Mechanics of Materials Solutions Chapter09 Probs38 53

9.38 A wooden beam is fabricated from one 2 × 8 and two 2 × 4 pieces of dimension lumber to form the I-beam cross section shown in Fig. P9.38. The flanges of the beam are fastened to the web with nails that can safely transmit a force of 100 lb in direct shear. If the beam is simply supported and carries a 1000-lb load at the center of a 12-ft span, determine: (a) the horizontal force transferred from each flange to the web in a 12in. long segment of the beam. (b) the maximum spacing s spacing s (along (along the length of the beam) required for the nails. (c) the maximum horizontal shear stress in the I-beam.

Fig. P9.38

Solution Moment of inertia about the z axis: axis: Shape Width b Height h (in.) (in.) top flange 4 2 web 2 8 bottom flange 4 2

d = y = yi – y y (in.) 5 0 −5

I C C 4 (in. ) 2.6667 85.3333 2.6667

d²A 4 (in. ) 200.0000 0.0000 200.0000 4

Moment of inertia about the z the z axis axis (in. ) =

I C C + d²A 4 (in. ) 202.6667 85.3333 202.6667 490.6667

Maximum shear force For P For P = = 1,000 lb, V = P = P /2 /2 = 500 lb (a) Horizontal force transferred from each flange (in a 12-in. length): Q = (4 in. in.)( )(2 2 in.) in.)(5 (5 in.) in.) = 40 in. in.3

q=

VQ I

=

(500 (500 lb)(40 lb)(40 in.3 ) 490.6667 in.4

=

40.761 lb/in.

F (40.76 761 1 lb/in b/in.) .)(1 (12 2 in.) in.) = 489 489 lb lb H = (40.

Ans.

(b) Maximum nail spacing: q s ≤ n f V f ∴ s ≤

n f V f q

=

(1 nail)(100 lb/nail) 40.761 lb/in.

=

2.45 in.

Ans.

(c) Maximum horizontal shear stress: Q = (4 in. in.)( )(2 2 in.) in.)(5 (5 in.) in.) + (2 in. in.)( )(4 4 in.) in.)(2 (2 in.) in.) = 56 in. in.3 τ =

(500 (500 lb)( lb)(56 56 in. in.3 ) (490.666 (490.6667 7 in. in.4 )(2 in.) in.)

=

28.5 psi

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 Copyright Act without the permission of the copyright copyright owner is unlawful. unlawful.

9.39 A wooden beam is fabricated from one 2 × 8 and two 2 × 4 pieces of dimension lumber to form the I-beam cross section shown in Fig. P9.39. The I-beam will be used as a simply supported beam to carry a concentrated load P at the center of a 20-ft span. The wood has an allowable bending stress of 1,200 psi and an allowable shear stress of 90 psi. The Th e flanges of the beam are fastened to the web with nails that can c an safely transmit a force of 100 lb in direct shear. (a) If the nails are uniformly spaced at an interval of s of s = 3.5 in. along the span, what is the maximum concentrated load P load P that that can be supported by the beam? Demonstrate that the bending and shear stresses produced by P are are acceptable. (b) Determine the magnitude of load P that produces the allowable bending stress in the span (i.e., σ b = 1,200 psi). What nail spacing s spacing s is required to support this load magnitude?

Fig. P9.39

Solution Moment of inertia about the z axis: axis: Shape Width b Height h (in.) (in.) top flange 4 2 web 2 8 bottom flange 4 2

d = y = yi – y y (in.) 5 0 −5

I C C 4 (in. ) 2.6667 85.3333 2.6667

I C C + d²A 4 (in. ) 202.6667 85.3333 202.6667

d²A 4 (in. ) 200.0000 0.0000 200.0000 4


490.6667

Maximum shear force For P For P = = 1,000 lb, V = P = P /2 /2 = 500 lb (a) Maximum concentrated load P : Q = (4 in. in.)( )(2 2 in. in.)( )(5 5 in. in.)) = 40 in. in.3

q s ≤ n f V f ∴q ≤

q=

=

s

=

(1 nail)(100 lb/nail) 3.5 in.

=

28.571 lb/in.

VQ I

∴V =

V

n f V f

P 2

q I Q

=

(28.571 lb/in.)(49 lb/in.)(490.6667 0.6667 in.4 ) 40 in.3 ∴ P max =

701 lb

= 350.5

lb Ans.


(b) Magnitude of load P that that produces the allowable bending stress in the span: M c ≤ 1, 200 psi σ x = I ∴ M ≤

M max

=

(1, 200 psi)(490 psi)(490.6667 .6667 in.4 ) 6 in.

lb-in.

P L 4

∴ P max =

4 M max L

=

4(98,133.34 lb-in.)

Required nail spacing s: P 1,635.56 lb V max = max = 2 2

q=

= 98,133.34

VQ

=

= 817.78

(817.7 (817.78 8 lb)(40 lb)(40 in.3 ) 490.6667 in.4

I q s ≤ n f V f ∴ s ≤

(20 ft)(12 in./ft)

n f V f q

=

=

(1 nail)(100 lb) 66.67 lb/in.

,635.56 = 1,63

lb lb = 1,636 ,636 lb lb

Ans.

lb 66.67 lb/in.

=

1.500 in.

Ans.


