PROBLEM 13.31
The extruded aluminum beam has a uniform wall thickness of
1
8
in.
Knowing that
the vertical shear in the beam is 2 kips, determine the corresponding shearing
stress at each of the five points indicated.
SOLUTION
3 34
11
(2.50)(2.50) (2.125)(2.25) 1.23812 in
12 12
I=−=
Add symmetric points c’, b’, and a’.
0
e
Q=
3
1.125
(0.125)(1.125) 0.079102 in
2
d
Q
= =
0.125 in.
d
t=
24
(0.125) (1.1875) 0.097657 in
cd
QQ=+=
0.25 in.
c
t=
3
(2)(1.0625)(0.125)(1.1875) 0.41309 in
bc
QQ=+=
0.25 in.
b
t=
3
1.25
(2)(0.125)(1.25) 0.60840 in
2
ab
QQ
=+=
0.25 in.
a
t=
(2)(0.60840)
(1.23812)(0.25)
a
aa
VQ
It
t
= =
3.93 ksi
a
t
=
(2)(0.41309)
(1.23812)(0.25)
b
bb
VQ
It
t
= =
2.67 ksi
b
t
=
(2)(0.097657)
(1.23812)(0.25)
c
cc
VQ
It
t
= =
0.631ksi
c
t
=
(2)(0.079102)
(1.23812)(0.125)
d
dd
VQ
It
t
= =
1.02 ksi
d
t
=
e
ee
VQ
It
t
=
0
e
t
=
consent of McGraw–Hill Education.
PROBLEM 13.32
The extruded aluminum beam has a uniform wall thickness of
1
8
in.
Knowing
that the vertical shear in the beam is 2 kips, determine the corresponding shearing
stress at each of the five points indicated.
SOLUTION
3 34
11
(2.50)(2.50) (2.125)(2.25) 1.23812 in
12 12
I=−=
0.125 in.t=
at all sections.
2 kipsV=
0
a
Q=
a
a
VQ
It
t
=
0
a
t
=
3
1.25
(0.125)(1.25) 0.097656 in
2
b
Q
= =
(2)(0.097656)
(1.23812)(0.125)
b
b
VQ
It
t
= =
1.262 ksi
b
t
=
2
(1.0625)(0.125)(1.1875) 0.25537 in.
cb
QQ=+=
(2)(0.25537)
(1.23812)(0.125)
c
cVQ
It
t
= =
3.30 ksi
c
t
=
2
2 (0.125) (1.1875) 0.52929
dc
QQ=+=
(2)(0.52929)
(1.23812)(0.125)
d
d
VQ
It
t
= =
6.84 ksi
d
t
=
1.125
(0.125)(1.125) 0.60839
2
ed
QQ
=+=
(2)(0.60839)
(1.23812)(0.125)
e
VQ
It
t
= =
7.86 ksi
e
t
=
consent of McGraw–Hill Education.
PROBLEM 13.33
Knowing that a given vertical shear V causes a maximum shearing
stress of 75 MPa in the hat–shaped extrusion shown, determine the
corresponding shearing stress at (a) point a, (b) point b.
SOLUTION
PROBLEM 13.34
Knowing that a given vertical shear V causes a maximum shearing stress of
50 MPa in a thin–walled member having the cross section shown, determine
the corresponding shearing stress (a) at point a, (b) at point b, (c) at point c.
SOLUTION
33
33
33
33
(12)(30)(25 10 15) 18 10 mm
(40)(10)(25 5) 12 10 mm
2 (12)(10)(25 5) 45.6 10 mm
25
(12)(25) 49.35 10 mm
2
12 mm
10 mm
a
b
ca b
mc
a cm
b
Q
Q
QQ Q
QQ
t tt
t
= ++ =×
= +=×
= + + += ×
=+=×
= = =
=
50 MPa
m
τ
=
(a)
18 12 0.36474
49.35 12
a am
m ma
Qt
Qt
τ
τ
= ⋅= ⋅=
18.23 MPa
a
τ
=
(b)
12 12 0.29179
49.35 10
b bm
m mb
Qt
Qt
τ
τ
= ⋅= ⋅=
14.59 MPa
b
τ
=
(c)
45.6 12 0.92401
49.35 12
c cm
m mc
Qt
Qt
τ
τ
= ⋅= ⋅=
46.2 MPa
c
τ
=
PROBLEM 13.35
The vertical shear is 1200 lb in a beam having the cross section
shown. Knowing that d = 4 in., determine the shearing stress at
(a) point a, (b) point b.
