. (a) If the bending stress at point K is 5.4 ksi (C), what is the bending stress at point H? State whether the normal stress at H is tension or compression. (b) If the allowable bending stress is 30 ksi, what is the magnitude of the maximum bending moment Mz that can be supported by the beam?

FIGURE P8.10

Solution The y coordinates at H and K are: d 17 in. yH = − t f = − 1.0 in. = 7.5 in.

2 2 d 17 in. yK = − + a = − + 3.5 in. = −5.0 in.

2 2 (a) Bending stress at H: Since the bending stress is linearly distributed over the depth of the section, we can calculate the bending stress at H from the principle of similar triangles:

H yH

=

K yK

 H =

yH 7.5 in. K = ( −5.4 ksi ) = 8.10 ksi (T) yK −5.0 in.

Ans.

Moment of inertia about z axis:

(10 in.)(17 in.) − (10 in. − 0.60 in.) 17 in. − 2 (1.0 in.) = 1, 450.417 in.4 I = 3

3

z

12

12

(b) Maximum bending moment Mz: M c x = z Iz M =

 xIz c

( 30 kips/in. )(1, 450.417 in. ) = 5,119.117 kip  in. = 427 kip  ft 2

=

17 in. / 2

4

Ans.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.11 The cross-sectional dimensions of the beam shown in Figure P8.11 are ro = 115 mm and ri = 95 mm. Given Mz = 16 kN·m,  = 30°, and  = 55°, what are the bending stresses at points H and K?

FIGURE P8.11

Solution Moment of inertia about the z axis: Iz =



 r − r ) = (115 mm ) − ( 95 mm )  = 73,395, 458 mm (  4 4 4 o

4

4

4

4

i

The y coordinates at H and K are: yH = ro sin  = (115 mm ) sin 30 = 57.5 mm

yK = −ri sin  = − ( 95 mm ) sin 55 = −77.819 mm Bending stress at H: My x = − Iz =−

(16 kN  m )( 57.5 mm )(1, 000 N/kN )(1,000 mm/m ) 73,395,458 mm 4

= −12.535 MPa = 12.54 MPa (C)

Ans.

Bending stress at K: My x = − Iz =−

(16 kN  m )( −77.819 mm )(1, 000 N/kN )(1,000 mm/m ) 73,395,458 mm 4

= 16.964 MPa = 16.96 MPa (T)

Ans.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.12 The cross-sectional dimensions of the beam shown in Figure P8.12 are a = 5.0 in., b = 6.0 in., d = 4.0 in., and t = 0.5 in. The internal bending moment about the z centroidal axis is Mz = −4.25 kip-ft. Determine: (a) the maximum tension bending stress in the beam. (b) the maximum compression bending stress in the beam.

FIGURE P8.12

Solution

(a) Centroid location in y direction: (reference axis at bottom of shape) yi Shape Area Ai (from bottom) yi Ai 2 (in. ) (in.) (in.3) (1) 2.50 3.75 9.375 (2) 4.00 2.00 8.00 (3) 4.50 0.25 1.125 11.00 18.50 3 yi Ai 18.50 in. y= = = 1.6818 in. (measured upward from bottom edge of section) Ai 11.00 in.2 Moment of inertia about the z axis: d = y − yi Shape IC 4 (in. ) (in.)

(1) 0.05208 –2.0682 (2) 5.33333 –0.31820 (3) 0.09375 1.43180 Moment of inertia about the z axis (mm4) =

d²A (in.4) 10.69363 0.40500 9.22523

IC + d²A (in.4) 10.74571 5.73834 9.31898 25.80303

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

The y coordinates at the top and bottom surfaces of the section are: ytop = d − y = 4.0 in. − 1.6818 in. = 2.31820 in.

ybottom = − y = −1.6818 in.

Bending stress at top surface: The maximum tensile bending stress occurs at the top surface of the beam for a negative bending moment: My x = − Iz =−

( −4.25 kip  ft )( 2.31820 in.)(12 in./ft ) = 4.582 ksi = 4.58 ksi (T) 25.80303 in.4

Ans.

Bending stress at bottom surface: The maximum compressive bending stress occurs at the bottom surface of the beam for a negative bending moment: My x = − Iz =−

( −4.25 kip  ft )( −1.6818 in.)(12 in./ft ) = −3.324 ksi = 3.32 ksi (C) 25.80303 in.4

Ans.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.13 Two uniformly distributed loads of w = 3,600 lb/ft act on the simply supported beam shown in Figure P8.13a. The beam spans are a = 8 ft and L = 16 ft. The beam cross section shown in Figure P8.13b has dimensions of b1 = 16 in., d1 = 6 in., b2 = 10 in., and d2 = 10 in. Calculate the maximum tensile and compressive bending stresses produced in segment BC of the beam.