9.40 A wooden box beam is fabricated from four boards, which are fastened together with nails, as shown in Fig. P9.40b P9.40b. The nails are installed at a spacing of s of s = = 125 mm (Fig. P9.40a P9.40a), and each nail can provide a resistance of V f = 500 N. In service, the box beam will be installed so that bending occurs about the z the z axis. axis. Determine the maximum shear force V that V that can be supported by the box beam based on the shear capacity of the nailed connections.

Fig. P9.40a P9.40a

Fig. P9.40b P9.40b

Solution Moment of inertia I z : (200 (200 mm)( mm)(25 250 0 mm) mm)3 I z = 12

−

(120 120 mm) mm)(2 (200 00 mm)3 12

180, 416, 416, 667 = 180,

mm4

First moment of area Q: Q = (200 (200 mm)( mm)(25 25 mm) mm)(1 (112. 12.5 5 mm) mm) = 562,500 562,500 mm mm3 Shear flow q based on nail shear force: q s ≤ n f V f ∴q ≤

n f V f s

=

(2 nails)(500 N/nail) 125 mm

=8

N/mm

Maximum shear force V : VQ q I z (8 N/mm) N/mm)(18 (180,4 0,416,6 16,667 67 mm4 ) q= ∴V = = I z Q 562,500 mm3

=

2, 56 566 N = 2.57 kN

Ans.


9.41 A wooden box beam is fabricated from four boards, which are fastened together with screws, as shown in Fig. P9.41b P9.41b. Each screw can provide a resistance of 800 of 800 N. In service, the box beam will be installed so that bending occurs about the z the z axis, axis, and the maximum shear force in the beam will be 7 kN. Determine the maximum permissible spacing interval s interval s for for the screws (see Fig. P9.41a P9.41a).

Fig. P9.41a P9.41a

Fig. P9.41b P9.41b

Solution Moment of inertia I z : (190 190 mm) mm)(2 (250 50 mm) mm)3 I z = 12

−

(140 140 mm mm)(15 )(150 0 mm) mm)3 12

=

208,020,833 mm4

First moment of area Q: Q = (140 140 mm)( mm)(50 50 mm) mm)(1 (100 00 mm) mm) = 700,000 700,000 mm3 Shear flow q based on beam shear force V : VQ (7,000 (7,000 N)(75 N)(750,0 0,000 00 mm3 ) q= = = 23.5553 N/mm I z 208,020,833 mm4 Maximum spacing interval s: q s ≤ n f V f ∴ s ≤

n f V f q

=

(2 screws)(800 N/screw) 23.5553 N/mm

=

67.9 mm

Ans.


9.42 A wooden beam is fabricated by nailing together three pieces of dimension lumber, as shown in Fig. P9.42a P9.42a. The cross-sectional dimensions of the beam are shown in Fig. P9.42b P9.42b. The beam must support an internal shear force of V = = 600 lb. (a) Determine the maximum horizontal shear stress in the cross section for V = = 600 lb. (b) If each nail can provide 120 lb of horizontal resistance, determine the maximum spacing s spacing s for the nails. (c) If the three boards were connected by glue instead of nails, what minimum shear strength would be necessary for the glued joints?

Fig. P9.42a P9.42a

Fig. P9.42b P9.42b

Solution Centroid location in y direction:

Shape

Width b (in.) 2 4 2

left board flange board right board

y

=

Σ yi Ai Σ A i

=

184 in.3 40 in.2

= =

Height h (in.) 8 2 9

4.6 in in.

Area A Area Ai 2 (in. ) 16 8 16 40

yi (from bottom) (in.) 4 7 4

yi Ai 3 (in. ) 64 56 64 184

(from bottom of shape to centroid)

3.4 in in.

(from top of shape to centroid)

Moment of inertia about the z axis: axis:

Shape left board flange board right board

I C C 4 (in. ) 85.3333 2.6667 85.3333

d = y = yi –

d²A 4 (in.) (in. ) –0.60 5.7600 2.40 46.0800 –0.60 5.7600 4 Moment of inertia about the z the z axis axis (in. ) =

I C C + d²A 4 (in. ) 91.0933 48.7467 91.0933 230.9333

(a) Maximum horizontal shear stress: At neutral axis: Q = 2(2 in.) in.)(4 (4.6 .6 in.) in.)(4 (4.6 .6 in. in./2 /2)) = 42.3 42.32 2 in. in.3 τ =

(600 (600 lb)(42 lb)(42.32 .32 in. in.3 ) (230.9333 (230.9333 in.4 )(4 in.) in.)

=

27.5 psi

Ans.


Shear flow q based on beam shear force V : Q = (4 in. in.))(2 in.) n.)(3.4 3.4 in. in. − 1 in.) in.) = 19.2 19.20 0 in. in.3

q=

VQ I z

=

(600 lb)( lb)(19.2 19.20 0 in.3 ) 230.9333 in.4

=

49.8845 lb/in.

(b) Maximum spacing interval s: q s ≤ n f V f ∴ s ≤

n f V f q

=

(2 nails)(120 nails)(120 lb/nail) 49.8845 lb/in.

(c) Glue joint shear stress: 49.8845 lb/in. τ = = 12.47 psi 2(2 in.)

=

4.81 in.

Ans.

Ans.


9.43 A wooden beam is fabricated by gluing four dimension lumber boards, each 40-mm wide and 90-mm deep, to a 32 × 400 plywood web, as shown in Fig. P9.43. Determine the maximum allowable shear force and the maximum allowable bending moment that this section can carry if the allowable bending stress is 6 MPa, the allowable shear stress in the plywood is 640 kPa, and the allowable shear stress in the glued joints is 250 kPa.

