assignment 6 solution
CIV 207 – Winter 2009 Friday, February 13th
Assignment #6
Complete the first three questions. Submit your work to Box #5 on the 4th floor of the MacDonald building by 12 noon on Tuesday March 3rd. No late submissions will be accepted. The second set of three questions are for practice only. Groups of up to 3 members are permitted. Clearly print your names and student numbers on your submission. Ensure that the submission is properly stapled.
8.16. The cross-sectional dimensions of a beam are shown. The internal bending moment about the z centroidal axis is Mz = +270 lb-ft. Determine: (a) the maximum tension bending stress in the beam. (b) the maximum compression bending stress in the beam.
8.22. A flanged wooden shape is used to support the loads shown on the beam. The dimensions of the shape are shown in Fig. b. Consider the entire 18-ft length of the beam and determine: (a) the maximum tension bending stress at any location along the beam, and (b) the maximum compression bending stress at any location along the beam.
8.50 Two steel plates, each 4 in. wide and 0.25 in. thick, reinforce a wood beam that is 3 in. wide and 8 in. deep. The steel plates are attached to the vertical sides of the wood beam in a position such that the composite shape is symmetric about the z axis, as shown in the sketch of the beam cross section. Determine the maximum bending stresses produced in both the wood and the steel if a bending moment of Mz = +50 kip-in is applied about the z axis. Assume Ewood = 2,000 ksi and Esteel = 30,000 ksi.
For practice
8.17
8.36
8.45
8.16 The cross-sectional dimensions of a beam are shown in Fig. P8.16. The internal bending moment about the z centroidal axis is Mz = +270 lb-ft. Determine: (a) the maximum tension bending stress in the beam. (b) the maximum compression bending stress in the beam.
Solution
Fig. P8.16
Centroid location in y direction:
yi Area Ai (from bottom) yi Ai 2 (in. ) (in.) (in.3) bottom flange 0.40625 0.06250 0.02539 left web 0.28125 1.25000 0.35156 left top flange 0.09375 2.43750 0.22852 right web 0.28125 1.25000 0.35156 right top flange 0.09375 2.43750 0.22852 1.18555 in.3 1.15625 in.2 Σyi Ai 1.18555 in.3 y= = = 1.0253 in. (measured upward from bottom edge of bottom flange) 1.15625 in.2 ΣAi Shape
Moment of inertia about the z axis: d = yi – y Shape d²A IC 4 (in.) (in.4) (in. ) bottom flange 0.000529 -0.962838 0.376617 left web 0.118652 0.224662 0.014196 left top flange 0.000122 1.412162 0.186956 right web 0.118652 0.224662 0.014196 right top flange 0.000122 1.412162 0.186956 Moment of inertia about the z axis (in.4) =
IC + d²A (in.4) 0.377146 0.132848 0.187079 0.132848 0.187079 1.016999
(a) Maximum tension bending stress: For a positive bending moment of Mz = +270 lb-ft, the maximum tension bending stress will occur at the bottom surface of the cross section (i.e., y = −1.0253 in.). From the flexure formula, the bending stress at the bottom of the cross section is: My (270 lb-ft)( − 1.0253 in.)(12 in./ft) Ans. =− = 3, 266.446 psi = 3, 270 psi (T) σx = − Iz 1.016999 in.4 (b) Maximum compression bending stress: The maximum compression bending stress will occur at the top surface of the cross section (i.e., y = 2.50 in. − 1.0253 in. = 1.4747 in.). From the flexure formula, the bending stress at the top of the cross section is: My (270 lb-ft)(1.4747 in.)(12 in./ft) σx = − =− = −4, 698.164 psi = 4, 700 psi (C) Ans. Iz 1.016999 in.4 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.
8.17 Two vertical forces are applied to a simply supported beam (Fig. P8.17a) having the cross section shown in Fig. P8.17b. Determine the maximum tension and compression bending stresses produced in segment BC of the beam.
Fig. P8.17a
Fig. P8.17b
Solution Centroid location in y direction:
Shape top flange stem
y=
Σyi Ai ΣAi
=
Area Ai (mm2) 3,000.0 1,440.0 4,440 mm2
yi (from bottom) (mm) 167.5 80.0
yi Ai (mm3) 502,500.0 115,200.0 617,700 mm3
617, 700 mm3 = 139.1216 mm (measured upward from bottom edge of stem) 4,440 mm 2
Moment of inertia about the z axis: d = yi – y Shape d²A IC 4 (mm) (mm4) (mm ) top flange 56,250.00 28.38 2,415,997.08 stem 3,072,000.00 −59.12 5,033,327.25 Moment of inertia about the z axis (mm4) =
IC + d²A (mm4) 2,472,247.08 8,105,327.25 10,577,574.32
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.