SOLUTION
3 24
1
1(4)(0.5) (4)(0.5)(3.75) 28.167 in
12
I=+=
34
2
1(5)(4) 106.67 in
3
I= =
4
12
4 2 326 inIII=+=
(a)
11 2 2
2
a
Q Ay Ay
= +
3
(2)(4)(0.5)(3.75) (5)(4)(2) 55 in= +=
5 in.
a
t=
(1200)(55) 40.5 psi
(326)(5)
a
aa
VQ
It
t
= = =
(b)
4
11
(4)(0.5)(3.75) 7.5 in
b
Q Ay= = =
0.5 in.
b
t=
(1200)(7.5) 55.2 psi
(326)(0.5)
b
bb
VQ
It
t
= = =
consent of McGraw–Hill Education.
PROBLEM 13.36
The vertical shear is 1200 lb in a beam having the cross section
shown. Determine (a) the distance d for which
t
a =
t
b, (b) the
corresponding shearing stress at points a and b.
SOLUTION
2
1
0.5 in ,Ad=
1
3.75 in.,y=
0.5 in.
b
t=
2
2
(5)(4) 20 in ,A= =
2
2 in.,y=
5 in.
a
t=
3
11
1.875 in
b
Q Ay d= =
1.875 3.75
0.5
b
bb
VQ V d Vd
It I I
t
= = =
22
2 (20)(2) (2)(1.875 )
ab
Q Ay Q d= += +
40 3.75d= +
5 in.
a
t=
(a)
(40 3.75 ) 8 0.75 3.75
(5)
a
ab
a
VQ V d V Vd Vd
It I I I I
tt
+
== =+==
8 0.75 3.75dd+=
82.6667 in.
3
d= =
2.67 ind=
.
(b)
3 24
1
1(2.6667)(0.5) (2.6667)(0.5)(3.75) 18.780 in
12
I=+=
34
2
1(0.5)(4) 106.667 in
3
I= =
4
12
4 2 288.46 inIII=+=
(3.75)(1200)(2.6667)
3.75 288.46
ab
Vd
I
tt
= = =
41.6 psi
a
t
=
consent of McGraw–Hill Education.
PROBLEM 13.37
A beam consists of three planks connected as shown by steel bolts with a
longitudinal spacing of 225 mm. Knowing that the shear in the beam is
vertical and equal to 6 kN and that the allowable average shearing stress in
each bolt is 60 MPa, determine the smallest permissible bolt diameter that
can be used.
SOLUTION
2
26 4
64
PROBLEM 13.38
Four L102 × 102 × 9.5 steel angle shapes and a 12 × 400–mm 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
PROBLEM 13.39
Three planks are connected as shown by bolts of
3
8
-in.
diameter spaced every
6 in. along the longitudinal axis of the beam. For a vertical shear of 2.5 kips,
determine the average shearing stress in the bolts.
SOLUTION
Locate neutral axis.
2
(2)(2)(10) (10)(4) 80 inAΣ= + =
3
(2)(2)(10)(5) (10)(4)(8) 520 inAyΣ= + =
520 6.5 in.
80
Ay
YA
Σ
= = =
Σ
32
3 24
1
2 (2)(10) (2)(10)(1.5)
12
1(10)(4) (10)(4)(1.5) 566.67 in
12
I
= +
++ =
3
(2)(10)(1.5) 30 inQ= =
(2.5)(30)(6) 0.79411kips
566.67
VQs
F qs I
= = = =
2
22
bolt bolt
30.110447 in
4 48
Ad
ππ
= = =
bolt bolt
0.79411 7.19 ksi
0.110447
F
A
t
= = =
consent of McGraw–Hill Education.
PROBLEM 13.31
The extruded aluminum beam has a uniform wall thickness of
1
8
in.
Knowing that
the vertical shear in the beam is 2 kips, determine the corresponding shearing
stress at each of the five points indicated.