Fig. P9.43

Solution Moment of inertia I z : (112 112 mm) mm)(4 (400 00 mm) mm)3 I z = 12

−

(80 mm)( mm)(22 220 0 mm) mm)3 12

= 526,346,667

mm4

Maximum allowable bending moment: M c σ

=

I

∴ M max =

I

σ

c

=

(6 N/m N/mm 2 )(52 )(526,3 6,346 46,667 ,667 mm4 ) 200 mm

15,790 0,400 = 15,79

N-m N-mm = 15.7 15.79 9 kNkN-m m

Ans.

Maximum allowable shear force: Consider maximum shear stress, which occurs at the neutral axis: Q = (32 (32 mm) mm)(2 (200 00 mm)( mm)(10 100 0 mm) mm) + 2(40 2(40 mm)( mm)(90 90 mm)( mm)(20 200 0 mm − 90 mm/2 mm/2)) = 1,756,000 ,756,000 mm3 τ =

VQ

V =

I t

τ

It

Q

=

(0.640 N/mm) N/mm)(526,34 (526,346,667 6,667 mm4 )(32 mm) mm) 1, 756,000 mm3

=

6,138.7 N = 6.14 kN

(a)

Consider shear stress in glue joints: Q = (40 (40 mm mm)(90 )(90 mm)(200 200 mm − 90 mm/2) = 558 558,000 ,000 mm mm3 The shear stress in the glue joints can be found from the shear flow across the glue joint divided by the width of the glue joint; thus, q VQ / I τ glue

=

∴V =

=

tglue

t glue

I t glue

τ glue

Q

=

(0.250 N/mm) N/mm)(526,3 (526,346,667 46,667 mm4 )(90 mm) mm) 558,000 mm3

=

21,22 ,224 N = 21.2 kN

(b)

Compare results (a) and (b) to find that the maximum allowable shear force for the section is: V max

=

6.14 kN

Ans.


9.44 A wooden beam is fabricated from one 2 × 12 and two 2 × 8 dimension lumber boards to form the double-tee cross section shown in Fig. P9.44. The beam flange is fastened to the stem with nails. Each nail can safely transmit a force of 150 lb in direct shear. The allowable shear stress of the wood is 75 psi. (a) If the nails are uniformly spaced at an interval of s of s = 4 in. along the span, what is the maximum internal shear force V that V that can be supported by the double-tee cross section? (b) What nail spacing s s would be necessary to develop the full strength of strength of the double-tee shape in shear? ( Full Full strength means that the maximum horizontal shear stress in the double-tee shape equals the allowable shear stress of the wood.)

Fig. P9.44


Shape

Width b (in.) 12 2 2

top flange left stem right stem

y

=

Σ yi Ai Σ A i

=

344 in.3

=

56 in.2

Height h (in.) 2 8 8

6.14 6.1429 29 in. in.

yi (from bottom) (in.) 9 4 4

Area A Area Ai 2 (in. ) 24 16 16 56

yi Ai 3 (in. ) 216 64 64 344

(fro (from m bot botto tom m of of sha shape pe to to cent centro roid id))

3.8571 71 in. in. = 3.85

(from rom top top of sha shape to cent entroi roid)


Shape

I C C 4 (in. ) 8.0000 85.3333 85.3333

left board flange board right board

d = y = yi –

d²A 4 (in.) (in. ) 2.8571 195.9184 –2.1429 73.4694 –2.1429 73.4694 4 Moment of inertia about the z the z axis axis (in. ) =

I C C + d²A 4 (in. ) 203.9184 158.8027 158.8027 521.5238

(a) Maximum shear force for s = 4 in.: Q = (12 in.) in.)(2 (2 in.) in.)(3 (3.8 .857 571 1 in. in. − 2 in. in./2 /2)) = 68.5 68.570 704 4 in. in.3

q= q=

(2 nails)(150 lb/nail)

=

4 in. VQ I

∴V =

qI Q

=

75 lb/in. (75 lb/in. lb/in.)(5 )(521. 21.5238 5238 in.4 ) 68.5704 in.3

=

570 lb

Ans.

(a) Full strength of double tee shape (in shear): At neutral axis: Q = 2(2 in.) in.)(6. (6.1429 1429 in.)( in.)(6.14 6.1429 29 in./2) in./2) = 75.4704 75.4704 in.3 τ

=

VQ I t

∴V max =

τ

It

Q

=

(75 (75 psi) psi)(5 (521 21.5 .523 238 8 in. in.4 )(2 × 2 in. in.)) 75.4704 in.3

=

2,073.1 lb


Consider nailed portion (i.e., top flange only) to establish required nail spacing: V Q (2,073.1 lb)(68.5704 lb)(68.5704 in.3 ) = = 272.57 lb/in. q= I 521.5238 in.4 q s ≤ n f V f ∴ smax ≤

n f V f q

=

(2 nails)(150 lb/nail) 272.57 lb/in.

=

1.101 in.

Ans.


9.45 A box beam is fabricated from two plywood webs that are secured to dimension lumber boards at its top and bottom flanges (Fig. P9.45b P9.45b). The beam supports a concentrated load of P of P = = 5,000 lb at the center of a 15-ft span (Fig. P9.45a P9.45a). Bolts (⅜-in. diameter) connect the plywood webs and the lumber flanges at a spacing of s of s = 12 in. along the span. Supports A Supports A and and C can C can be idealized as a pin and a roller, respectively. Determine: (a) the maximum horizontal shear stress in the plywood webs. (b) the shear stress in the bolts. (c) the maximum bending stress in the lumber flange s.