Shear-force and bending-moment diagrams:
The maximum moment occurs between B and C. The moment magnitude is 12 kN-m. Maximum tension bending stress: For a positive bending moment, the maximum tension bending stress will occur at the bottom surface of this cross section. From the flexure formula, the bending stress at the bottom of the tee stem is: My (12 kN-m)( − 139.1216 mm)(1,000 N/kN)(1,000 mm/m) Ans. =− = 157.8 MPa (T) σx = − Iz 10.5776 × 106 mm 4 Maximum compression bending stress: The maximum compression bending stress will occur at the top of the flange: My σx = − Iz
(12 kN-m)(175 mm − 139.1216 mm)(1,000 N/kN)(1,000 mm/m) 10.5776 × 106 mm 4 = −40.7 MPa = 40.7 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.
8.22 A flanged wooden shape is used to support the loads shown on the beam in Fig. P8.22a. The dimensions of the shape are shown in Fig. P8.22b. Consider the entire 18-ft length of the beam and determine: (a) the maximum tension bending stress at any location along the beam, and (b) the maximum compression bending stress at any location along the beam.
Fig. P8.22a
Fig. P8.22b
Solution Centroid location in y direction: Area Ai (in.2) 20.0 20.0 12.0 52.0 in.2
Shape top flange web bottom flange
y=
Σyi Ai ΣAi
=
yi (from bottom) (in.) 13.0 7.0 1.0
yi Ai (in.3) 260.0 140.0 12.0 412.0 in.3
412.0 in.3 = 7.9231 in. (from bottom of shape to centroid) 52.0 in.2 = 6.0769 in.
(from top of shape to centroid)
Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) top flange 6.6667 5.0769 515.5030 web 166.6667 −0.9231 17.0414 bottom flange 4.0000 −6.9231 575.1479 Moment of inertia about the z axis (in.4) =
IC + d²A (in.4) 522.1696 183.7081 579.1479 1,285.0256
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.
Shear-force and bending-moment diagrams Maximum bending moments positive M = 14,851 lb-ft negative M = −8,400 lb-ft Bending stresses at max positive moment (14,851 lb-ft)(6.0769 in.)(12 in./ft) σx = − 1,285.0256 in.4 = 843 psi (C) (14,851 lb-ft)( − 7.9231 in.)(12 in./ft) σx = − 1,285.0256 in.4 = 1, 099 psi (T) Bending stresses at max negative moment (−8, 400 lb-ft)(6.0769 in.)(12 in./ft) σx = − 1,285.0256 in.4 = 477 psi (T) (−8, 400 lb-ft)( − 7.9231 in.)(12 in./ft) σx = − 1,285.0256 in.4 = 622 psi (C) (a) Maximum tension bending stress = 1,099 psi (T)
Ans.
(b) Maximum compression bending stress = 843 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.
8.36 A cantilever timber beam (Fig. P8.36a) with a span of L = 3 m supports a uniformly distributed load of w. The beam width is b = 300 mm and the beam height is h = 200 mm (Fig. P8.36b). The allowable bending stress of the wood is 6 MPa. Determine the magnitude of the maximum load w that may be carried by the beam.
Fig. P8.36a
Fig. P8.36b
Solution Section modulus for solid rectangular section I bh3 /12 bh 2 (300 mm)(200 mm) 2 S= = = = = 2 × 106 mm3 c h/2 6 6 Maximum allowable bending moment: M σx ≥ ∴ M allow ≤ σ x S = (6 N/mm 2 )(2 ×106 mm3 ) = 12 ×106 N-mm S Maximum bending moment in cantilever span: The maximum bending moment magnitude in the cantilever beam occurs at support A: wL2 M max = 2 Maximum distributed load: wL2 ≤ M allow 2 2 M allow 2(12 ×106 N-mm) ∴ wallow ≤ = = 2.67 N/mm = 2.67 kN/m [(3 m)(1,000 mm/m)]2 L2
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.
8.45 An aluminum [E = 10,000 ksi] bar is bonded to a steel [E = 30,000 ksi] bar to form a composite beam (Fig. P8.45b). The composite beam is subjected to a bending moment of M = +300 lb-ft about the z axis (Fig. P8.45a). Determine: (a) the maximum bending stresses in the aluminum and steel bars. (b) the stress in the two materials at the joint where they are bonded together.