FIGURE P8.13a

FIGURE P8.13b

Solution

Centroid location in y direction: (reference axis at bottom of shape) yi Shape Area Ai (from bottom) yi Ai 2 (in. ) (in.) (in.3) (1) 96.00 13.00 1,248.00 (2) 100.00 5.00 500.00 196.00 1,748.00 yi Ai 1, 748.00 in.3 (measured upward from bottom edge of section) y= = = 8.918 in. Ai 196.00 in.2 Moment of inertia about the z axis: d = y − yi Shape IC d²A 4 (in. ) (in.) (in.4) (1) 288.000 –4.082 1,599.334 (2) 833.333 3.918 1,535.360 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 1,887.334 2,368.694 4,256.027

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Shear-force and bending-moment diagrams

The maximum bending moment in segment BC of the beam is –115,200 lb·ft.

Bending stresses: The y coordinates at the top and bottom surfaces of the section are: ytop = d1 + d 2 − y = 6 in. + 10 in. − 8.918 in. = 7.082 in.

ybottom = − y = −8.918 in.

Bending stress at top surface: The bending stress at the top surface of the beam in segment BC is: My x = − Iz =−

( −115, 200 lb  ft )( 7.082 in.)(12 in./ft ) = 2,300.185 psi 4,256.027 in.4

Bending stress at bottom surface: The bending stress at the bottom surface of the beam in segment BC is: My x = − Iz =−

( −115, 200 lb  ft )( −8.918 in.)(12 in./ft ) = −2,896.775 psi 4,256.027 in.4

Maximum bending stresses: The maximum tensile bending stress is thus:  x = 2,300 psi (T)

Ans.

and the maximum compressive bending stress is:  x = 2,900 psi (C)

Ans.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.14 An extruded polymer beam is subjected to a bending moment M as shown in Figure P8.14a. The length of the beam is L = 800 mm. The cross-sectional dimensions (Figure P8.14b) of the beam are b1 = 34 mm, d1 = 100 mm, b2 = 20 mm, d2 = 20 mm, and a = 7 mm. For this material, the allowable tensile bending stress is 16 MPa, and the allowable compressive bending stress is 12 MPa. Determine the largest moment M that can be applied as shown to the beam.

FIGURE P8.14a

FIGURE P8.14b

Solution Centroid location in y direction: Shape

yi (from bottom) (mm) 50 83

Area Ai yi Ai 2 (mm ) (mm3) Outer rectangle 3,400 170,000 Cutout square –400 –33,200 2 3,000 mm 136,800 yi Ai 136,800 mm3 y= = = 45.6 mm (measured upward from bottom edge of shape) Ai 3,000 mm 2 Moment of inertia about the z axis: d = y − yi Shape IC d²A 4 (mm ) (mm) (mm4) Outer rectangle 2,833,333.33 –4.4 65,824 Cutout square –13,333.33 –37.4 –559,504 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 2,899,157.33 –572,837.33 2,326,320

Bending stresses: The y coordinates at the top and bottom surfaces of the cross section are: ytop = d1 − y = 100 mm − 45.6 mm = 54.4 mm

ybottom = − y = −45.6 mm Bending moment based on allowable tensile bending stress: For the indicated moment direction (i.e., counterclockwise), tensile bending stress will be produced above the neutral axis. The largest tensile

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

bending stress will be found at the top surface of the beam. Solve the flexure formula for M given that the bending stress at ytop must be less than or equal to 16 MPa: My −   allow Iz M  −

 allow I z ytop

(16 N/mm )( 2,326,320 mm ) = −684, 211.8 N  mm 2

=−

4

54.4 mm

Bending moment based on allowable compressive bending stress: For the indicated moment direction (i.e., counterclockwise), compressive bending stress will be produced below the neutral axis. The largest compressive bending stress will be found at the bottom surface of the beam. Solve the flexure formula for M given that the bending stress at ybottom must be less than or equal to 12 MPa: My −   allow Iz M  −

 allow I z ybottom

( −12 N/mm )( 2,326,320 mm ) = −612,189.5 N  mm =− 2

4

−45.6 mm

Largest moment M: The largest bending moment that can be applied as shown in Figure P8.14a is:

M max = 612,189.5 N  mm = 612 N  m

Ans.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.