Fig. P9.45a P9.45a

Fig. P9.45b P9.45b

Solution Moment of inertia about the z axis: axis: = yi – d = y Shape I C C 4

left web top flange bottom flange right web

(in. ) 576 16 16 576

d²A 4 (in. ) 0 1,200 1,200 0

(in.) 0 10 –10 0

4

I C C + d²A 4 (in. ) 576 1,216 1,216 576


3,584

Maximum shear force: For P For P = = 5,000 lb, V = P = P /2 /2 = 2,500 lb Maximum bending moment: For P For P = = 5,000 lb, M lb, M = PL = PL/4 /4 = 18,750 lb-ft = 225,000 lb-in. (a) Maximum horizontal shear stress (in plywood webs): Q = (3 in.)(4 in.)(10 in.)

2(0.5 +2(0.5 τ =

VQ I t

=

in.) in.)(1 (12 2 in. in.)( )(6 6 in. in.)) = 192 192 in. in.3

(2,500 lb)( lb)(192 192 in.3 ) (3,584 in.4 )(2 × 0.5 in. in.))

=

133.9 psi

Ans.


(b) Bolt shear stress: Consider the dimension lumber boards that comprise the top flange. Qflange = (3 in.) in.)(4 (4 in.) in.)(1 (10 0 in.) in.) = 120 120 in. in.3

q=

VQflange

(2,500 lb)( lb)(120 120 in.3 )

= = 83.7054 lb/in. I 3,584 in.4 Determine the force carried by one bolt: q s ≤ n f V f ∴V f ≥

qs n f

=

(83.7054 lb/in.)(12 in.) 1 bolt

= 1,004.4643

lb/bolt

The bolt cross-sectional area is: A bolt

=

π

(0.375 in in.)2

=

0.110447 in in.2

4 Each bolt acts in double shear; therefore, the shear stress in each bolt is: 1, 004.4643 lb/bolt τ bolt = ,547.284 ps psi = 4,55 ,550 ps psi = 4,54 2(0.1 2(0.110 10447 447 in. in.2 ) (c) Maximum bending stress in lumber flanges: M c (225, 000 lb-in.)(12 lb-in.)(12 in.) σ = = = 753 psi I 3,584 in.4

Ans.

Ans.


9.46 A box beam is fabricated from two plywood webs that are secured to dimension lumber boards at its top and bottom flanges (Fig. P9.46b P9.46b). The lumber has an allowable bending stress of 1,500 psi. The plywood has an allowable shear stress of 300 psi. The ⅜-in. diameter bolts have an allowable shear stress stress of 6,000 psi, and they are spaced at intervals intervals o s = 9 in. The beam span is L L = 15 ft (Fig. P9.46a P9.46a). Support A can be assumed to be pinned and support C can can be idealized as a roller. (a) Determine the maximum load P that can be applied to the beam at midspan. (b) Report the bending stress in the lumber, the shear stress in the plywood, and an d the shear stress in the bolts at the load P load P determined determined in part (a).

Fig. P9.46a P9.46a

Fig. P9.46b P9.46b

Solution Moment of inertia about the z axis: axis: d = y = yi – Shape I C C 4

left web top flange bottom flange right web

(in. ) 576 16 16 576

I C C + d²A 4 (in. ) 576 1,216 1,216 576

d²A 4 (in. ) 0 1,200 1,200 0

(in.) 0 10 –10 0

4


3,584

Maximum shear force: V = P = P /2 /2 Maximum bending moment: M = PL = PL/4 /4 (a) Determine maximum load P : Consider maximum bending stress: M c σ = ≤ 1,500 psi I ∴ M ≤

M =

(1,500 psi)(3,584 psi)(3,584 in. in.4 )

PL 4

∴ P max ≤

12 in. ≤

=

448,000 lb-in. lb-in.

448,000 lb-in.

4(448,000 lb-in.) lb-in.) (15 ft)(12 in./ft)

=

9,956 lb

(a)


Consider maximum horizontal shear stress (in plywood webs): Q = (3 in. in.))(4 in. in.)( )(10 10 in. in.)) + 2(0. 2(0.5 5 in.) in.)((12 in. in.)( )(6 6 in. in.)) = 192 192 in.3 τ =

VQ I t

≤ 300

(300 (300 psi) psi)(3 (3,5 ,584 84 in. in.4 )(2 × 0.5 0.5 in. in.))

∴V max ≤

V max

=

psi

192 in.3

= 5,600

lb

P 2

∴ Pmax ≤

2V max

=

2(5, 600 lb) = 11, 200 lb

(b)

Consider bolt shear stress: The bolt cross-sectional area is: A bolt

=

π

(0.375 in in.)2

=

0.110447 in in.2

4 Each bolt acts in double shear; therefore, the maximum shear force that can be carried by one bolt is: V bolt = 2(0. 2(0.11 1104 0447 47 in. in.2 )(6,0 )(6,000 00 psi psi)) = 1,325.3 ,325.364 64 lb lb Determine the shear flow that can be allowed based on the bolt shear stress: q s ≤ n f V f ∴q ≥

n f V f

(1 bolt)(1,325.364 lb/bolt)

= 147.263 lb/in. s 9 in. Consider the dimension lumber boards that comprise the top flange. Qflange = (3 in.) in.)(4 (4 in.) in.)(1 (10 0 in.) in.) = 120 120 in. in.3

q=

VQflange I

∴V max =

V max

=

=

q I Qflange

=

(147.263 lb/in.) lb/in.)(3,584 (3,584 in. in.4 ) 120 in.3

=

4,398.245 lb

P 2

∴ Pmax ≤

2V max

=

2(4, 398.245 lb) = 8, 796 lb

(c)

Compare the three values obtained for P for P max max in Eqs. (a), (b), and (c) to find P max

,796 = 8,79

lb = 8.80 kips

Ans.