Fig. P8.45a
Fig. P8.45b
Solution Denote the aluminum as material (1) and denote the steel as material (2). The modular ratio is: E 30, 000 ksi n= 2 = =3 E1 10,000 ksi Transform the steel bar (2) into an equivalent amount of aluminum (1) by multiplying its width by the modular ratio: b2, trans = 3(1.50 in.) = 4.50 in. Thus, for calculation purposes, the 1.50 in. × 0.50 in. steel bar is replaced by an aluminum bar that is 4.50-in. wide and 0.50-in. thick. Centroid location of the transformed section in the vertical direction
Shape aluminum bar (1) transformed steel bar (2)
y=
Σyi Ai ΣAi
=
Width b (in.) 1.50 4.50
Height h (in.) 0.50 0.50
Area Ai (in.2) 0.75 2.25 3.00
yi (from bottom) (in.) 0.25 0.75
yi Ai (in.3) 0.1875 1.6875 1.8750
1.8750 in.3 = 0.6250 in. (measured upward from bottom edge of section) 3.00 in.2
Moment of inertia about the horizontal centroidal axis d = yi – y Shape IC 4 (in. ) (in.) aluminum bar (1) 0.015625 –0.375 transformed steel bar (2) 0.046875 0.125 Moment of inertia about the z axis =
d²A (in.4) 0.105469 0.035156
IC + d²A (in.4) 0.121094 0.082031 0.203125 in.4
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.
(a) Maximum bending stress in aluminum bar (1) From the flexure formula, the maximum bending stress in aluminum bar (1) is: My (300 lb-ft)( − 0.6250 in.)(12 in./ft) σ1 = − =− = 11, 080 psi (T) I 0.203125 in.4
Ans.
(a) Maximum bending stress in steel bar (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the maximum bending stress in steel bar (2) is: My (300 lb-ft)(1.000 in. − 0.6250 in.)(12 in./ft) Ans. σ2 = − = −(3) = 19,940 psi (C) I 0.203125 in.4 (b) Bending stress in aluminum bar (1) at interface My (300 lb-ft)(0.50 in. − 0.6250 in.)(12 in./ft) σ1 = − =− = 2, 220 psi (T) I 0.203125 in.4 (b) Maximum bending stress in steel bar (2) at interface My (300 lb-ft)(0.50 in. − 0.6250 in.)(12 in./ft) σ2 = − = −(3) = 6, 650 psi (T) I 0.203125 in.4
Ans.
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.
8.50 Two steel plates, each 4 in. wide and 0.25 in. thick, reinforce a wood beam that is 3 in. wide and 8 in. deep. The steel plates are attached to the vertical sides of the wood beam in a position such that the composite shape is symmetric about the z axis, as shown in the sketch of the beam cross section (Fig. P8.50). Determine the maximum bending stresses produced in both the wood and the steel if a bending moment of Mz = +50 kip-in is applied about the z axis. Assume Ewood = 2,000 ksi and Esteel = 30,000 ksi.
Fig. P8.50
Solution Let the wood be denoted as material (1) and the steel plates as material (2). The modular ratio is: E 30, 000 ksi n= 2 = = 15 E1 2,000 ksi Transform the steel plates (2) into an equivalent amount of wood (1) by multiplying the plate thicknesses by the modular ratio: b2, trans = 15(0.25 in.) = 3.75 in. (each). Thus, for calculation purposes, each 4 in. × 0.25 in. steel plate is replaced by a wood board that is 4-in. tall and 3.75-in. wide. Centroid location: Since the transformed section is doubly symmetric, the centroid location is found from symmetry. Moment of inertia about the z centroidal axis Shape IC (in.4) wood beam (1) 128 two transformed steel plates (2) 40 Moment of inertia about the z axis =
d = yi – y (in.) 0 0
d²A (in.4) 0 0
Bending stress in wood beam (1) From the flexure formula, the maximum bending stress in wood beam (1) is: M c (50 kip-in.)(4 in.) σ1 = z = = 1.190 ksi = 1,190 psi 168 in.4 Iz
IC + d²A (in.4) 128 40 168 in.4
Ans.
Bending stress in steel plates (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the maximum bending stress in the steel plates (2) is: M c (50 kip-in.)(2 in.) Ans. σ 2 = n z = (15) = 8.93 ksi = 8,930 psi Iz 168 in.4
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.
