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9.11 A 1.6-m long cantilever beam supports a concentrated load of 7.2 kN, as shown below. The beamis made of a rectangular timber having a width of 120 mm and a depth of 280 mm. Calculate themaximum horizontal shear stresses at points located 35 mm, 70 mm, 105 mm, and 140 mm below the top surface of the beam. From these results, plot a graph showing the distribution of shear stresses fromtop to bottom of the beam.

Fig. P9.11a Cantilever beam Fig. P9.11b Cross-sectional dimensions

Solution Shear force in cantilever beam: V = 7.2 kN = 7,000 N Shear stress formula:

V QI t

τ =

Section properties:

3

6 4(120 mm)(280 mm) 219.52 10 mm12

I = = ×

t = 120 mm

Distance below top surface of beam y Q τ

35 mm 105 mm 514,500 mm3 140.6 kPa

70 mm 70 mm 882,000 mm3 241 kPa

105 mm 35 mm 1,102,500 mm3 301 kPa

140 mm 0 mm 1,176,000 mm3 321 kPa

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9.12 A 14-ft long simply supported timber beam carries a 6-kip concentrated load at midspan, as shown in Fig. P9.12a. The cross-sectional dimensions of the timber are shown in Fig. P9.12b. (a) At section a–a, determine the magnitude of the shear stress in the beam at point H. (b) At section a–a, determine the magnitude of the shear stress in the beam at point K. (c) Determine the maximum horizontal shear stress that occurs in the beam at any location within the14-ft span length. (d) Determine the maximum tension bending stress that occurs in the beam at any location within the14-ft span length.

Fig. P9.12a Simply supported timber beam Fig. P9.12b Cross-sectional

dimensions

Solution Section properties:

3

4(6 in.)(15 in.) 1,687.5 in. 6 in.12

I t= = =

(a) Shear stress at H:

3

3

4

(6 in.)(3 in.)(6 in.) 108 in.

(3,000 lb)(108 in. ) 32.0 psi(1,687.500 in. )(6 in.)

Q

V QI t

τ

= =

=

= = Ans.

(b) Shear stress at K:

3

3

4

(6 in.)(1 in.)(7 in.) 42 in.

(3,000 lb)(42 in. ) 12.44 psi(1,687.500 in. )(6 in.)

Q

V QI t

τ

= =

=

= = Ans.

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(c) Maximum shear stress at any location:

3max

3

4

(6 in.)(7.5 in.)(3.75 in.) 168.75 in.

(3,000 lb)(168.75 in. ) 50.0 psi(1,687.500 in. )(6 in.)

Q

V QI t

τ

= =

= = = Ans.

(d) Maximum bending stress at any location:

max

4

21 kip-ft 21,000 lb-ft

(21,000 lb-ft)(7.5 in.)(12 in./ft) 1,120 psi (T) and (C)1,687.500 in.x

M

M cI

σ

= =

= = = Ans.

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9.13 A 4-m long simply supported timber beam carries a uniformly distributed load of 7 kN/m, as shownin Fig. P9.13a. The cross-sectional dimensions of the beam are shown in Fig. P9.13b. (a) At section a–a, determine the magnitude of the shear stress in the beam at point H. (b) At section a–a, determine the magnitude of the shear stress in the beam at point K. (c) Determine the maximum horizontal shear stress that occurs in the beam at any location within the 4-m span length. (d) Determine the maximum compression bending stress that occurs in the beam at any location withinthe 4-m span length.

Fig. P9.13a Simply supported timber beam Fig. P9.13b Cross-sectional

dimensions

Solution Section properties:

3

6 4(100 mm)(300 mm) 225 10 mm 100 mm12

I t= = × =

(a) Shear stress magnitude at H:

3

3

6 4

(100 mm)(90 mm)(105 mm)945,000 mm

(7,000 N)(945,000 mm )(225 10 mm )(100 mm)

294 kPa

Q

V QI t

τ

==

=

=×

= Ans. (b) Shear stress magnitude at K:

3

3

6 4

(100 mm)(75 mm)(112.5 mm)843,750 mm

(7,000 N)(843,750 mm )(225 10 mm )(100 mm)

263 kPa

Q

V QI t

τ

==

=

=×

= Ans.

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(c) Maximum shear stress at any location:

3max

3

6 4

(100 mm)(150 mm)(75 mm) 1,125,000 mm

(14,000 N)(1,125,000 mm ) 700 kPa(225 10 mm )(100 mm)

Q

V QI t

τ

= =

= = =×

Ans.

(d) Maximum compression bending stress at any location:

max

6 4

14 kN-m

(14 kN-m)(150 mm)(1,000 N/kN)(1,000 mm/m)225 10 mm

9.3333 MPa 9,330 kPa (C)

x

M

M yI

σ

=

= − = −×

= − = Ans.

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9.14 A 5-m long simply supported timber beam carries two concentrated loads, as shown in Fig. P9.14a. The cross-sectional dimensions of the beam are shown in Fig. P9.14b. (a) At section a–a, determine the magnitude of the shear stress in the beam at point H. (b) At section a–a, determine the magnitude of the shear stress in the beam at point K. (c) Determine the maximum horizontal shear stress that occurs in the beam at any location within the 5-m span length. (d) Determine the maximum compression bending stress that occurs in the beam at any location withinthe 5-m span length.

