Mechanics of Materials Solutions Chapter08 Probs42 51

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8.42 A composite beam is fabricated by bolting two 3 in. wide × 12 in. deep timber planks to the sidesof a 0.50 in. × 12 in. steel plate (Fig. 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 , which are applied at the quarter points of the span (Fig. P8.42a).(a) Determine the maximum bending stresses produced in the timber planks and the steel plate 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 . (You may neglect

the weight of the beam in your calculations.)

Fig. P8.42a

Fig. P8.42b

Solution

Let the timber be denoted as material (1) and the steel plate as material (2). The modular ratio is:

2

1

30,000 ksi16.6667

1,800 ksi

E n

E = = =

Transform the steel plate (2) into an equivalent amount of wood (1) by multiplying its width by themodular 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

Shape I C d = yi – d²A I C + d²A

(in.4) (in.) (in.

4) (in.

4)

timber (1) 864 0 0 864

transformed steel plate (2) 1,200 0 0 1,200

Moment of inertia about the z axis = 2,064 in.4

Bending moment in beam for P = 3 kips

The bending moment in the simply supported beam with two 3-kip concentrated loads is:max (3 kips)(5 ft) 15 kip-ft 180 kip-in. M = = =

Bending stress in timber (1) From the flexure formula, the maximum bending stress in timber (1) is:

1 4

(180 kip-in.)( 6 in.)0.5233 ksi 523 psi

2,064 in.

My

I σ

±= − = − = ± = 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:

2 4

(180 kip-in.)( 6 in.)(16.6667) 8.7209 ksi 8,720 psi

2,064 in.

Myn

I σ

±= − = − = ± = Ans.

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:4

11 max

(1.200 ksi)(2,064 in. )412.80 kip-in.

6 in.

I My M

I y

σ σ = ∴ ≤ = =

If the allowable bending stress in the steel is 165 MPa, then the maximum bending moment that may besupported by the beam is:

4

22 max

(24.00 ksi)(2,064 in. )495.36 kip-in.

(16.667)(6 in.)

I Myn M

I n y

σ σ = ∴ ≤ = =

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 412.80 kip-in. The

maximum concentrated load magnitude P that can be supported is found from:

max

max

(5 ft)

412.80 kip-in.6.88 kips

5 ft (5 ft)(12 in./ft)

M P

M P

=

∴ = = = Ans.





8.43 The cross section of a composite beam thatconsists 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 momentof 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 bending stresses in the 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 (1) and the fiberglass as material (2). The modular ratio is:

2

1

30 GPa3

10 GPa

E n

E = = =

Transform the fiberglass faces into an equivalent amount of particleboard by multiplying their width by

the modular ratio: b2, trans = 3(50 mm) = 150 mm. Thus, for calculation purposes, the 50 mm × 3 mmfiberglass faces are replaced by particleboard faces that are 150-mm wide and 3-mm thick.


Shape I C d = yi – y d²A I C + d²A

(mm4) (mm) (mm

4) (mm

4)

transformed fiberglass top face 337.50 11.5 59,512.50 59,850.00

particleboard core 33,333.33 0 0 33,333.33

transformed fiberglass bot face 337.50 –11.5 59,512.50 59,850.00

Moment of inertia about the z axis = 153,033.33 mm4

Bending stress in particleboard core (1) From the flexure formula, the maximum bending stress in the particleboard core is:

1 4

(35 N-m)( 10 mm)(1,000 mm/m)2.29 MPa

153,033.33 mm

My

I σ

±= − = − = 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:

2 4

(35 N-m)( 13 mm)(1,000 mm/m)(3) 8.92 MPa

153,033.33 mm

Myn

I σ

±= − = − = Ans.

Bending stress in fiberglass (2) at interface At the interface between the particleboard and the fiberglass, y = ±10 mm:

2 4

(35 N-m)( 10 mm)(1,000 mm/m)(3) 6.86 MPa

153,033.33 mm

Myn

I σ

±= − = − = Ans.





8.44 A composite beam is made of two brass [ E =100 GPa] plates bonded to an aluminum [ E = 75

GPa] bar, as shown in Fig. P8.44. The beam is

subjected to a bending moment of 1,750 N-m actingabout the z 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:

2

1

100 GPa1.3333

75 GPa

E n

E = = =

Transform the brass plates into an equivalent amount of aluminum by multiplying their width by themodular ratio: b2, trans = 1.3333(50 mm) = 66.6666 mm. Thus, for calculation purposes, the 50 mm × 10

mm brass plates are replaced by aluminum plates that are 66.6666-mm wide and 10-mm thick.


