8.42 A composite beam is fabricated by bolting two 3 in. wide × 12 in. deep timber planks to the sides of a 0.50 in. × 12 in. steel plate (Fig. P8.42b P8.42b). The moduli of elasticity of the timber and the steel are 1,800 ksi and 30,000 ksi, respectively. The simply supported beam spans a distance of 20 ft and carries two concentrated loads P loads P , which are applied at the quarter points of the span (Fig. P8.42a P8.42a). (a) Determine the maximum bending stresses produced in the timber planks and the steel plate if P if P = =3 kips. (b) Assume that the allowable bending stresses of the timber and the steel are 1,200 psi and 24,000 psi, respectively. Determine the largest acceptable magnitude for concentrated loads P loads P . (You may neglect the weight of the beam in your calculations.)
Fig. P8.42a P8.42a Fig. P8.42b P8.42b
Solution Let the timber be denoted as material material (1) and the steel plate plate as material (2). The modular ratio is: E 30,000 ksi n= 2 = = 16.6667 E 1 1,800 ksi Transform the steel plate (2) into an equivalent amount of wood (1) by multiplying its width by the modular ratio: b2, trans = 16.6667(0.50 in.) = 8.3333 in. Thus, for calculation purposes, the 12 in. × 0.50 in. steel plate is replaced by a wood board that is 12 in. deep and 8.3333-in. thick. Moment of inertia about the horizontal centroidal axis d = y = yi – Shape I C C 4
(in. ) timber (1) 864 transformed steel plate (2) 1,200 Moment of inertia about the z the z axis axis =
(in.) 0 0
d²A 4 (in. ) 0 0
I C + d²A C + 4 (in. ) 864 1,200 4 2,064 in.
Bending moment in beam for P = = 3 kips The bending moment in the simply supported beam with two 3-kip concentrated loads is: M max = (3 kips kips)( )(5 5 ft) = 15 kipkip-fft = 180 180 kip kip--in. Bending stress in timber (1) From the flexure formula, the maximum bending stress in timber (1) is: My (180 (180 kipkip-in in.) .)(( ± 6 in.) in.) =− = ±0.5233 ksi = 523 psi σ 1 = − I 2,064 in.4
Ans.
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Bending stress in steel plate (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the maximum bending stress in steel plate (2) is: My (180 (180 kipkip-in in.) .)(( ± 6 in.) in.) Ans. σ 2 = − n = −(16.6667 ) = ±8.7209 ksi = 8, 720 psi I 2,064 in.4 Determine maximum P If the allowable bending stress in the timber is 1,200 psi, then the maximum bending moment that may be supported by the beam is: σ 1 I (1.200 (1.200 ksi)(2 ksi)(2,, 064 in.4 ) My σ 1 = ∴ M max ≤ = = 412.80 kip-in. I y 6 in. If the allowable bending stress in the steel is 165 MPa, then the maximum bending moment that may be supported by the beam is: σ 2 I My (24.00 (24.00 ksi)(2 ksi)(2,, 064 in.4 ) ∴ M max ≤ = = 495.36 kip-in. σ 2 = n I ny (16.667)(6 in.)
Note: The negative signs were omitted in the previous two equations because only the moment magnitude is of interest here. From these two results, the maximum moment that the beam can support support is 412.80 kip-in. kip-in. maximum concentrated load magnitude P magnitude P that that can be supported is found from: M max = (5 ft)P ∴ P =
M max 5 ft ft
=
412.80 kip-in. (5 ft ft)(12 in./ft)
=
6.88 kips
The
Ans.
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8.43 The cross section of a composite beam that consists of 3-mm-thick fiberglass faces bonded to a 20-mm-thick particleboard core is shown in Fig. P8.43. The beam is subjected to a bending moment of 35 N-m acting about the z axis. The elastic moduli for the fiberglass and the particleboard are 30 GPa and 10 GPa, respectively. Determine: (a) the maximum maximum bending stresses stresses in the fiberglass fiberglass faces and the particleboard core. (b) the stress in the fiberglass at the joint where the two materials are bonded together.
