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```PROBLEM 6.1
Three boards, each of 1.5 × 3.5-in. rectangular cross section, are nailed
together to form a beam that is subjected to a vertical shear of 250 lb.
Knowing that the spacing between each pair of nails is 2.5 in., determine
the shearing force in each nail.
SOLUTION
I =
1 3
1
(3.5)(4.5)3 = 26.578 in 4
bh =
12
12
A = (3.5)(1.5) = 5.25 in 2
y1 = 1.5 in.
Q = Ay1 = 7.875 in 3
q=
VQ (250)(7.875)
=
= 74.074 lb/in
I
26.578
qs = 2 Fnail
Fnail =
qs
(74.074)(2.5)
=
2
2
Fnail = 92.6 lb
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 6.7
A columm is fabricated by connecting the rolled-steel members shown by
bolts of 34 -in. diameter spaced longitudinally every 5 in. Determine the
average shearing stress in the bolts caused by a shearing force of 30 kips
parallel to the y axis.
SOLUTION
Geometry:
f =
d
2
+ (t w )C
s
10.0
+ 0.303 = 5.303 in.
2
x = 0.534 in.
=
y1 = f − x = 5.303 − 0.534 = 4.769 in.
Determine moment of inertia.
Part
C8 × 13.7
S 10 × 25.4
C8 × 13.7
Σ
d (in.)
4.769
0
4.769
A(in 2 )
4.04
7.4
4.04
91.88
0
91.88
183.76
I (in 4 )
1.52
123
1.52
126.04
I = ΣAd 2 + ΣI = 183.76 + 126.04 = 309.8 in 4
Q = A y1 = (4.04)(4.769) = 19.267 in 3
VQ (30)(19.267)
=
= 1.8658 kip/in
309.8
I
1
1
= qs =
(1.8658)(5) = 4.664 kips
2
2
q=
Fbolt
Abolt =
τ bolt =
π
4
2
=
d bolt
π 3
4 4
2
= 0.4418 in 2
4.664
Fbolt
=
= 10.56 ksi
Abolt 0.4418
τ bolt = 10.56 ksi
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 6.8
The composite beam shown is fabricated by connecting two W6 × 20 rolled-steel
members, using bolts of 58 -in. diameter spaced longitudinally every 6 in. Knowing
that the average allowable shearing stress in the bolts is 10.5 ksi, determine the largest
allowable vertical shear in the beam.
SOLUTION
W6 × 20: A = 5.87 in 2 , d = 6.20 in., I x = 41.4 in 4
y =
Composite:
1
d = 3.1 in.
2
I = 2[41.4 + (5.87)(3.1)2 ]
= 195.621 in 4
Q = A y = (5.87)(3.1) = 18.197 in 3
Bolts:
d =
Abolt =
5
in., τ all = 10.5 ksi, s = 6 in.
8
π 5
4 8
2
= 0.30680 in 2
Fbolt = τ all Abolt = (10.5)(0.30680) = 3.2214 kips
Shear:
q=
2Fbolt
(2)(3.2214)
=
= 1.07380 kips/in
s
6
q =
VQ
I
V =
Iq
(195.621)(1.0780)
=
Q
18.197
V = 11.54 kips
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 6.31
Several wooden planks are glued together to form the box beam shown. Knowing
that the beam is subjected to a vertical shear of 3 kN, determine the average
shearing stress in the glued joint (a) at A, (b) at B.
SOLUTION
IA =
1 3
1
(60)(20)3 + (60)(20)(50) 2
12
12
= 3.04 × 106 mm 4
1 3
1
(60)(20)3 = 0.04 × 106 mm 4
bh =
12
12
1 3
1
(20)(120)3 = 2.88 × 106 mm 4
IC =
bh =
12
12
IB =
I = 2I A + I B + 2 I C = 11.88 × 106 mm 4 = 11.88 × 10−6 m 4
QA = Ay = (60)(20)(50) = 60 × 103 mm3 = 60 × 10−6 m3
t = 20 mm + 20 mm = 40 mm = 40 × 10−3 m
(a)
τA =
VQA
(3 × 103 )(60 × 10−6 )
=
= 379 × 103 Pa
It
(11.88 × 10−6 )(40 × 10−3 )
QB = 0
(b)
τB =
VQB
=0
It
τ A = 379 kPa
τB = 0
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 6.33
The built-up wooden beam shown is subjected to a vertical shear of
8 kN. Knowing that the nails are spaced longitudinally every 60 mm
at A and every 25 mm at B, determine the shearing force in the nails
(a) at A, (b) at B. (Given: I x = 1.504 × 109 mm 4.)
SOLUTION
I x = 1.504 × 109 mm 4 = 1504 × 10−6 m 4
s A = 60 mm = 0.060 m
sB = 25 mm = 0.025 m
(a)
QA = Q1 = A1 y1 = (50)(100)(150) = 750 × 103 mm3
= 750 × 10−6 m3
FA = q A s A
=
(8 × 103 )(750 × 10−6 )(0.060)
VQ1s A
=
I
1504 × 10−6
FA = 239 N
(b)
Q2 = A2 y2 = (300)(50)(175) = 2625 × 103 mm3
QB = 2Q1 + Q2 = 4125 × 103 mm3
= 4125 × 10−6 m3
FB = qB sB =
VQB sB
(8 × 103 )(4125 × 10−6 )(0.025)
=
I
1504 × 10−6
FB = 549 N
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 6.37
Knowing that a given vertical shear V causes a maximum shearing stress of 75
MPa in an extruded beam having the cross section shown, determine the shearing
stress at the three points indicated.
SOLUTION
τ =
VQ
It
is proportional to Q/t.
Qc = (30)(10)(75)
Point c:
= 22.5 × 103 mm3
tc = 10 mm
Qc /tc = 2250 mm 2
Qb = Qc + (20)(50)(55)
Point b:
= 77.5 × 103 mm3
tb = 20 mm
Qb /tb = 3875 mm 2
Qa = 2Qb + (120)(30)(15)
Point a:
= 209 × 103 mm3
ta = 120 mm
Qa /ta = 1741.67 mm 2
τ m = τ b = 75 MPa
(Q/t ) m occurs at b.
τa
Qa /ta
τa
1741.67 mm
2
=
=
τb
Qb /tb
=
τc
Qc /tc
75 MPa
τa
=
2
3875 mm
2250 mm 2
τ a = 33.7 MPa
τ b = 75.0 MPa
τ c = 43.5 MPa