(b) Bending stress in lumber flanges for P max max: PL (8,796 lb)(15 ft) M = = = 32,985 lb-ft 4 4 M c (32,985 lb-ft)( lb-ft)(12 in.)(12 in.)(12 in./ft) σ

=

I

=

=

3,584 in.4

1,325 psi

Ans.

Maximum shear stress in plywood webs: P 8,796 lb = 4,398 lb V = = 2 2 τ =

VQ I t

=

(4,398 lb)( lb)(192 192 in.3 ) (3,5 3,584 in in.4 )(2 × 0.5 0.5 in in.)

=

236 psi

Ans.

Bolt shear stress: τ bolt

=

6,000 psi

Ans.


9.47 A wooden beam is fabricated from three boards, which are fastened together with screws, as shown in Fig. P9.47b P9.47b. The screws are uniformly spaced along the span of the beam at intervals of 150 mm (see Fig. P9.47a P9.47a). In service, the beam will be positioned so that bending occurs about the z axis. The maximum bending moment in the beam is M is M z = = −4.50 kN-m, and the maximum shear force in the beam is V y = −2.25 kN. Determine: (a) the magnitude of the maximum horizontal shear stress in the beam. (b) the shear force in each screw. (c) the magnitude of the maximum bending stress in the beam.

Fig. P9.47a P9.47a

Fig. P9.47b P9.47b


Shape

Width b (mm) 40 140 40

left board bottom board right board

y

=

Σ yi Ai Σ A i

=

1,408,000 mm3 20,000 mm2

Height h (mm) 180 40 180

=

70.4 70.4 mm

109.6 6 = 109.

Area A Area Ai (mm2) 7,200 5,600 7,200 20,000

yi (from bottom) (mm) 90 20 90

yi Ai (mm3) 648,000 112,000 648,000 1,408,000

(from from bott bottom om of sha shape to ce centr ntroid) oid)

mm mm

(from rom top of shape hape to cent entroi roid)


Shape left board bottom board right board

I C C 4 (mm ) 19,440,000.00 746,666.67 19,440,000.00

d = y = yi –

d²A 4 (mm) (mm ) 19.60 2,765,952.00 –50.40 14,224,896.00 19.60 2,765,952.00 4 Moment of inertia about the z the z axis axis (mm ) =

I C C + d²A 4 (mm ) 22,205,952.00 14,971,562.67 22,205,952.00 59,383,466.67

(a) Maximum horizontal shear stress: At neutral axis: Q = 2(40 mm)(1 mm)(109. 09.6 6 mm) mm)(10 (109.6 9.6 mm/2) mm/2) = 480,486 480,486.4 .4 mm3 τ =

VQ I t

=

(2,250 N)(48 N)(480,48 0,486.4 6.4 mm3 ) (59, (59,38 383, 3,46 466. 6.67 67 mm4 )(2 )(2 × 40 mm)

=

0.2276 MP MPa = 228 kP kPa

Ans.


(b) Shear force in each screw Consider bottom board, which is held in place by two screws:

Q = (140 140 mm mm)(40 (40 mm mm)(70 )(70.4 .4 mm − 40 mm/2) /2) = 282 282,240 ,240 mm mm3 q=

VQ I

=

(2,250 N)(28 N)(282,240 2,240 mm3 ) 59,383,466.67 mm4

= 10.6939

N/mm

q s ≤ n f V f ∴V f =

qs n f

=

(10.6939 N/mm)(150 mm) 2 screws

=

802 N per screw

(c) Maximum bending stress: (−4.50 × 106 N-mm)(109.6 mm) M z y σ x

=−

I z

=−

59,383,466.67 mm4

=

8.31 MPa (T)

Ans.

Ans.


9.48 A wooden beam is fabricated by bolting together three members, as shown in Fig. P9.48a P9.48a. The cross-sectional dimensions are shown in Fig. P9.48b P9.48b. The 8-mm-diameter bolts are spaced at intervals of s of s = = 200 mm along the x the x axis of the beam. If the internal shear force in the beam is V = V = 7 kN, determine the shear stress in each bolt.

