PROBLEM 13.50
A square box beam is made of two
20 80-mm
×
planks and two
20 120-mm×
planks nailed together as shown. Knowing that the
spacing between the nails is
30s=
mm and that the vertical shear in
the beam is
1200 N,V=
determine (a) the shearing force in each nail,
consent of McGraw–Hill Education.
PROBLEM 13.51
The composite beam shown is fabricated by connecting two
W6 20×
rolled–steel
members, using bolts of
5
8
-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
24
W6 20: 5.87 in , 6.20 in., 41.4 in
13.1 in.
2
x
AdI
yd
×= = =
= =
Composite:
2
4
3
2[41.4 (5.87)(3.1) ]
195.621 in
(5.87)(3.1) 18.197 in
I
Q Ay
= +
=
= = =
Bolts:
all
22
bolt
bolt all bolt
bolt
5 in., 10.5 ksi, 6 in.
8
50.30680 in
48
(10.5)(0.30680) 3.2214 kips
2 (2)(3.2214) 1.07380 kips/in.
6
ds
A
FA
F
qs
t
p
t
= = =
= =
= = =
= = =
Shear:
(195.621)(1.0780)
18.197
VQ Iq
qV
IQ
= = =
11.54 kipsV=
consent of McGraw–Hill Education.
PROBLEM 13.52
For the beam and loading shown, consider section
n–n and determine (a) the largest shearing stress in
that section, (b) the shearing stress at point a.
SOLUTION
At section n–n,
80 kN
V=
Consider cross section as composed of rectangles of types , , and .
3 2 64
1
3 2 64
2
3 34
3
64
123
64
1(12)(80) (12)(80)(90) 8.288 10 mm
12
1(180)(16) (180)(16)(42) 5.14176 10 mm
12
1(16)(68) 419.24 10 mm
12
4 2 2 44.274 10 mm
44.274 10 m
I
I
I
III I
−
=+=×
=+=×
= = ×
=++= ×
= ×
(a) Calculate Q at neutral axis.
34
1
34
2
34
3
(12)(80)(90) 86.4 10 mm
(180)(16)(42) 120.96 10 mm
(16)(34)(17) 9.248 10 mm
Q
Q
Q
= = ×
= = ×
= = ×
3 3 63
12 3
36 6
63
2 2 312.256 10 mm 312.256 10 m
(80 10 )(312.256 10 ) 17.63 10 Pa
(44.274 10 )(2 16 10 )
Q QQ Q
VQ
It
t
−
−
−−
= ++ = × = ×
××
= = = ×
× ××
17.63 MPa
t
=
(b) At point a,
3 4 64
1
86.4 10 mm 86.4 10 mQQ
−
==×=×
36 6
63
(80 10 )(86.4 10 ) 13.01 10 Pa
(44.274 10 )(12 10 )
VQ
It
t
−
−−
××
= = = ×
××
13.01 MPa
t
=
consent of McGraw–Hill Education.
PROBLEM 13.53
A timber beam AB of length L and rectangular cross section carries a
uniformly distributed load w and is supported as shown. (a) Show that
the ratio
/
mm
τσ
of the maximum values of the shearing and normal
stresses in the beam is equal to
2/,hL
where h and L are, respectively,
the depth and the length of the beam. (b) Determine the depth h and
the width b of the beam, knowing that
5 m,
L=
8w=
kN/m,
1.08 MPa,
m
τ
=
and
12 MPa.
m
σ
=
SOLUTION
2
AB
wL
RR= =
From shear diagram,
|| 4
m
wL
V=
(1)
For rectangular section,
A bh=
(2)
33
28
m
m
V wL
A bh
τ
= =
(3)
From bending moment diagram,
2
|| 32
mwL
M=
(4)
For a rectangular cross section,
2
1
6
S bh=
(5)
2
2
|| 3
16
m
mM wL
Sbh
σ
= =
(6)
(a) Dividing Eq. (3) by Eq. (6),
2
m
m
h
L
τ
σ
=
(b) Solving for h,
63
6
(5)(1.08 10 ) 225 10 m
2(2)(12 10 )
m
m
L
h
τ
σ
−
×
= = = ×
×
225 mmh=
Solving Eq. (3) for b,
3
36
3 (3)(8 10 )(5)
8(8)(225 10 )(1.08 10 )
m
wL
bh
τ
−
×
= = ××
3
61.7 10 m
−
= ×
61.7 mmb=
consent of McGraw–Hill Education.