Fig. P8.43
Solution Let the particleboard be denoted as material material (1) and the fiberglass as material (2). The modular ratio is: E 30 GPa =3 n= 2 = E 1 10 GPa Transform the fiberglass faces into an equivalent amount of particleboard by multiplying their width by the modular ratio: b2, trans = 3(50 mm) mm) = 150 mm. Thus, for calculation purposes, the 50 mm × 3 mm fiberglass faces are replaced by particleboard faces that are 150-mm wide and 3-mm thick. Moment of inertia about the horizontal centroidal axis d = y = yi – y y Shape I C C 4
(mm ) transformed fiberglass top face 337.50 particleboard core 33,333.33 transformed fiberglass bot face 337.50 Moment of inertia about the z the z axis axis =
(mm) 11.5 0 –11.5
d²A 4 (mm ) 59,512.50 0 59,512.50
Bending stress in particleboard core (1) From the flexure formula, the maximum bending ben ding stress in the particleboard core is: My (35 N-m) N-m)(( ± 10 mm)( mm)(1,00 1,000 0 mm/ mm/m) m) =− = 2.29 MPa σ 1 = − I 153, 153, 033.33 mm4
I C C + d²A 4 (mm ) 59,850.00 33,333.33 59,850.00 4 153,033.33 mm
Ans.
Bending stress in fiberglass faces (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the maximum bending stress in the fiberglass faces (2) is: My (35 N-m) N-m)(( ± 13 mm)(1 mm)(1,00 ,000 0 mm/m mm/m)) = −(3) = 8.92 MPa Ans. σ 2 = − n I 153,033.33 53, 033.33 mm4 Bending stress in fiberglass (2) at interface At the interface between the particleboard and the fiberglass, y fiberglass, y = = ±10 mm: My (35 (35 N-m) N-m)(( ± 10 mm)( mm)(1, 1,000 000 mm/ mm/m) m) σ 2 = − n = −(3) = 6.86 MPa 4 I 153,033.33 53, 033.33 mm
Ans.
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8.44 A composite beam is made of two brass [ E = 100 GPa] plates bonded to an aluminum [ E E = 75 GPa] bar, as shown in Fig. P8.44. The beam is subjected to a bending moment of 1,750 N-m acting about the z the z axis. axis. Determine: (a) the maximum bending stresses in the brass plates and the aluminum bar. (b) the stress in the brass at the joints where the two materials are bonded together.
Fig. P8.44
Solution Let the aluminum be denoted as material (1) and the brass as material (2). The modular ratio is: E 100 GPa n= 2 = = 1.3333 E 1 75 GPa Transform the brass plates into an equivalent amount of aluminum by multiplying their width by the modular ratio: b2, trans = 1.3333(50 mm) = 66.6666 mm. Thus, for calculation purposes, purposes, the 50 mm × 10 mm brass plates are replaced by aluminum plates that are 66.6666-mm wide and 10-mm thick. Moment of inertia about the horizontal centroidal axis d = y = yi – y y Shape I C C 4
(mm ) transformed top brass plate 5,555.55 aluminum bar 112,500.00 transformed bot brass plate 5,555.55 Moment of inertia about the z the z axis axis =
(mm) 20 0 –20
d²A 4 (mm ) 266,666.40 0 266,666.40
Bending stress in aluminum bar (1) From the flexure formula, the maximum bending stress in the aluminum bar is: My (1,750 (1,750 N-m) N-m)(( ± 15 mm) mm)(1, (1,000 000 mm/m) mm/m) σ 1 = − =− = 40.0 MPa I 656,943.90 mm4
I C C + d²A 4 (mm ) 272,221.95 112,500.00 272,221.95 4 656,943.90 mm
Ans.
Maximum bending stress in brass plates (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the maximum bending stress in the brass plates (2) is: My (1,7 (1,750 50 N-m) N-m)(( ± 25 mm) mm)(1 (1,00 ,000 0 mm/m mm/m)) = −(1.3333) = 88.8 MPa Ans. σ 2 = − n I 656,943.90 mm4 Bending stress in brass plates (2) at interface At the interface between the brass plates and the aluminum bar, y bar, y = = ±15 mm: My (1,750 (1,750 N-m) N-m)(( ± 15 mm)( mm)(1,0 1,000 00 mm/m mm/m)) σ 2 = − n = −(1.3333) = 53.3 MPa 4 I 656,943.90 mm
Ans.