Assignment #6
Complete the first three questions. Submit your work to Box #5 on the 4th floor of the MacDonald building by 12 noon on Tuesday March 3rd. No late submissions will be accepted. The second set of three questions are for practice only. Groups of up to 3 members are permitted. Clearly print your names and student numbers on your submission. Ensure that the submission is properly stapled.
8.16. The cross-sectional dimensions of a beam are shown. The internal bending moment about the z centroidal axis is Mz = +270 lb-ft. Determine: (a) the maximum tension bending stress in the beam. (b) the maximum compression bending stress in the beam.
8.22. A flanged wooden shape is used to support the loads shown on the beam. The dimensions of the shape are shown in Fig. b. Consider the entire 18-ft length of the beam and determine: (a) the maximum tension bending stress at any location along the beam, and (b) the maximum compression bending stress at any location along the beam.
8.50 Two steel plates, each 4 in. wide and 0.25 in. thick, reinforce a wood beam that is 3 in. wide and 8 in. deep. The steel plates are attached to the vertical sides of the wood beam in a position such that the composite shape is symmetric about the z axis, as shown in the sketch of the beam cross section. Determine the maximum bending stresses produced in both the wood and the steel if a bending moment of Mz = +50 kip-in is applied about the z axis. Assume Ewood = 2,000 ksi and Esteel = 30,000 ksi.
For practice
8.17
8.36
8.45
8.16 The cross-sectional dimensions of a beam are shown in Fig. P8.16. The internal bending moment about the z centroidal axis is Mz = +270 lb-ft. Determine: (a) the maximum tension bending stress in the beam. (b) the maximum compression bending stress in the beam.
Solution
Fig. P8.16
Centroid location in y direction:
yi Area Ai (from bottom) yi Ai 2 (in. ) (in.) (in.3) bottom flange 0.40625 0.06250 0.02539 left web 0.28125 1.25000 0.35156 left top flange 0.09375 2.43750 0.22852 right web 0.28125 1.25000 0.35156 right top flange 0.09375 2.43750 0.22852 1.18555 in.3 1.15625 in.2 Σyi Ai 1.18555 in.3 y= = = 1.0253 in. (measured upward from bottom edge of bottom flange) 1.15625 in.2 ΣAi Shape
Moment of inertia about the z axis: d = yi – y Shape d²A IC 4 (in.) (in.4) (in. ) bottom flange 0.000529 -0.962838 0.376617 left web 0.118652 0.224662 0.014196 left top flange 0.000122 1.412162 0.186956 right web 0.118652 0.224662 0.014196 right top flange 0.000122 1.412162 0.186956 Moment of inertia about the z axis (in.4) =
IC + d²A (in.4) 0.377146 0.132848 0.187079 0.132848 0.187079 1.016999
(a) Maximum tension bending stress: For a positive bending moment of Mz = +270 lb-ft, the maximum tension bending stress will occur at the bottom surface of the cross section (i.e., y = −1.0253 in.). From the flexure formula, the bending stress at the bottom of the cross section is: My (270 lb-ft)( − 1.0253 in.)(12 in./ft) Ans. =− = 3, 266.446 psi = 3, 270 psi (T) σx = − Iz 1.016999 in.4 (b) Maximum compression bending stress: The maximum compression bending stress will occur at the top surface of the cross section (i.e., y = 2.50 in. − 1.0253 in. = 1.4747 in.). From the flexure formula, the bending stress at the top of the cross section is: My (270 lb-ft)(1.4747 in.)(12 in./ft) σx = − =− = −4, 698.164 psi = 4, 700 psi (C) Ans. Iz 1.016999 in.4 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.
8.17 Two vertical forces are applied to a simply supported beam (Fig. P8.17a) having the cross section shown in Fig. P8.17b. Determine the maximum tension and compression bending stresses produced in segment BC of the beam.
Fig. P8.17a
Fig. P8.17b
Solution Centroid location in y direction:
Shape top flange stem
y=
Σyi Ai ΣAi
=
Area Ai (mm2) 3,000.0 1,440.0 4,440 mm2
yi (from bottom) (mm) 167.5 80.0
yi Ai (mm3) 502,500.0 115,200.0 617,700 mm3
617, 700 mm3 = 139.1216 mm (measured upward from bottom edge of stem) 4,440 mm 2
Moment of inertia about the z axis: d = yi – y Shape d²A IC 4 (mm) (mm4) (mm ) top flange 56,250.00 28.38 2,415,997.08 stem 3,072,000.00 −59.12 5,033,327.25 Moment of inertia about the z axis (mm4) =
IC + d²A (mm4) 2,472,247.08 8,105,327.25 10,577,574.32
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.