Shape I C d = yi – y d²A I C + d²A

(mm4) (mm) (mm

4) (mm

4)

transformed top brass plate 5,555.55 20 266,666.40 272,221.95

aluminum bar 112,500.00 0 0 112,500.00

transformed bot brass plate 5,555.55 –20 266,666.40 272,221.95

Moment of inertia about the z axis = 656,943.90 mm4

Bending stress in aluminum bar (1) From the flexure formula, the maximum bending stress in the aluminum bar is:

1 4

(1,750 N-m)( 15 mm)(1,000 mm/m)40.0 MPa

656,943.90 mm

My

I σ

±= − = − = 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:

2 4

(1,750 N-m)( 25 mm)(1,000 mm/m)(1.3333) 88.8 MPa

656,943.90 mm

Myn

I

σ ±

= − = − = Ans.

Bending stress in brass plates (2) at interface At the interface between the brass plates and the aluminum bar, y = ±15 mm:

2 4

(1,750 N-m)( 15 mm)(1,000 mm/m)(1.3333) 53.3 MPa

656,943.90 mm

Myn

I σ

±= − = − = Ans.





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:

2

1

30,000 ksi3

10,000 ksi

E n

E = = =

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 Width b Height h Area Ai

yi (from bottom) yi Ai

(in.) (in.) (in.2

) (in.) (in.3

)aluminum bar (1) 1.50 0.50 0.75 0.25 0.1875

transformed steel bar (2) 4.50 0.50 2.25 0.75 1.6875

3.00 1.8750

3

2

1.8750 in.0.6250 in.

3.00 in.

i i

i

y A y

A

Σ= = =

Σ (measured upward from bottom edge of section)



(in.4) (in.) (in.4) (in.4)aluminum bar (1) 0.015625 –0.375 0.105469 0.121094

transformed steel bar (2) 0.046875 0.125 0.035156 0.082031

Moment of inertia about the z axis = 0.203125 in.4





(a) Maximum bending stress in aluminum bar (1) From the flexure formula, the maximum bending stress in aluminum bar (1) is:

1 4

(300 lb-ft)( 0.6250 in.)(12 in./ft)11,080 psi (T)

0.203125 in.

My

I σ

−= − = − = 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:

2 4

(300 lb-ft)(1.000 in. 0.6250 in.)(12 in./ft)(3) 19,940 psi (C)

0.203125 in.

My

I σ

−= − = − = Ans.

(b) Bending stress in aluminum bar (1) at interface

1 4

(300 lb-ft)(0.50 in. 0.6250 in.)(12 in./ft)2,220 psi (T)

0.203125 in.

My

I σ

−= − = − = Ans.

(b) Maximum bending stress in steel bar (2) at interface

2 4

(300 lb-ft)(0.50 in. 0.6250 in.)(12 in./ft)(3) 6,650 psi (T)

0.203125 in.

My

I σ

−= − = − = Ans.





8.46 An aluminum [ E = 10,000 ksi] bar is bonded to a steel [ E = 30,000 ksi] bar to form a composite beam (Fig. 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 that can be applied to the beam.


Solution

Denote the aluminum as material (1) and denote the steel as material (2). The modular ratio is:

2

1

30,000 ksi310,000 ksi

E n E = = =

Transform the steel bar (2) into an equivalent amount of aluminum (1) by multiplying its width by themodular 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.




(in.) (in.) (in.2) (in.) (in.

3)

aluminum bar (1) 1.50 0.50 0.75 0.25 0.1875

transformed steel bar (2) 4.50 0.50 2.25 0.75 1.68753.00 1.8750

3

2

1.8750 in.0.6250 in.

3.00 in.

i i

i

y A y

A

Σ= = =




(in.4) (in.) (in.

4) (in.

4)

aluminum bar (1) 0.015625 –0.375 0.105469 0.121094

transformed steel bar (2) 0.046875 0.125 0.035156 0.082031






(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:4

11

(20 ksi)(0.203125 in. )6.50 kip-in.

0.6250 in.

I My M

I y

σ σ ≥ − ∴ ≤ − = − =

− (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 magnitudethat be applied to the cross section is:

4

22

(30 ksi)(0.203125 in. )5.4167 kip-in.

(3)(1.00 in. 0.6250 in.)

I Myn M

I n y

σ σ ≥ − ∴ ≤ − = − =

− (b)

Maximum bending moment magnitude

From the values obtained in Eqs. (a) and (b), the maximum bending moment that can be applied to thecross section is

max 5.4167 kip-in. 451 lb-ft M = = Ans.





8.47 Two steel [ E = 30,000 ksi] plates are securelyattached to a Southern pine [ E = 1,800 ksi] timber

to form a composite beam (Fig. P8.47). The

allowable bending stress for the steel plates is24,000 psi and the allowable bending stress for the

Southern pine is 1,200 psi. Determine the maximum

bending moment that can be applied about thehorizontal axis of the beam.