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8.45 An aluminum [ E E = = 10,000 ksi] bar is bonded to a steel [ E = E = 30,000 ksi] bar to form a composite beam (Fig. P8.45b P8.45b). The composite beam is subjected to a bending moment of M of M = = +300 lb-ft about the z axis axis (Fig. P8.45a P8.45a). Determine: (a) the maximum bending stresses in the aluminum an d steel bars. (b) the stress in the two materials at the joint where they are bonded together.
Fig. P8.45a P8.45a
Fig. P8.45b 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 E 1 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. 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
Width b (in.) 1.50 4.50
aluminum bar (1) transformed steel bar (2)
y
=
Σ yi Ai Σ Ai
=
1.8750 in.3 3.00 in.2
=
Height h Height h (in.) 0.50 0.50
yi Ai 3 (in. ) 0.1875 1.6875 1.8750
0.6250 in. (measured upward from bottom edge of section) section)
Moment of inertia about the horizontal centroidal axis = yi – d = y Shape I C C 4
aluminum bar (1) transformed steel bar (2)
Area A Area Ai 2 (in. ) 0.75 2.25 3.00
yi (from bottom) (in.) 0.25 0.75
(in. ) 0.015625 0.046875
Moment of inertia about the z the z axis axis =
(in.) –0.375 0.125
d²A 4 (in. ) 0.105469 0.035156
I C C + d²A 4 (in. ) 0.121094 0.082031 4
0.203125 in.
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(a) Maximum bending stress in aluminum bar (1) From the flexure formula, the maximum bending b ending stress in aluminum bar (1) is: My (300 (300 lblb-ft) ft)(( − 0.6250 0.6250 in.)(1 in.)(12 2 in./ft in./ft)) σ 1 = − 11, 080 psi (T) =− = 11, 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 (300 lb-ft lb-ft)(1 )(1.000 .000 in. in. − 0.6250 0.6250 in.)(12 in.)(12 in./ in./ft) ft) Ans. σ 2 = − = −(3) = 19, 940 psi (C) I 0.203125 in.4 (b) Bending stress in aluminum bar (1) at interface My (300 (300 lb-ft) lb-ft)(0. (0.50 50 in. in. − 0.6250 0.6250 in.)( in.)(12 12 in./f in./ft) t) σ 1
=−
I
=−
0.203125 in.4
=
2,220 psi (T)
(b) Maximum bending stress in steel bar (2) at interface My (300 (300 lb-ft lb-ft)(0 )(0.50 .50 in. in. − 0.6250 0.6250 in.)(1 in.)(12 2 in./ft in./ft)) σ 2 = − = −(3) = 6, 650 psi (T ) I 0.203125 in.4
Ans.
Ans.
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8.46 An aluminum [ E E = = 10,000 ksi] bar is bonded to a steel [ E = E = 30,000 ksi] bar to form a composite beam (Fig. P8.46b P8.46b). The allowable bending stresses for the aluminum and steel bars are 20 ksi and 30 ksi, respectively. Determine the maximum bending moment M moment M that that can be applied to the beam.
Fig. P8.46a P8.46a
Fig. P8.46b P8.46b
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 E 1 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. in. Thus, for calculation purposes, the 1.50 in. 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
Width b (in.) 1.50 4.50
aluminum bar (1) transformed steel bar (2)
y
=
Σ yi Ai Σ Ai
=
1.8750 in.3 3.00 in.2
=
Height h Height h (in.) 0.50 0.50
yi Ai 3 (in. ) 0.1875 1.6875 1.8750
section) 0.6250 in. (measured upward from bottom edge of section)
Moment of inertia about the horizontal centroidal axis d = y = yi – Shape I C C 4
aluminum bar (1) transformed steel bar (2)
Area A Area Ai 2 (in. ) 0.75 2.25 3.00
yi (from bottom) (in.) 0.25 0.75
(in. ) 0.015625 0.046875
Moment of inertia about the z the z axis axis =
(in.) –0.375 0.125
d²A 4 (in. ) 0.105469 0.035156
I C C + d²A 4 (in. ) 0.121094 0.082031 4
0.203125 in.