Shear-force and bending-moment diagrams:
The maximum moment occurs between B and C. The moment magnitude is 12 kN-m. Maximum tension bending stress: For a positive bending moment, the maximum tension bending stress will occur at the bottom surface of this cross section. From the flexure formula, the bending stress at the bottom of the tee stem is: My (12 kN-m)( − 139.1216 mm)(1,000 N/kN)(1,000 mm/m) Ans. =− = 157.8 MPa (T) σx = − Iz 10.5776 × 106 mm 4 Maximum compression bending stress: The maximum compression bending stress will occur at the top of the flange: My σx = − Iz
(12 kN-m)(175 mm − 139.1216 mm)(1,000 N/kN)(1,000 mm/m) 10.5776 × 106 mm 4 = −40.7 MPa = 40.7 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.
8.22 A flanged wooden shape is used to support the loads shown on the beam in Fig. P8.22a. The dimensions of the shape are shown in Fig. P8.22b. Consider the entire 18-ft length of the beam and determine: (a) the maximum tension bending stress at any location along the beam, and (b) the maximum compression bending stress at any location along the beam.
Fig. P8.22a
Fig. P8.22b
Solution Centroid location in y direction: Area Ai (in.2) 20.0 20.0 12.0 52.0 in.2
Shape top flange web bottom flange
y=
Σyi Ai ΣAi
=
yi (from bottom) (in.) 13.0 7.0 1.0
yi Ai (in.3) 260.0 140.0 12.0 412.0 in.3
412.0 in.3 = 7.9231 in. (from bottom of shape to centroid) 52.0 in.2 = 6.0769 in.
(from top of shape to centroid)
Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) top flange 6.6667 5.0769 515.5030 web 166.6667 −0.9231 17.0414 bottom flange 4.0000 −6.9231 575.1479 Moment of inertia about the z axis (in.4) =
IC + d²A (in.4) 522.1696 183.7081 579.1479 1,285.0256
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.
Shear-force and bending-moment diagrams Maximum bending moments positive M = 14,851 lb-ft negative M = −8,400 lb-ft Bending stresses at max positive moment (14,851 lb-ft)(6.0769 in.)(12 in./ft) σx = − 1,285.0256 in.4 = 843 psi (C) (14,851 lb-ft)( − 7.9231 in.)(12 in./ft) σx = − 1,285.0256 in.4 = 1, 099 psi (T) Bending stresses at max negative moment (−8, 400 lb-ft)(6.0769 in.)(12 in./ft) σx = − 1,285.0256 in.4 = 477 psi (T) (−8, 400 lb-ft)( − 7.9231 in.)(12 in./ft) σx = − 1,285.0256 in.4 = 622 psi (C) (a) Maximum tension bending stress = 1,099 psi (T)
Ans.
(b) Maximum compression bending stress = 843 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.
8.36 A cantilever timber beam (Fig. P8.36a) with a span of L = 3 m supports a uniformly distributed load of w. The beam width is b = 300 mm and the beam height is h = 200 mm (Fig. P8.36b). The allowable bending stress of the wood is 6 MPa. Determine the magnitude of the maximum load w that may be carried by the beam.
Fig. P8.36a
Fig. P8.36b
Solution Section modulus for solid rectangular section I bh3 /12 bh 2 (300 mm)(200 mm) 2 S= = = = = 2 × 106 mm3 c h/2 6 6 Maximum allowable bending moment: M σx ≥ ∴ M allow ≤ σ x S = (6 N/mm 2 )(2 ×106 mm3 ) = 12 ×106 N-mm S Maximum bending moment in cantilever span: The maximum bending moment magnitude in the cantilever beam occurs at support A: wL2 M max = 2 Maximum distributed load: wL2 ≤ M allow 2 2 M allow 2(12 ×106 N-mm) ∴ wallow ≤ = = 2.67 N/mm = 2.67 kN/m [(3 m)(1,000 mm/m)]2 L2
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.
8.45 An aluminum [E = 10,000 ksi] bar is bonded to a steel [E = 30,000 ksi] bar to form a composite beam (Fig. P8.45b). The composite beam is subjected to a bending moment of M = +300 lb-ft about the z axis (Fig. P8.45a). Determine: (a) the maximum bending stresses in the aluminum and steel bars. (b) the stress in the two materials at the joint where they are bonded together.