Fig. P8.47

Solution

Denote the timber as material (1) and denote the steel as material (2). The modular ratio is:

2

1

30,000 ksi16.6667

1,800 ksi

E n

E = = =

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.



(in.4) (in.) (in.

4) (in.

4)

transformed steel plate at top 0.1736 8.125 2,200.52 2,200.694

timber (1) 3,413.3333 0 0 3,413.333

transformed steel plate at bottom 0.1736 –8.125 2,200.52 2,200.694

Moment of inertia about the z axis = 7,814.72 in.4

(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:4

11

(1.200 ksi)(7,814.72 in. )1,172.208 kip-in.

8 in.

I My M

I y

σ σ ≥ − ∴ ≤ − = − =

± (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:4

22

(24 ksi)(7,814.72 in. )1,364.021 kip-in.

(16.6667)( 8.25 in.)

I Myn M

I n y

σ σ ≥ − ∴ ≤ − = − =

±

(b)

Maximum bending moment magnitude

From the values obtained in Eqs. (a) and (b), the maximum bending moment that can be applied to thecross section is

max 1,172.208 kip-in. 97.7 kip-ft M = = Ans.





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 = 12 GPa] timber, 200 mm wide by 360 mm deep, that is

reinforced on its lower surface by a steel [ E = 200 GPa] plate that is 150 mm wide by 12 mm thick (Fig.

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 theweight of the beam in your calculations.)


Solution

Let the timber be denoted as material (1) and the steel plate as material (2). The modular ratio is:

2

1

200 GPa16.6667

12 GPa

E n

E = = =

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 purposes, the 150 mm ×12 mm steel plate is replaced by a wood board that is 2,500-mm wide and 12-mm thick.


Shape Width b Height h Area Ai yi

(from bottom) yi Ai

(mm) (mm) (mm2) (mm) (mm

3)

timber (1) 200 360 72,000 192 13,824,000

transformed steel plate (2) 2,500 12 30,000 6 180,000

102,000 14,004,000

3

2

14,004,000 mm137.294 mm

102,000 mm

i i

i

y A y

A

Σ= = =


Moment of inertia about the horizontal centroidal axisShape I C d = yi – d²A I C + d²A

(mm4) (mm) (mm4) (mm4)

timber (1) 777,600,000 54.71 215,476,817 993,076,817

transformed steel plate (2) 360,000 –131.29 517,144,360 517,504,360

Moment of inertia about the z axis =1,510,581,176 mm

4

= 1.5106 ×109 mm

4





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:

2 26

max

(12 kN/m)(5 m)37.5 kN-m 37.5 10 N-mm

8 8

wL M = = = = ×

Bending stress in timber (1) From the flexure formula, the maximum bending stress in timber (1) is:

6

1 9 4(37.5 10 N-mm)(372 mm 137.294 mm) 5.83 MPa (C)1.5106 10 mm

My I

σ × −

= − = − =×

Ans.

Bending stress in steel plate (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, themaximum bending stress in steel plate (2) is:

6

2 9 4

(37.5 10 N-mm)( 137.294 mm)(16.6667) 56.8 MPa (T)

1.5106 10 mm

My

I σ

× −= − = − =

× 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:2 9 4

611 max

(9 N/mm )(1.5106 10 mm )57.925 10 N-mm

(372 mm 137.294 mm)

I My M

I y

σ σ

×= ∴ ≤ = = ×

−

If the allowable bending stress in the steel is 165 MPa, then the maximum bending moment that may be

supported by the beam is:2 9 4

622 max

(165 N/mm )(1.5106 10 mm )108.926 10 N-mm

(16.6667)(137.294 mm)

I Myn M

I n y

σ σ

×= ∴ ≤ = = ×

Note: The negative signs were omitted in the previous two equations because only the momentmagnitude is of interest here.

From these two results, the maximum moment that the beam can support is 57.925×106 N-mm. Themaximum distributed load magnitude w that can be supported is found from:

2

max

6

max

2 2

8

8 8(57.925 10 N-mm)(1 m/1000 mm)18,536 N/m 18.54 kN/m

(5 m)

wL M

M w

L

=

×∴ = = = = Ans.





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. The

elastic modulus of the wood 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 at midspan (Fig. P8.49a).(a) Determine the maximum bending stresses produced in the timber and the CFRP 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 . (You may neglect theweight of the beam in your calculations.)

Fig. P8.49a

Fig. P8.49b

Solution

Denoted the timber as material (1) and denote the CFRP as material (2). The modular ratio is:

2

1

112 GPa9.3333

12 GPa

E n

E = = =

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.