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(a) Maximum bending moment magnitude based on allowable aluminum stress Based on an allowable bending stress of 20 ksi for the aluminum, the maximum bending moment magnitude that be applied to the cross section is: σ 1
My
≥−
I
∴ M ≤ −
I
σ 1
y
= −
(20 ksi) ksi)(0.2 (0.20312 03125 5 in.4 ) −0.6250
in.
=
6.50 kip-in.
(a)
Maximum bending moment magnitude based on allowable steel stress Based on an allowable bending stress of 30 ksi for the steel, the maximum bending moment magnitude that be applied to the cross section is: σ 2
My ≥ −n I
∴ M ≤ −
I
σ 2
ny
= −
(30 ksi) ksi)(0. (0.2031 203125 25 in.4 ) (3)( (3)(1. 1.00 00 in. in. − 0.625 0.6250 0 in.) in.)
= 5.4167
kip-in.
(b)
Maximum bending moment magnitude From the values obtained in Eqs. (a) and (b), the maximum bending moment that can be applied to the cross section is
M max
=
5.41 5.4167 67 ki kip-in p-in.. = 451 451 lb-f b-ft
Ans.
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8.47 Two steel [ E = = 30,000 ksi] plates are securely attached to a Southern pine [ E E = 1,800 ksi] timber to form a composite beam (Fig. P8.47). The allowable bending stress for the steel plates is 24,000 psi and the allowable bending stress for the Southern pine is 1,200 psi. Determine the maximum bending moment that can be applied about the horizontal axis of the beam.
Fig. P8.47
Solution Denote the timber as material material (1) and denote the steel as material (2). The modular ratio is: E 30,000 ksi n= 2 = = 16.6667 E 1 1,800 ksi Transform the steel plates into an equivalent amount of timber by multiplying their width by the modular ratio: b2, trans = 16.6667(8 in.) = 133.3333 in. Thus, for calculation purposes, the 8 in. × 0.25 in. steel plates can be replaced by wood plates that are 133.3333-in. wide and 0.25-in. thick. Moment of inertia about the horizontal centroidal axis d = y = yi – Shape I C C 4
(in. ) 0.1736 3,413.3333 0.1736
transformed steel plate at top timber (1) transformed steel plate at bottom
(in.) 8.125 0 –8.125
d²A 4 (in. ) 2,200.52 0 2,200.52
I C C + d²A 4 (in. ) 2,200.694 3,413.333 2,200.694 4
Moment of inertia about the z the z axis axis =
7,814.72 in.
(a) Maximum bending moment magnitude based on allowable Southern pine stress Based on an allowable bending stress of 1,200 psi for the Southern pine timber, the maximum bending moment magnitude that be applied to the cross section is: σ 1
My
≥−
I
∴ M ≤ −
I
σ 1
y
= −
(1.200 (1.200 ksi)(7 ksi)(7,814.72 ,814.72 in. in.4 ) ±8
in.
= 1,172.208
kip-in.
(a)
Maximum bending moment magnitude based on allowable steel stress Based on an allowable bending stress of 24,000 psi for the steel plates, the maximum bending moment magnitude that be applied to the cross section is: σ 2
My ≥ −n I
∴ M ≤ −
I
σ 2
ny
= −
(24 ksi)(7,814 ksi)(7,814.72 .72 in. in.4 ) (16.6 16.6667 667)( )(±8.25 8.25 in.) in.)
= 1, 364.021 kip-in.
(b)
Maximum bending moment magnitude From the values obtained in Eqs. (a) and (b), the maximum bending moment that can be applied to the cross section is
M max
,172.2 .208 08 = 1,172
kip kip--in. in. = 97.7 97.7 kipkip-ft ft
Ans.