PROBLEM 13.54
For the beam and loading shown, consider
section n–n and determine the shearing stress
at (a) point a, (b) point b.
SOLUTION
12 kips
AB
RR= =
Draw shear diagram.
12 kipsV=
Determine section properties.
Part
2
(in )A
(in.)y
3
(in )
Ay
d(in.)
24
(in )Ad
4
(in )I
4 4 16 2 16 5.333
8 1 8 −1 8 2.667
Σ 12 24 24 8.000
24
24 2 in.
12
32 in
Ay
YA
I Ad I
Σ
= = =
Σ
=Σ +Σ =
(a)
23
1 in 3.5 in. 3.5 in
1 in.
a
A y Q Ay
t
= = = =
=
(12)(3.5)
(32)(1)
a
a
VQ
It
t
= =
1.313 ksi
a
t
=
(b)
23
2 in 3 in. 6 in
1 in.
b
A y Q Ay
t
= = = =
=
(12)(6)
(32)(1)
b
b
VQ
It
t
= =
2.25 ksi
b
t
=
consent of McGraw–Hill Education.
PROBLEM 13.55
Two W8 × 31 rolled–steel sections may be welded at A and B in either
of the two ways shown in order to form a composite beam. Knowing
that for each weld the allowable shearing force is 3000 lb per inch of
weld, determine for each arrangement the maximum allowable vertical
PROBLEM 13.56
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
32 3 2
64
3 3 64
3 3 64
6 4 64
11
(60)(20) (60)(20)(50)
12 12
3.04 10 mm
11
(60)(20) 0.04 10 mm
12 12
11
(20)(120) 2.88 10 mm
12 12
2 2 11.88 10 mm 11.88 10 m
A
B
C
AB C
I bh Ad
I bh
I bh
III I
−
= += +
= ×
= = = ×
= = = ×
= ++ = × = ×
3 3 63
3
(60)(20)(50) 60 10 mm 60 10 m
20 mm 20 mm 40 mm 40 10 m
A
Q Ay
t
−
−
== =×=×
=+==×
(a)
36 3
63
(3 10 )(60 10 ) 379 10 Pa
(11.88 10 )(40 10 )
A
A
VQ
It
t
−
−−
××
= = = ×
××
0
B
Q=
379 kPa
A
t
=
(b)
0
B
B
VQ
It
t
= =
0
B
t
=
PROBLEM 13.57
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.
94
(Given: 1.504 10 mm .)
x
I= ×
SOLUTION
9 4 64
1.504 10 mm 1504 10 m
60 mm 0.060 m
25 mm 0.025 m
x
A
B
I
s
s
−
=×=×
= =
= =
(a)
33
1 11
63
(50)(100)(150) 750 10 mm
750 10 m
A
Q Q Ay
−
= = = = ×
= ×
36
16
(8 10 )(750 10 )(0.060)
1504 10
A AA
A
F qs
VQ s
I
−
−
=
××
= = ×
239 N
A
F=
(b)
33
2 22
33
12
63
(300)(50)(175) 2625 10 mm
2 4125 10 mm
4125 10 m
B
Q Ay
Q QQ
−
= = = ×
= += ×
= ×
36
6
(8 10 )(4125 10 )(0.025)
1504 10
BB
B BB
VQ s
F qs I
−
−
××
= = = ×
549 N
B
F=
consent of McGraw–Hill Education.
PROBLEM 13.58
An extruded beam has the cross section shown and a uniform wall thickness
of 0.20 in. Knowing that a given vertical shear V causes a maximum shearing
stress
9 ksi,
τ
=
determine the shearing stress at the four points indicated.
SOLUTION
3
3
3
3
3
(0.2)(0.5)(0.5 0.25) 0.125 in
(0.2)(0.5)(0.3 0.25) 0.055 in
(1.4)(0.2)(0.9) 0.432 in
2 2 (3.0)(0.2)(0.9) 0.900 in.
(0.2)(0.8)(0.4) 0.964 in
a
b
c ab
d ab
md
Q
Q
QQQ
Q QQ
QQ
= −=
= +=
=++ =
=++ =
=+=
VQ
It
τ
=
Since V, I, and t are constant,
τ
is proportional to Q.