Fig. P9.48a P9.48a

Fig. P9.48b P9.48b


Shape

Width b (mm) 40 40 40

left board center board right board

y

=

Σ yi Ai Σ A i

=

3,636,000 mm3 19,200 mm2

Height h (mm) 90 300 90

= 189.375 = 110.625

Area A Area Ai 2 (mm ) 3,600 12,000 3,600 19,200


yi Ai 3 (mm ) 918,000 1,800,000 918,000 3,636,000

mm


mm



Shape left board center board right board

I C C 4 (mm ) 2,430,000 90,000,000 2,430,000

d = y = yi –

d²A 4 (mm) (mm ) 65.625 15,503,906.25 -39.375 18,604,687.50 65.625 15,503,906.25 4 Moment of inertia about the z the z axis axis (mm ) =

I C C + d²A 4 (mm ) 17,933,906.25 108,604,687.50 17,933,906.25 144,472,500.00


Shear force in each bolt Consider left board, which is held in place by the bolt: Q = (40 (40 mm)(90 )(90 mm)(110. 110.62 625 5 mm − 90 mm/2) /2) = 236 236,250 mm3

q=

(7,000 (7,000 N)(23 N)(236,250 6,250 mm mm3 )

VQ

= = 11.4468 N/mm I 144,472,500 mm4 Note that this value of q is the shear flow that must be transmitted by one surface of the bolt cross section. The cross-sectional area of the bolt is:

A bolt

=

π

(8 mm)2

=

50.2655 mm2

4 Relate the shear flow and the bolt shear stress with Eq. (9.14): q s ≤ n f τ f Af ∴τ f =

qs n f Af

=

(11.4468 N/mm)(200 mm) (1 bolt bolt surface)( surface)(50.2655 50.2655 mm2 )

=

45.5 MPa

Ans.


9.49 A wooden beam is fabricated by bolting together three members, as shown in Fig. P9.49a P9.49a. The cross-sectional dimensions are shown in Fig. P9.49b P9.49b. The allowable shear stress of the wood is 850 kPa, and the allowable shear stress of the 10mm-diameter bolts is 40 MPa. Determine: (a) the maximum internal shear force V that that the cross section can withstand based on the allowable shear stress in the wood. (b) the maximum bolt spacing s spacing s required required to develop the internal shear force computed in part (a).

Fig. P9.49a P9.49a

Fig. P9.49b P9.49b


Shape

Width b (mm) 40 40 40

left board center board right board

y

=

Σ yi Ai Σ A i

=

3,636,000 mm3 19,200 mm2

Height h (mm) 90 300 90

= 189.375 = 110.625

Area A Area Ai 2 (mm ) 3,600 12,000 3,600 19,200


yi Ai 3 (mm ) 918,000 1,800,000 918,000 3,636,000

mm


mm



Shape left board center board right board

I C C 4 (mm ) 2,430,000 90,000,000 2,430,000

d = y = yi –

d²A 4 (mm) (mm ) 65.625 15,503,906.25 -39.375 18,604,687.50 65.625 15,503,906.25 4 Moment of inertia about the z the z axis axis (mm ) =

I C C + d²A 4 (mm ) 17,933,906.25 108,604,687.50 17,933,906.25 144,472,500.00


Consider maximum horizontal shear stress: Q = (40 mm) mm)(18 (189.3 9.375 75 mm)( mm)(189 189.375 .375 mm/2) mm/2) = 717, 717, 257.813 257.813 mm3 τ =

VQ I t

≤ 850

∴V max ≤

kP kPa = 0.850 MP MPa

(0.8 (0.850 50 N/mm N/mm2 )(14 )(144, 4,472 472,5 ,500 00 mm4 )(40 )(40 mm) 717,257.813 717, 257.813 mm mm3

=

6, 84 848.4 N = 6.85 kN

Ans.

Maximum bolt spacing Consider left board, which is held in place by the bolt: Q = (40 (40 mm)(90 )(90 mm)(110. 110.62 625 5 mm − 90 mm/2) /2) = 236 236,250 mm3

q=

(6,848 (6,848.4 .4 N)(23 N)(236,250 6,250 mm3 )

VQ

= = 11.1989 N/mm I 144,472,500 mm4 Note that this value of q is the shear flow that must be transmitted by one surface of the bolt cross section. The cross-sectional area area of the bolt is:

A bolt

=

π

(10 mm mm)2

=

78.5398 mm mm2

4 Relate the shear flow and the bolt shear stress with Eq. (9.14): q s ≤ n f τ f Af ∴ s ≤

n f τ f Af q

=

(1 bolt bolt surfac surface)( e)(40 40 N/mm N/mm2 )(78.53 )(78.5398 98 mm mm2 ) 11.1989 N/mm

=

281 mm

Ans.


9.50 A cantilever flexural member is fabricated by bolting two identical coldrolled steel channels back-to-back, as shown in Fig. P9.50a P9.50a. The cantilever beam has a span of L of L = = 1,600 mm and supports a concentrated load of P of P = = 600 N. The cross-sectional dimensions of the built-up shape are shown in Fig. P9.50b. P9.50b. The The effect of the rounded corners can be neglected in determining the section properties for the built-up shape. (a) If 4-mm-diameter bolts are installed at intervals of s of s = 75 mm, determine the shear stress produced in the bolts. (b) If the allowable shear stress in the bolts is 96 MPa, determine the minimum bolt diameter required if a spacing of s of s = = 400 mm is used.