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P8.48a). 8.48 A simply supported composite beam 5 m long carries a uniformly distributed load w (Fig. P8.48a The beam is constructed of a Southern pine [ E = E = 12 GPa] timber, 200 mm wide by 360 mm deep, that is reinforced on its lower surface by a steel [ E E = = 200 GPa] plate that is 150 mm wide by 12 mm thick (Fig. P8.48b P8.48b). (a) Determine the maximum bending stresses produced in the timber and the steel if w = 12 kN/m. (b) Assume that the allowable bending stresses of the timber and the steel are 9 MPa and 165 MPa, respectively. Determine the largest acceptable magnitude for distributed load w. (You may neglect the weight of the beam in your calculations.)
Fig. P8.48a P8.48a
Fig. P8.48b P8.48b
Solution Let the timber be denoted as material (1) and the steel plate as material (2). The modular ratio is: E 200 GPa n= 2 = = 16.6667 E 1 12 GPa Transform the steel plate (2) into an equivalent amount of wood (1) by multiplying its width by the modular ratio: b2, trans = 16.6667(150 mm) = 2,500 mm. Thus, for calculation calculation purposes, the 150 mm × 12 mm steel plate is replaced by a wood board that is 2,500-mm wide and 12-mm thick. Centroid location of the transformed section in the vertical direction
Shape timber (1) transformed steel plate (2)
y
=
Σ yi Ai Σ Ai
=
Width b (mm) 200 2,500
14,004,000 mm3 102,000 mm2
=
Height h Height h (mm) 360 12
yi Ai 3 (mm ) 13,824,000 180,000 14,004,000
137.294 mm (measured upward from bottom edge of section)
Moment of inertia about the horizontal centroidal axis d = y = yi – Shape I C C 4
timber (1) transformed steel plate (2)
Area A Area Ai 2 (mm ) 72,000 30,000 102,000
yi (from bottom) (mm) 192 6
(mm ) 777,600,000 360,000
Moment of inertia about the z the z axis axis =
(mm) 54.71 –131.29
d²A (mm4) 215,476,817 517,144,360
I C C + d²A (mm4) 993,076,817 517,504,360 4 1,510,581,176 mm 9 4 = 1.5106 ×10 mm
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Bending moment in beam for w = 12 kN/m The bending moment in the simply supported beam with a uniformly distributed load of 12 kN/m is:
M max
=
wL2 8
=
(12 kN/m)(5 m)2 8
= 37.5
kN-m = 37.5 × 106 N-mm
Bending stress in timber (1) From the flexure formula, the maximum bending bend ing stress in timber (1) is: σ 1
My
=−
I
=−
(37.5 ×106 N-mm)(372 mm mm − 137. 37.294 mm mm) 1.5106 × 109 mm4
=
5.83 MPa (C)
Ans.
Bending stress in steel plate (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the maximum bending stress in steel plate (2) is: σ 2
My
=−
I
= −(16.6667)
(37.5 × 106 N-mm)( − 137.294 mm) 1.5106 × 109 mm4
=
56.8 MPa (T)
Ans.
Determine maximum w If the allowable bending stress in the timber is 9 MPa, then the maximum bending moment that may be supported by the beam is: My σ 1 I (9 N/mm 2 )(1.5106 × 109 mm4 ) 6 σ 1 = ∴ M max ≤ = = 57.925 × 10 N-mm I y (372 (372 mm − 137. 137.29 294 4 mm) mm) If the allowable bending stress in the steel is 165 MPa, then the maximum bending moment that may be supported by the beam is: My σ 2 I (165 N/ N/mm2 )(1.5106 × 109 mm4 ) 6 σ 2 = n ∴ M max ≤ = = 108.926 × 10 N-mm I ny (16.6667)(137.294 (16.6667)(137.294 mm)
Note: The negative signs were omitted in the previous two equations because only the moment magnitude is of interest here. 6
From these two results, the maximum moment that the beam can support is 57.925×10 N-mm. The maximum distributed load magnitude w that can be supported is found from: wL2 M max = 8 ∴w =
8 M max L2
=
8(57 8(57.9 .925 25× 106 N-mm N-mm)( )(1 1 m/10 m/1000 00 mm) mm) (5 m) 2
,536 = 18,53
N/ N/m = 18.54 kN kN/m
Ans.