9
0.125 0.055 0.432 0.900 0.964 0.964
abcdm
τττττ
= = = = =
1.167 ksi; 0.513 ksi; 4.03 ksi; 8.40 ksi
a b cd
τ τ ττ
= = = =
PROBLEM 13.50
A square box beam is made of two
20 80-mm
×
planks and two
20 120-mm×
planks nailed together as shown. Knowing that the
spacing between the nails is
30s=
mm and that the vertical shear in
the beam is
1200 N,V=
determine (a) the shearing force in each nail,
consent of McGraw–Hill Education.
PROBLEM 13.51
The composite beam shown is fabricated by connecting two
W6 20×
rolled–steel
members, using bolts of
5
8
-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
24
W6 20: 5.87 in , 6.20 in., 41.4 in
13.1 in.
2
x
AdI
yd
×= = =
= =
Composite:
2
4
3
2[41.4 (5.87)(3.1) ]
195.621 in
(5.87)(3.1) 18.197 in
I
Q Ay
= +
=
= = =
Bolts:
all
22
bolt
bolt all bolt
bolt
5 in., 10.5 ksi, 6 in.
8
50.30680 in
48
(10.5)(0.30680) 3.2214 kips
2 (2)(3.2214) 1.07380 kips/in.
6
ds
A
FA
F
qs
t
p
t
= = =
= =
= = =
= = =
Shear:
(195.621)(1.0780)
18.197
VQ Iq
qV
IQ
= = =
11.54 kipsV=
consent of McGraw–Hill Education.
PROBLEM 13.52
For the beam and loading shown, consider section
n–n and determine (a) the largest shearing stress in
that section, (b) the shearing stress at point a.
SOLUTION
At section n–n,
80 kN
V=
Consider cross section as composed of rectangles of types , , and .
3 2 64
1
3 2 64
2
3 34
3
64
123
64
1(12)(80) (12)(80)(90) 8.288 10 mm
12
1(180)(16) (180)(16)(42) 5.14176 10 mm
12
1(16)(68) 419.24 10 mm
12
4 2 2 44.274 10 mm
44.274 10 m
I
I
I
III I
−
=+=×
=+=×
= = ×
=++= ×
= ×
(a) Calculate Q at neutral axis.
34
1
34
2
34
3
(12)(80)(90) 86.4 10 mm
(180)(16)(42) 120.96 10 mm
(16)(34)(17) 9.248 10 mm
Q
Q
Q
= = ×
= = ×
= = ×
3 3 63
12 3
36 6
63
2 2 312.256 10 mm 312.256 10 m
(80 10 )(312.256 10 ) 17.63 10 Pa
(44.274 10 )(2 16 10 )
Q QQ Q
VQ
It
t
−
−
−−
= ++ = × = ×
××
= = = ×
× ××
17.63 MPa
t
=
(b) At point a,
3 4 64
1
86.4 10 mm 86.4 10 mQQ
−
==×=×
36 6
63
(80 10 )(86.4 10 ) 13.01 10 Pa
(44.274 10 )(12 10 )
VQ
It
t
−
−−
××
= = = ×
××
13.01 MPa
t
=
consent of McGraw–Hill Education.
PROBLEM 13.53
A timber beam AB of length L and rectangular cross section carries a
uniformly distributed load w and is supported as shown. (a) Show that
the ratio
/
mm
τσ
of the maximum values of the shearing and normal
stresses in the beam is equal to
2/,hL
where h and L are, respectively,
the depth and the length of the beam. (b) Determine the depth h and
the width b of the beam, knowing that
5 m,
L=
8w=
kN/m,
1.08 MPa,
m
τ
=
and
12 MPa.
m
σ
=
SOLUTION
2
AB
wL
RR= =
From shear diagram,
|| 4
m
wL
V=
(1)
For rectangular section,
A bh=
(2)
33
28
m
m
V wL
A bh
τ
= =
(3)
From bending moment diagram,
2
|| 32
mwL
M=
(4)
For a rectangular cross section,
2
1
6
S bh=
(5)
2
2
|| 3
16
m
mM wL
Sbh
σ
= =
(6)
(a) Dividing Eq. (3) by Eq. (6),
2
m
m
h
L
τ
σ
=
(b) Solving for h,
63
6
(5)(1.08 10 ) 225 10 m
2(2)(12 10 )
m
m
L
h
τ
σ
−
×
= = = ×
×
225 mmh=
Solving Eq. (3) for b,
3
36
3 (3)(8 10 )(5)
8(8)(225 10 )(1.08 10 )
m
wL
bh
τ
−
×
= = ××
3
61.7 10 m
−
= ×
61.7 mmb=
consent of McGraw–Hill Education.