Fig. P9.50a P9.50a Fig. P9.50b P9.50b

Solution Centroid location in y direction for a single channel shape:

Shape

Width b (mm) 3 59 3

left element center element right element

y

=

Σ yi Ai Σ A i

=

Height h (mm) 40 3 40

5,065.5 mm3 417 mm 2

= 12.1475

Area A Area Ai 2 (mm ) 120 177 120 417

mm

yi (from bottom) (mm) 20 1.5 20

yi Ai 3 (mm ) 2400 265.5 2400 5065.5

(from back of channel to centroid)

Moment of inertia (both channels):

⎡ (3 mm)(40 mm)3 I = 2 ⎢ 2 3 ⎣

+

(65 mm − 2(3 mm))(3 mm)3 ⎤

4

⎥ = 257, 062 mm ⎦

3

Shear flow: Q = (12.1 12.147 475 5 mm) mm)(4 (417 17 mm mm2 ) = 5,065. ,065.51 51 mm3

q=

VQ

I Bolt area: A bolt

=

=

π

(600 N)(5,065. N)(5,065.51 51 mm3 ) 257,062 mm4

(4 mm)2

4 (a) Bolt shear stress: q s ≤ n f τ f Af ∴τ f =

qs n f Af

=

= 12.5664

= 11.8232

N/mm

mm2

(11.8232 N/mm)(75 mm) (1 bolt bolt surface)( surface)(12.5664 12.5664 mm2 )

=

70.6 MPa

Ans.


(b) Minimum bolt diameter for s = 400 mm: q s ≤ n f τ f Af ∴ A f ≥

A bolt D bolt

=

≥

qs n f τ f

π

4

D2

=

≥

(11.8232 N/mm)(400 mm) 2

(1 bolt surface)( surface)(96 96 N/mm N/mm )

=

49.2633 mm2

49.2633 mm2

4(49 4(49.2 .263 633 3 mm2 )

=

7.92 mm

Ans.

π


9.51 A W310 × 60 steel beam (see Appendix B) in an existing structure is to be strengthened by adding a 250 mm wide by 16 mm thick cover plate to its lower flange, as shown in Fig. P9.51. The cover plate is attached to the lower flange by pairs of 24-mm-diameter bolts spaced at intervals of s along the beam span. Bending occurs about the z the z centroidal centroidal axis. (a) If the allowable bolt shear stress is 96 MPa, determine the maximum bolt spacing interval s interval s required required to support an internal shear force in the beam of V = = 50 kN. (b) If the allowable bending stress is 150 MPa, determine the allowable bending moment for the existing W310 × 60 shape, the allowable bending moment for the W310 × 60 with the added cover plate, and the percentage increase in moment capacity that is gained by adding the cover plate.

Fig. P9.51


Shape

Width b (mm) 250

W310 × 60 cover plate

y

=

Σ yi Ai Σ A i

=

Height h (mm) 16

1, 292,850 mm3 11,550 mm2

= 111.935 =

mm

Area A Area Ai 2 (mm ) 7,550 4,000 11,550

yi (from bottom) (mm) 167 8

yi Ai 3 (mm ) 1,260,850 32,000 1,292,850


206.065 mm



Shape

I C C 4 (mm ) 128,000,000 85,333.33

W310 × 60 cover plate

d = y = yi –

d²A 4 (mm) (mm ) 55.065 22,892,764 –103.935 43,209,937 4 Moment of inertia about the z the z axis axis (mm ) =

I C C + d²A 4 (mm ) 150,892,764 43,295,270 194,188,035

(a) Maximum bolt spacing Consider the cover plate, which is connected to the W310 × 60 shape with two bolts: Q = (250 (250 mm mm)(16 )(16 mm mm)(111 (111.9 .935 35 mm − 16 mm mm/2) /2) = 415 415,740 ,740 mm3

q=

(50,000 (50,000 N)(4 N)(415 15,740 ,740 mm3 )

VQ

= = 107.0457 N/mm I 194,188,035 mm4 The cross-sectional area of a 24-mm-diameter bolt is:

A bolt

=

π

(24 mm)2

=

452.389 mm2

4 Relate the shear flow and the bolt shear stress with Eq. (9.14): q s ≤ n f τ f Af ∴ s ≤

n f τ f Af q

=

(2 bol bolts ts)( )(96 96 N/m N/mm m2 )(45 )(452. 2.389 389 mm mm2 ) 107.0457 N/mm

=

811 mm

Ans.


(b) Allowable bending moment for W310 × 60 shape (without cover plate): M c σ = ≤ 150 MPa I ∴ M allow =

(150 150 N/mm N/mm2 )(12 )(128,0 8,000 00,00 ,000 0 mm mm4 ) (302 mm/2)

127,15 152 2,318 ,318 = 127,

N-m N-mm = 127. 127.2 2 kNkN-m

Ans.

Allowable bending moment for W310 × 60 shape (with cover plate): M c σ = ≤ 150 MPa I ∴ M allow =

(150 150 N/mm N/mm2 )(19 )(194,1 4,188 88,03 ,035 5 mm mm4 ) (206.065 mm)

= 141 141,354 ,354,452 ,452

N-m N-mm = 141. 141.4 4 kNkN-m

Percentage increase in moment capacity: 141 141,354, ,354, 452 NN-mm − 127,1 127,152 52,318 ,318 N-m N-mm m % increase = (100%) = 11.17% 127,152,318 N-mm

Ans.

Ans.