Fig. P9.14a Simply supported timber beam Fig. P9.14b Cross-sectional

dimensions

Solution Section properties:

3

6 4(150 mm)(450 mm) 1,139.1 10 mm 150 mm12

I t= = × =

(a) Shear stress magnitude at H:

3

3

6 4

(150 mm)(150 mm)(150 mm)3,375,000 mm

(39,200 N)(3,375,000 mm )(1,139.1 10 mm )(150 mm)

774 kPa

Q

V QI t

τ

==

=

=×

= Ans. (b) Shear stress magnitude at K:

3

3

6 4

(150 mm)(100 mm)(175 mm)2,625,000 mm

(39,200 N)(2,625,000 mm )(1,139.1 10 mm )(150 mm)

602 kPa

Q

V QI t

τ

==

=

=×

= Ans.

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(c) Maximum shear stress at any location:

3max

3

6 4

(150 mm)(225 mm)(112.5 mm) 3,796,875 mm

(39, 200 N)(3,796,875 mm ) 871 kPa(1,139.1 10 mm )(150 mm)

Q

V QI t

τ

= =

= = =×

Ans.

(d) Maximum bending stress at any location:

max

6 4

39.2 kN-m

(39.2 kN-m)(225 mm)(1,000 N/kN)(1,000 mm/m)1,139.1 10 mm

7.74296 MPa 7,740 kPa (C)

x

M

M yI

σ

=

= − = −×

= − = Ans.

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9.15 A laminated wood beam consists of eight 2 in. × 6-in. planks glued together to form a section 6 in. wide by 16 in. deep, as shown in Fig. P9.15a. If the allowable strength of the glue in shear is 160 psi,determine: (a) the maximum uniformly distributed load w that can be applied over the full length of the beam if thebeam is simply supported and has a span of 20 ft. (b) the shear stress in the glue joint at H, which is located 4 in. above the bottom of the beam and 3 ftfrom the left support. Assume the beam is subjected to the load w determined in part (a). (c) the maximum tension bending stress in the beam when the load of part (a) is applied.

Fig. P9.15a Simply supported timber beam Fig. P9.15b Cross-sectional

dimensions

Solution Section properties:

3

4(6 in.)(16 in.) 2,048 in. 6 in.12

I t= = =

(a) Maximum Q: 3(6 in.)(8 in.)(4 in.) 192 in.Q = = Maximum shear force V:

4

3

160 psi

(160 psi)(2,048 in. )(6 in.) 10, 240 lb192 in.

V QI t

V

τ = ≤

∴ ≤ =

Maximum distributed load w:

max

max

10,240 lb2

2(10,240 lb) 1,024 lb/ft20 ft

wLV

w

= ≤

∴ ≤ = Ans.

(b) Shear force at x = 3 ft: 10,240 lb (1,024 lb/ft)(3 ft) 7,168 lbV = − =

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3

3

4

(6 in.)(4 in.)(6 in.) 144 in.(7,168 lb)(144 in. ) 84.0 psi(2,048 in. )(6 in.)

QV QI t

τ

= =

= = = Ans.

(c) Maximum tension bending stress at any location:

2 2

max

4

(1,024 lb/ft)(20 ft) 51,200 lb-ft8 8

(51,200 lb-ft)( 8 in.)(12 in./ft) 2, 400 psi (T)2,048 in.x

wLM

M yI

σ

= = =

−= − = − = 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.

9.16 A 5-ft long simply supported wood beam carries a concentrated load P at midspan, as shown in Fig. P9.16a. The cross-sectional dimensions of the beam are shown in Fig. P9.16b. If the allowable shear strength of the wood is 80 psi, determine the maximum load P that may be applied at midspan. Neglect the effects of the beam’s self weight.

Fig. P9.16a Simply supported timber beam Fig. P9.16b Cross-sectional dimensions

Solution Section properties:

3

4(6 in.)(10 in.) 500 in. 6 in.12

I t= = =

Maximum Q: 3(6 in.)(5 in.)(2.5 in.) 75 in.Q = = Maximum shear force V:

4

3

80 psi

(80 psi)(500 in. )(6 in.) 3, 200 lb75 in.

V QI t

V

τ = ≤

∴ ≤ =

Maximum concentrated load P:

max

max

3, 200 lb2

2(3, 200 lb) 6, 400 lb

PV

P

= ≤

∴ ≤ = Ans.

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9.17 A wood beam supports the loads shown in Fig. P9.17a. The cross-sectional dimensions of the beam are shown in Fig. P9.17b. Determine the magnitude and location of: (a) the maximum horizontal shear stress in the beam. (b) the maximum tension bending stress in the beam.

Fig. P9.17a Simply supported timber beam Fig. P9.17b Cross-sectional dimensions

Solution Section properties:

3 3

4(75 mm)(240 mm) (20 mm)(100 mm)2 89,733,333 mm12 12

I = + =

(a) Maximum shear force: Vmax = 9.54 kN = 9,540 N @ support A Check shear stress at neutral axis:

3

(75 mm)(120 mm)(60 mm)2(20 mm)(50 mm)(25 mm) 590,000 mm

Q =+ =

3

4

(9,540 N)(590,000 mm ) 545 kPa(89,733,333 mm )(115 mm)

V QI t

τ = = =

Check shear stress at top edge of cover plates: 3(75 mm)(70 mm)(85 mm) 446,250 mmQ = =

3

4

(9,540 N)(446,250 mm ) 633 kPa(89,733,333 mm )(75 mm)

V QI t

τ = = =

Maximum shear stress in beam: ,max 633 kPaHτ = Ans.

(b) Maximum bending moment: Mmax = 6.49 kN-m (between support A and point B) Maximum tension bending stress:

4

(6.49 kN-m)( 120 mm)(1,000 N/kN)(1,000 mm/m)89,733,333 mm

8.67905 MPa 8,680 kPa (T)

xM y

Iσ −= − = −

= = Ans.