Fig. P8.45a
Fig. P8.45b
Solution Denote the aluminum as material (1) and denote the steel as material (2). The modular ratio is: E 30, 000 ksi n= 2 = =3 E1 10,000 ksi Transform the steel bar (2) into an equivalent amount of aluminum (1) by multiplying its width by the modular ratio: b2, trans = 3(1.50 in.) = 4.50 in. Thus, for calculation purposes, the 1.50 in. × 0.50 in. steel bar is replaced by an aluminum bar that is 4.50-in. wide and 0.50-in. thick. Centroid location of the transformed section in the vertical direction
Shape aluminum bar (1) transformed steel bar (2)
y=
Σyi Ai ΣAi
=
Width b (in.) 1.50 4.50
Height h (in.) 0.50 0.50
Area Ai (in.2) 0.75 2.25 3.00
yi (from bottom) (in.) 0.25 0.75
yi Ai (in.3) 0.1875 1.6875 1.8750
1.8750 in.3 = 0.6250 in. (measured upward from bottom edge of section) 3.00 in.2
Moment of inertia about the horizontal centroidal axis d = yi – y Shape IC 4 (in. ) (in.) aluminum bar (1) 0.015625 –0.375 transformed steel bar (2) 0.046875 0.125 Moment of inertia about the z axis =
d²A (in.4) 0.105469 0.035156
IC + d²A (in.4) 0.121094 0.082031 0.203125 in.4
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.
(a) Maximum bending stress in aluminum bar (1) From the flexure formula, the maximum bending stress in aluminum bar (1) is: My (300 lb-ft)( − 0.6250 in.)(12 in./ft) σ1 = − =− = 11, 080 psi (T) I 0.203125 in.4
Ans.
(a) Maximum bending stress in steel bar (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the maximum bending stress in steel bar (2) is: My (300 lb-ft)(1.000 in. − 0.6250 in.)(12 in./ft) Ans. σ2 = − = −(3) = 19,940 psi (C) I 0.203125 in.4 (b) Bending stress in aluminum bar (1) at interface My (300 lb-ft)(0.50 in. − 0.6250 in.)(12 in./ft) σ1 = − =− = 2, 220 psi (T) I 0.203125 in.4 (b) Maximum bending stress in steel bar (2) at interface My (300 lb-ft)(0.50 in. − 0.6250 in.)(12 in./ft) σ2 = − = −(3) = 6, 650 psi (T) I 0.203125 in.4
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8.50 Two steel plates, each 4 in. wide and 0.25 in. thick, reinforce a wood beam that is 3 in. wide and 8 in. deep. The steel plates are attached to the vertical sides of the wood beam in a position such that the composite shape is symmetric about the z axis, as shown in the sketch of the beam cross section (Fig. P8.50). Determine the maximum bending stresses produced in both the wood and the steel if a bending moment of Mz = +50 kip-in is applied about the z axis. Assume Ewood = 2,000 ksi and Esteel = 30,000 ksi.
Fig. P8.50
Solution Let the wood be denoted as material (1) and the steel plates as material (2). The modular ratio is: E 30, 000 ksi n= 2 = = 15 E1 2,000 ksi Transform the steel plates (2) into an equivalent amount of wood (1) by multiplying the plate thicknesses by the modular ratio: b2, trans = 15(0.25 in.) = 3.75 in. (each). Thus, for calculation purposes, each 4 in. × 0.25 in. steel plate is replaced by a wood board that is 4-in. tall and 3.75-in. wide. Centroid location: Since the transformed section is doubly symmetric, the centroid location is found from symmetry. Moment of inertia about the z centroidal axis Shape IC (in.4) wood beam (1) 128 two transformed steel plates (2) 40 Moment of inertia about the z axis =
d = yi – y (in.) 0 0
d²A (in.4) 0 0
Bending stress in wood beam (1) From the flexure formula, the maximum bending stress in wood beam (1) is: M c (50 kip-in.)(4 in.) σ1 = z = = 1.190 ksi = 1,190 psi 168 in.4 Iz
IC + d²A (in.4) 128 40 168 in.4
Ans.
Bending stress in steel plates (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the maximum bending stress in the steel plates (2) is: M c (50 kip-in.)(2 in.) Ans. σ 2 = n z = (15) = 8.93 ksi = 8,930 psi Iz 168 in.4
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