9.52 A W410 × 60 steel beam (see Appendix B) is simply supported at its ends and carries a concentrated load P load P at at the center of a 7-m span. The W410 × 60 shape will be strengthened by adding two 250 mm wide by 16 mm thick cover plate to its flanges, as shown in Fig. P9.52. Each cover plate is attached to its flange by pairs of 20-mm-diameter bolts spaced at intervals of s of s along along the beam span. The allowable bending stress is 150 MPa, the allowable shear stress in the bolts is 96 MPa, and bending occurs about the z the z centroidal centroidal axis. (a) Based on the 150 MPa allowable bending stress, determine the maximum concentrated load P load P that that may be applied at the center o a 7-m span for a W410 × 60 steel beam without cover plates and with two cover plates. (b) Compute the ratio between P (with cover plates) and P (without cover plates). (c) Compute the section modulus of the W410 × 60 shape with two cover plates. Then, compute the ratio between the section moduli for the strengthened and the original cross sections. Compare this ratio to the ratio computed in part (b). (d) For the internal shear force V associated associated with the concentrated load P determined in part (a), compute the maximum spacing interval s required s required for the bolts that attach the cover plates to the Fig. P9.52 flanges.

Solution Moment of inertia about the z axis axis (with cover plates): d = y = yi – Shape I C C 4

top cover plate W410 × 60 bottom cover plate

(mm ) 85,333.33 216,000,000 85,333.33

(mm) 209 0 –209

d²A 4 (mm ) 174,724,000 0 174,724,000

I C C + d²A 4 (mm ) 174,809,333 216,000,000 174,809,333

4

565,618,666

Moment of inertia about the z the z axis axis (mm ) =

Maximum shear force: V = P = P /2 /2 Maximum bending moment: M = PL = PL/4 /4 (a) Determine maximum load P : Consider W410 × 60 without cover cove r plates: M c σ = ≤ 150 MPa I (150 150 N/mm N/mm2 )(21 )(216,00 6,000,0 0,000 00 mm3 ) ∴ M allow = (402 mm/2)

161,194,030 4,030 = 161,19 M =

PL 4

∴ P max ≤

≤ 161.194

N-mm N-mm = 161.1 161.194 94 kNkN-m

kN-m

4(161.194 kN-m) 7m

= 91.111

kN

=

92.1 kN

Ans.


Consider W410 × 60 with two cover plates: M c σ = ≤ 150 MPa I (150 150 N/mm N/mm 2 )(565 )(565,6 ,618 18,66 ,666 6 mm3 ) 390,980 80,645 ,645 NN-mm = 390. 390.98 981 1 kNkN-m m ∴ M allow = = 390,9 (434 mm/2) M =

PL 4

≤ 390.981

∴ P max ≤

kN-m

4(390.981 kN-m) 7m

=

223.418 kN = 223 kN

Ans.

(b) Ratio

ratio =

P (with cover plates tes) P (wi (with thou outt cove coverr plat plates es))

=

223.418 kN 91.1 91.111 11 kN

=

2.43

Ans.

(c) Section moduli comparison Section modulus of W410 × 60 without cover plates: 3 S = = 1,060,000 mm

Section modulus of W410 × 60 with two cover plates: I 565,618,666 mm4 3 S = = = 2,606,538 mm c (434 mm/2)

ratio =

2,606,538 mm3 1, 060,000 mm3

=

2.46

Ans.


9.53 A W310 × 60 steel beam (see Appendix B) has a C250 × 45 channel bolted to the top flange, as shown in Fig. P9.53. The beam is simply supported at its ends and carries a concentrated load o 70 kN at the center of an 8-m span. If pairs of bolts are spaced at 500-mm intervals along the beam, determine: (a) the shear force carried by each of the bolts. (b) the bolt diameter required if the shear shear stress in the bolts must be limited to 60 MPa.

Fig. P9.53


Shape

Area A Area Ai 2 (mm ) 7,550 5,680 13,230

W310 × 60 C250 × 45

y

=

Σ yi Ai Σ A i

yi (from bottom) (mm) 151 302 + 17.1 – 16.5 = 302.6

=

2,858,818 mm3 13,230 mm2

=

216.086 mm

= 103.014

yi Ai 3 (mm ) 1,140,050 1,718,768 2,858,818


mm



Shape W310 × 60 C250 × 45

I C C 4 (mm ) 128,000,000 1,640,000

d = y = yi – d²A 4 (mm) (mm ) –65.086 31,983,215 86.5140 42,512,938 4 Moment of inertia about the z the z axis axis (mm ) =

I C C + d²A 4 (mm ) 159,983,315 44,152,938 204,136,153

Maximum shear force: V = P = P /2 /2 = 70 kN/2 = 35 kN (a) Shear force in each bolt Consider the C250 × 45 shape, which is connected to the W310 × 60 shape with two bolts: Q = (5,68 (5,680 0 mm mm2 )(302. 302.6 6 mm mm − 216. 216.08 086 6 mm mm) =

q= =

491,399.5 mm3 VQ I (35, (35, 000 N)(491 N)(491,399.5 ,399.5 mm mm3 ) 204,136,153 mm4

= 84.2525

N/mm


Relate the shear flow and the bolt shear force with Eq. (9.13): q s ≤ n f V f ∴V f ≤

qs n f

=

(84.2525 N/mm)(500 mm) 2 bolts

(b) Required bolt diameter V bolt V bolt τ ≥ ∴ A bolt ≥ τ allow A bolt

A bolt

=

π

4

2 Dbolt

≥ 351.052

∴ D bolt ≥

=

=

21,063 N 60 N/mm

2

21,063 ,063 N = 21.1 kN pe per bo bolt

= 351.052

Ans.

mm2

mm2

4(35 4(351. 1.05 052 2 mm2 )

=

21.1 mm

Ans.

π


Mechanics of Materials Solutions Chapter09 Probs38 53

Recommend Documents