PROBLEM 13.54
For the beam and loading shown, consider
section n–n and determine the shearing stress
at (a) point a, (b) point b.
SOLUTION
12 kips
AB
RR= =
Draw shear diagram.
12 kipsV=
Determine section properties.
Part
2
(in )A
(in.)y
3
(in )
Ay
d(in.)
24
(in )Ad
4
(in )I
4 4 16 2 16 5.333
8 1 8 −1 8 2.667
Σ 12 24 24 8.000
24
24 2 in.
12
32 in
Ay
YA
I Ad I
Σ
= = =
Σ
=Σ +Σ =
(a)
23
1 in 3.5 in. 3.5 in
1 in.
a
A y Q Ay
t
= = = =
=
(12)(3.5)
(32)(1)
a
a
VQ
It
t
= =
1.313 ksi
a
t
=
(b)
23
2 in 3 in. 6 in
1 in.
b
A y Q Ay
t
= = = =
=
(12)(6)
(32)(1)
b
b
VQ
It
t
= =
2.25 ksi
b
t
=
consent of McGraw–Hill Education.
PROBLEM 13.55
Two W8 × 31 rolled–steel sections may be welded at A and B in either
of the two ways shown in order to form a composite beam. Knowing
that for each weld the allowable shearing force is 3000 lb per inch of
weld, determine for each arrangement the maximum allowable vertical
PROBLEM 13.56
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
32 3 2
64
3 3 64
3 3 64
6 4 64
11
(60)(20) (60)(20)(50)
12 12
3.04 10 mm
11
(60)(20) 0.04 10 mm
12 12
11
(20)(120) 2.88 10 mm
12 12
2 2 11.88 10 mm 11.88 10 m
A
B
C
AB C
I bh Ad
I bh
I bh
III I
−
= += +
= ×
= = = ×
= = = ×
= ++ = × = ×
3 3 63
3
(60)(20)(50) 60 10 mm 60 10 m
20 mm 20 mm 40 mm 40 10 m
A
Q Ay
t
−
−
== =×=×
=+==×
(a)
36 3
63
(3 10 )(60 10 ) 379 10 Pa
(11.88 10 )(40 10 )
A
A
VQ
It
t
−
−−
××
= = = ×
××
0
B
Q=
379 kPa
A
t
=
(b)
0
B
B
VQ
It
t
= =
0
B
t
=
PROBLEM 13.57
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.
94
(Given: 1.504 10 mm .)
x
I= ×
SOLUTION
9 4 64
1.504 10 mm 1504 10 m
60 mm 0.060 m
25 mm 0.025 m
x
A
B
I
s
s
−
=×=×
= =
= =
(a)
33
1 11
63
(50)(100)(150) 750 10 mm
750 10 m
A
Q Q Ay
−
= = = = ×
= ×
36
16
(8 10 )(750 10 )(0.060)
1504 10
A AA
A
F qs
VQ s
I
−
−
=
××
= = ×
239 N
A
F=
(b)
33
2 22
33
12
63
(300)(50)(175) 2625 10 mm
2 4125 10 mm
4125 10 m
B
Q Ay
Q QQ
−
= = = ×
= += ×
= ×
36
6
(8 10 )(4125 10 )(0.025)
1504 10
BB
B BB
VQ s
F qs I
−
−
××
= = = ×
549 N
B
F=
consent of McGraw–Hill Education.
PROBLEM 13.58
An extruded beam has the cross section shown and a uniform wall thickness
of 0.20 in. Knowing that a given vertical shear V causes a maximum shearing
stress
9 ksi,
τ
=
determine the shearing stress at the four points indicated.
SOLUTION
3
3
3
3
3
(0.2)(0.5)(0.5 0.25) 0.125 in
(0.2)(0.5)(0.3 0.25) 0.055 in
(1.4)(0.2)(0.9) 0.432 in
2 2 (3.0)(0.2)(0.9) 0.900 in.
(0.2)(0.8)(0.4) 0.964 in
a
b
c ab
d ab
md
Q
Q
QQQ
Q QQ
QQ
= −=
= +=
=++ =
=++ =
=+=
VQ
It
τ
=
Since V, I, and t are constant,
τ
is proportional to Q.
9
0.125 0.055 0.432 0.900 0.964 0.964
abcdm
τττττ
= = = = =
1.167 ksi; 0.513 ksi; 4.03 ksi; 8.40 ksi
a b cd
τ τ ττ
= = = =