• Author / Uploaded
• Jayesh Bankoti

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PROBLEM 6.5 The American Standard rolled-steel beam shown has been reinforced by attaching to it two 16 × 200-mm plates, using 18-mm-diameter bolts spaced longitudinally every 120 mm. Knowing that the average allowable shearing stress in the bolts is 90 MPa, determine the largest permissible vertical shearing force.

SOLUTION Calculate moment of inertia:

A (mm2 ) 3200 6650 3200

Part Top plate S310 × 52 Bot. plate Σ *d

=

d (mm) *

160.5 0

*

160.5

Ad 2 (106 mm4 ) 82.43 82.43 164.86

I (106 mm 4 ) 0.07 95.3 0.07 95.44

305 16 + = 160.5 mm 2 2

I = ΣAd 2 + ΣI = 260.3 × 106 mm 4 = 260.3 × 10−6 m 4 Q = Aplate d plate = (3200)(160.5) = 513.6 × 103 mm3 = 513.6 × 10−6 m3 Abolt =

π 4

2 d bolt =

π 4

(18 × 10−3 ) 2 = 254.47 × 10−6 m 2

Fbolt = τ all Abolt = (90 × 106 )(254.47 × 10−6 ) = 22.90 × 103 N qs = 2 Fbolt q=

VQ I

q = V =

2Fbolt (2)(22.90 × 103 ) = = 381.7 × 103 N/m −3 s 120 × 10

Iq (260.3 × 10−6 )(381.7 × 103 ) = = 193.5 × 103 N −6 Q 513.6 × 10

V = 193.5 kN 

PROBLEM 6.9 For the beam and loading shown, consider section n-nn and determine (a) the largest shearring stress in that section, (b) the shearinng stress at point a.

SOLUTION By symmetry, RA = RB

ΣFy = 0 RA + RB − 65 − 90 − 65 = 0 RA = RB = 110 kN

V = 110 kN at n-n.

From shear diagram Determine moment of inertia

A(mm 2 )

d (mm)

Top flange

3750

117.5

Web

2200

0

Bot. flange

3750

117.5

Part

Σ

(a)

Ad 2 (106 mm 4 ) I (106 mm m4 )

51.77 0

0.07 8.87

51.77

0.07

103.54

9.01

I = ΣAd 2 + Σ I = 112.55 × 1006 t = 10 mm

Q = Σ Ay = 501125 mm 3

Part

A(m mm 2 )

(1) (2)

y (mm)

Ay (mm3 )

37550

117.5

440625

11000

55

60500

Σ

501125

τ max =

V Qmax (110 × 103 )(501125) = = 50 N/mm 2 It (112.55 × 106 )(10) o or

50 MPa 

PROBLEM 6.9 (Continued)

(b) Part

A(mm 2 )

y (mm)

Ay (mm3 )

(1)

3750

117.5

440625

(2)

100

105

10500

Σ

451125 Q = Σ Ay = 451125 mm3

τ=

t = 10 mm

VQ (110 × 103 )(451125) = = 44 N/mm 2 It (112.55 × 106 )(10)

(i.e.) 44 MPa



PROBLEM 6.32 The built-up timber beam is subjected to a vertical shear of 5 kN. Knowing that the allowable shearing force in the nails is 300 N, determine the largest permissible spacing s of the nails.

SOLUTION 1 b1h13 + A1d12 12 1 = (50)(50)3 + (50)(50)(100)2 = 25520833 mm 4 12 1 1 I 2 = b2 h23 = (50)(250)3 = 65104167 mm 4 12 12 I = 4 I1 + I 2 = 167187499 mm 4 I1 =

Q = Q1 = A1 y1 = (50)(50)(100) = 250000 mm3 VQ (5000)(250000) = = 7.48 N/mm 167187499 I F 300 = qs s = nail = = 40.1 mm q 7.48

q=

Fnail



PROBLEM 6.43 Four L102 × 102 × 9.5 steel angle shapes and a 12 × 400-mm steel plate are bolted together to form a beam with the cross section shown. The bolts are of 22-mm diameter and are spaced longitudinally every 120 mm. Knowing that the beam is subjected to a vertical shear of 240 kN, determine the average shearing stress in each bolt.

SOLUTION Angle:

A = 1850 mm 2 , I = 1.83 × 106 mm 4 , y = 29 mm d = 200 − 29 = 171 mm I a = I + Ad 2 = 55.926 × 106 mm 4

Plate:

1 (12)(400)3 = 64 × 106 mm 4 12 I = 4 I a + I P = 287.7 × 106 mm 4 = 287.7 × 10−6 m 4

IP =

Q = (1850)(171) = 316.35 × 103 mm3 = 316.35 × 10−6 m3 VQ (240 × 103 )(316.35 × 10−6 ) = = 263.9 × 103 N/m −6 I 287.7 × 10 = qs = (263.9 × 103 )(120 × 10−3 ) = 31.668 × 103 N

q= Fbolt

τ bolt

π

π

d 2 = (22) 2 = 380.13 mm 2 = 380.13 × 10−6 m3 4 4 F 31.668 × 103 = bolt = = 83.3 × 106 Pa Abolt 380.13 × 10−6

Abolt =

τ bolt = 83.3 MPa 

PROBLEM 6.54 (a) Determine the shearing stress at point P of a thin-walled pipe of the cross section shown caused by a vertical shear V. (b) Show that the maximum shearing stress occurs for θ = 90° and is equal to 2V/A, where A is the cross-sectional area of the pipe.

SOLUTION A = 2π rm t

J = Arm2 = 2π rm3 t

sin θ

for a circular arc.

r=

θ

I=

1 J = π rm3t 2

AP = 2rθ t QP = AP r = 2rt sin θ (a)

τP =

(b)

τm =

VQP (V )(2rt sin θ ) = I (2t ) π rm3 t (2t )

(

2V sin π2 2π rmt

)

τP =

V sin θ  π rm t

τm =

2V  A

PROBLEM 6.89 Three boards, each 50 mm thick, are nailed together to form a beam that is subjected to a vertical shear. Knowing that the allowable shearing force in each nail is 600 N, determine the allowable shear if the spacing s between the nails is 75 mm.

SOLUTION 1 3 bh + Ad 2 12 1 = (150)(50)3 + (150)(50)(75)2 = 43.75 × 106 mm 4 12 1 1 I 2 = bh3 = (50)(100)3 = 4.17 × 106 mm 4 12 12 I 3 = I1 = 43.75 × 106 mm 4 I1 =

I = I1 + I 2 + I 3 = 91.67 × 106 mm 4

Q = A1 y1 = (150)(50)(75) = 562500 mm3 qs = Fnail (1)

Dividing Eq. (2) by Eq. (1),

q=

VQ I

(2)

1 VQ = s Fnail I

V=

Fnail I (600)(91.67 × 106 ) = = 1304 N Qs (562500)(75)