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8.49 A glue-laminated timber beam is reinforced by carbon fiber reinforced plastic (CFRP) material bonded to its bottom surface. The cross section of the composite beam is shown in Fig. P8.49b P8.49b. The elastic modulus of the wood is E is E = = 12 GPa and the elastic modulus of the CFRP is 112 GPa. The simply supported beam spans 6 m and carries a concentrated load P load P at at midspan (Fig. P8.49a P8.49a). (a) Determine the maximum bending stresses produced in the timber and the CFRP if P if P = = 4 kN. (b) Assume that the allowable bending stresses of the timber and the CFRP are 9 MPa and 1,500 MPa, respectively. Determine the largest acceptable magnitude for concentrated load P load P . (You may neglect the weight of the beam in your calculations.)
Fig. P8.49a P8.49a Fig. P8.49b P8.49b
Solution Denoted the timber as material (1) and denote the CFRP as material material (2). The modular ratio is: E 112 GPa n= 2 = = 9.3333 E 1 12 GPa Transform the CFRP into an equivalent amount of wood by multiplying its width by the modular ratio: b2, trans = 9.3333(40 mm) = 373.33 mm. Thus, for calculation purposes, the 40 mm × 3 mm CFRP is replaced by a wood board that is 373.33-mm wide and 3-mm thick. Centroid location of the transformed section in the vertical direction
Shape timber (1) transformed CFRP (2)
y
=
Σ yi Ai Σ Ai
=
Width b (mm) 90 373.33
2,881,680 mm3 23,620 mm2
=
Height h Height h (mm) 250 3
yi Ai (mm3) 2,880,000 1,680 2,881,680
122.00 mm (measured upward from bottom edge of section)
Moment of inertia about the horizontal centroidal axis d = y = yi – Shape I C C 4
timber (1) transformed CFRP (2)
Area A Area Ai (mm2) 22,500 1,120 23,620
yi (from bottom) (mm) 128 1.5
(mm ) 117,187,500 840
Moment of inertia about the z the z axis axis =
(mm) 6.00 –120.50
d²A 4 (mm ) 810,000 16,262,680
I C C + d²A 4 (mm ) 117,997,500 117,99 7,500 16,263,520 134,261,020 mm4 6 4 = 134.261 ×10 mm
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Bending moment in beam for P = = 4 kN The bending moment in the simply supported beam with a concentrated load of 4 kN at midspan is: PL (4 kN)(6 m) 6 M max = = = 6 kN-m = 6 × 10 N-mm 4 4 (a) Bending stress in timber (1) From the flexure formula, the maximum bending bend ing stress in timber (1) is: 6 My (6 × 10 N-mm)(253 mm − 122.00 mm mm) σ 1 = − =− = 5.85 MPa (C) I 134.261× 106 mm4
Ans.
(a) Bending stress in CFRP (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the maximum bending stress in the CFRP is: (6 ×106 N-mm)( − 122.00 mm) My = −(9.3333) = 50.9 MPa (T ) Ans. σ 2 = − I 134.261× 106 mm4 (b) Determine maximum P If the allowable bending stress in the timber is 9 MPa, then the maximum bending moment that may be supported by the beam is: My σ 1 I (9 N/mm 2 )(134.261× 106 mm4 ) 6 ∴ M max ≤ = = 9.224 × 10 N-mm σ 1 = I y (253 (253 mm − 122. 122.00 00 mm) If the allowable bending stress in the CFRP is 1,500 MPa, then the maximum bending moment that may be supported by the beam is: My σ 2 I (1,500 N/mm2 )(134.261× 106 mm4 ) 6 σ 2 = n ∴ M max ≤ = = 176.867 × 10 N-mm I ny (9.3333)(122.00 (9.3333)(122.00 mm)
Note: The negative signs were omitted in the previous two equations because only the moment magnitude is of interest here. 6
From these two results, the maximum moment that the beam can support is 9.224×10 N-mm. The maximum concentrated load magnitude P magnitude P that that can be supported is found from: PL M max = 4 4 M max 4(9. 4(9.22 224 4 × 106 N-mm N-mm)( )(1 1 m/1 m/100 000 0 mm) mm) P ∴ = = = 6,149 N = 6.15 kN Ans. L (6 m)
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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 M z = +50 kip-in is applied about the z the z axis. axis. Assume E Assume E wood wood = 2,000 ksi and E and E steel steel = 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 E 1 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, symmetric, the centroid location is is found from symmetry. Moment of inertia about the z centroidal centroidal axis
Shape
I C C 4 (in. ) wood beam (1) 128 two transformed steel plates (2) 40 Moment of inertia about the z the z axis axis =
= yi – d = y (in.) 0 0
d²A 4 (in. ) 0 0
Bending stress in wood beam (1) From the flexure formula, the maximum bending be nding stress in wood beam (1) is: M z c (50 kip-in.)(4 in.) ksi = 1,190 ps psi = = 1.190 ks σ 1 = I z 168 in.4
I C C + d²A 4 (in. ) 128 40 4 168 in.