(mm) (mm) (mm2) (mm) (mm3)

timber (1) 90 250 22,500 128 2,880,000

transformed CFRP (2) 373.33 3 1,120 1.5 1,680

23,620 2,881,680

3

2

2,881,680 mm122.00 mm

23,620 mm

i i

i

y A y

A

Σ= = =




(mm4) (mm) (mm

4) (mm

4)

timber (1) 117,187,500 6.00 810,000 117,997,500

transformed CFRP (2) 840 –120.50 16,262,680 16,263,520

Moment of inertia about the z axis =

134,261,020 mm4

= 134.261 ×106 mm

4





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:

6

max

(4 kN)(6 m)6 kN-m 6 10 N-mm

4 4

PL M = = = = ×

(a) Bending stress in timber (1) From the flexure formula, the maximum bending stress in timber (1) is:

6

1 6 4

(6 10 N-mm)(253 mm 122.00 mm)5.85 MPa (C)

134.261 10 mm

My

I σ

× −= − = − =

× Ans.

(a) Bending stress in CFRP (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, themaximum bending stress in the CFRP is:

6

2 6 4

(6 10 N-mm)( 122.00 mm)(9.3333) 50.9 MPa (T)

134.261 10 mm

My

I σ

× −= − = − =

× Ans.

(b) Determine maximum P

If the allowable bending stress in the timber is 9 MPa, then the maximum bending moment that may besupported by the beam is:

2 6 461

1 max

(9 N/mm )(134.261 10 mm )9.224 10 N-mm

(253 mm 122.00 mm)

I My M

I y

σ σ

×= ∴ ≤ = = ×

−

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:2 6 4

622 max

(1,500 N/mm )(134.261 10 mm )176.867 10 N-mm

(9.3333)(122.00 mm)

I Myn M

I n y

σ σ

×= ∴ ≤ = = ×

Note: The negative signs were omitted in the previous two equations because only the momentmagnitude is of interest here.

From these two results, the maximum moment that the beam can support is 9.224×106 N-mm. The

maximum concentrated load magnitude P that can be supported is found from:

max

6

max

4

4 4(9.224 10 N-mm)(1 m/1000 mm)6,149 N 6.15 kN

(6 m)

PL M

M P

L

=

×∴ = = = = Ans.





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 suchthat 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 isapplied about the z axis. Assume E wood = 2,000 ksi

and E 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:

2

1

30,000 ksi15

2,000 ksi

E n

E = = =

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


(in.4) (in.) (in.

4) (in.

4)

wood beam (1) 128 0 0 128

two transformed steel plates (2) 40 0 0 40

Moment of inertia about the z axis = 168 in.4

Bending stress in wood beam (1) From the flexure formula, the maximum bending stress in wood beam (1) is:

1 4

(50 kip-in.)(4 in.)1.190 ksi 1,190 psi

168 in.

z

z

M c

I σ = = = = Ans.

Bending stress in steel plates (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, themaximum bending stress in the steel plates (2) is:

2 4(50 kip-in.)(2 in.)(15) 8.93 ksi 8,930 psi

168 in.

z

z

M cn I

σ = = = = Ans.





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. 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 , which act at the quarter-points of thespan (Fig. 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

Fig. P8.51b

Solution

Denoted the timber as material (1) and denote the CFRP as material (2). The modular ratio is:

2

1

23,800 ksi14

1,700 ksi

E n

E = = =

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 awood board that is 42-in. wide and 0.125-in. thick.




(in.) (in.) (in.2) (in.) (in.

3)

timber (1) 5.5 12 66 6.125 404.25

transformed CFRP (2) 42.0 0.125 5.25 0.0625 0.3281

71.25 404.5781

3

2

404.5781 in.5.6783 in.

71.25 in.

i i

i

y A y

A

Σ= = =




(in.4) (in.) (in.

4) (in.

4)

timber (1) 792 0.4467 13.1703 805.170

transformed CFRP (2) 0.00684 –5.6158 165.5697 165.577





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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:4

11 max

(2.40 ksi)(970.747 in. )361.393 kip-in.

(12.125 in. 5.6783 in.)

I My M

I y

σ σ = ∴ ≤ = =

−

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:

4

22 max (175 ksi)(970.747 in. ) 2,137 kip-in.

(14)(5.6783 in.) I Myn M

I n yσ σ = ∴ ≤ = =

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 that can be supported is found from:

max

max

(6 ft)

30.116 kip-ft5.02 kips

6 ft 6 ft

M P

M P

=

∴ = = = Ans.

Mechanics of Materials Solutions Chapter08 Probs42 51

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