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 z c (50 kip-in.)(2 in.) Ans. = (15) = 8.93 ksi = 8, 930 psi σ 2 = n I z 168 in.4
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8.51 A glue-laminated timber beam is reinforced by carbon fiber reinforced plastic (CFRP) material bonded to its bottom surface. The cross section of the composite beam is shown in Fig. P8.51b P8.51b. The elastic modulus of the wood is 1,700 ksi and the elastic modulus of the CFRP is 23,800 ksi. The simply supported beam spans 24 ft and carries two concentrated loads P loads P , which act at the quarter-points of the span (Fig. P8.51a P8.51a). The allowable bending stresses of the timber and the CFRP are 2,400 psi and 175,000 psi, respectively. Determine the largest acceptable magnitude for the concentrated loads P . (You may neglect the weight of the beam in your calculations.)
Fig. P8.51a P8.51a Fig. P8.51b P8.51b
Solution Denoted the timber as material (1) and denote the CFRP as material (2). The modular ratio is: E 23, 23, 800 ksi n= 2 = = 14 E 1 1,700 ksi Transform the CFRP into an equivalent amount of wood by multiplying its width by the modular ratio: b2, trans = 14(3 in.) = 42 in. Thus, for calculation purposes, the 3 in. × 0.125 in. CFRP is replaced by a wood board that is 42-in. wide and 0.125-in. thick. Centroid location of the transformed section in the vertical direction
Shape timber (1) transformed CFRP (2)
y
=
Σ yi Ai Σ Ai
=
404.5781 in.3 71.25 in.2
Width b (in.) 5.5 42.0
=
Height h Height h (in.) 12 0.125
Area A Area Ai 2 (in. ) 66 5.25 71.25
yi (from bottom) (in.) 6.125 0.0625
yi Ai 3 (in. ) 404.25 0.3281 404.5781
5.6783 in. (measured upward from bottom edge of section)
Moment of inertia about the horizontal centroidal axis d = y = yi – Shape I C C 4
(in. ) timber (1) 792 transformed CFRP (2) 0.00684 Moment of inertia about the z the z axis axis =
(in.) 0.4467 –5.6158
d²A 4 (in. ) 13.1703 165.5697
I C + d²A C + 4 (in. ) 805.170 165.577 4 970.747 in.
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Determine maximum P If the allowable bending stress in the timber is 2,400 psi, then the maximum bending moment that may be supported by the beam is: σ 1 I My (2.40 (2.40 ksi)(9 ksi)(970. 70.747 747 in.4 ) ∴ M max ≤ = = 361.393 kip-in. σ 1 = I y (12. (12.12 125 5 in. in. − 5.67 5.6783 83 in. in.)) If the allowable bending stress in the CFRP is 175,000 psi, then the maximum bending moment that may be supported by the beam is: σ 2 I My (175 (175 ksi)(97 ksi)(970.7 0.747 47 in.4 ) ∴ M max ≤ = = 2,137 kip-in. σ 2 = n I ny (14)(5.6783 in.)
Note: The negative signs were omitted in the previous two equations because only the moment magnitude is of interest here. From these two results, the maximum moment that the beam can support is 351.393 kip-in. = 30.116 kip-ft. The maximum concentrated load magnitude P magnitude P that that can be supported is found from: M max = (6 ft)P ∴ P =
M max 6 ft
=
30.116 kip-ft 6 ft
=
5.02 kips
Ans.
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