SOLUTION
3 34
11
(2.50)(2.50) (2.125)(2.25) 1.23812 in
12 12
I=−=
Add symmetric points c’, b’, and a’.
0
e
Q=
3
1.125
(0.125)(1.125) 0.079102 in
2
d
Q
= =
0.125 in.
d
t=
24
(0.125) (1.1875) 0.097657 in
cd
QQ=+=
0.25 in.
c
t=
3
(2)(1.0625)(0.125)(1.1875) 0.41309 in
bc
QQ=+=
0.25 in.
b
t=
3
1.25
(2)(0.125)(1.25) 0.60840 in
2
ab
QQ
=+=
0.25 in.
a
t=
(2)(0.60840)
(1.23812)(0.25)
a
aa
VQ
It
t
= =
3.93 ksi
a
t
=
(2)(0.41309)
(1.23812)(0.25)
b
bb
VQ
It
t
= =
2.67 ksi
b
t
=
(2)(0.097657)
(1.23812)(0.25)
c
cc
VQ
It
t
= =
0.631ksi
c
t
=
(2)(0.079102)
(1.23812)(0.125)
d
dd
VQ
It
t
= =
1.02 ksi
d
t
=
e
ee
VQ
It
t
=
0
e
t
=
consent of McGraw–Hill Education.
PROBLEM 13.32
The extruded aluminum beam has a uniform wall thickness of
1
8
in.
Knowing
that the vertical shear in the beam is 2 kips, determine the corresponding shearing
stress at each of the five points indicated.
SOLUTION
3 34
11
(2.50)(2.50) (2.125)(2.25) 1.23812 in
12 12
I=−=
0.125 in.t=
at all sections.
2 kipsV=
0
a
Q=
a
a
VQ
It
t
=
0
a
t
=
3
1.25
(0.125)(1.25) 0.097656 in
2
b
Q
= =
(2)(0.097656)
(1.23812)(0.125)
b
b
VQ
It
t
= =
1.262 ksi
b
t
=
2
(1.0625)(0.125)(1.1875) 0.25537 in.
cb
QQ=+=
(2)(0.25537)
(1.23812)(0.125)
c
cVQ
It
t
= =
3.30 ksi
c
t
=
2
2 (0.125) (1.1875) 0.52929
dc
QQ=+=
(2)(0.52929)
(1.23812)(0.125)
d
d
VQ
It
t
= =
6.84 ksi
d
t
=
1.125
(0.125)(1.125) 0.60839
2
ed
QQ
=+=
(2)(0.60839)
(1.23812)(0.125)
e
VQ
It
t
= =
7.86 ksi
e
t
=
consent of McGraw–Hill Education.
PROBLEM 13.33
Knowing that a given vertical shear V causes a maximum shearing
stress of 75 MPa in the hat–shaped extrusion shown, determine the
corresponding shearing stress at (a) point a, (b) point b.
SOLUTION
PROBLEM 13.34
Knowing that a given vertical shear V causes a maximum shearing stress of
50 MPa in a thin–walled member having the cross section shown, determine
the corresponding shearing stress (a) at point a, (b) at point b, (c) at point c.
SOLUTION
33
33
33
33
(12)(30)(25 10 15) 18 10 mm
(40)(10)(25 5) 12 10 mm
2 (12)(10)(25 5) 45.6 10 mm
25
(12)(25) 49.35 10 mm
2
12 mm
10 mm
a
b
ca b
mc
a cm
b
Q
Q
QQ Q
QQ
t tt
t
= ++ =×
= +=×
= + + += ×
=+=×
= = =
=
50 MPa
m
τ
=
(a)
18 12 0.36474
49.35 12
a am
m ma
Qt
Qt
τ
τ
= ⋅= ⋅=
18.23 MPa
a
τ
=
(b)
12 12 0.29179
49.35 10
b bm
m mb
Qt
Qt
τ
τ
= ⋅= ⋅=
14.59 MPa
b
τ
=
(c)
45.6 12 0.92401
49.35 12
c cm
m mc
Qt
Qt
τ
τ
= ⋅= ⋅=
46.2 MPa
c
τ
=
PROBLEM 13.35
The vertical shear is 1200 lb in a beam having the cross section
shown. Knowing that d = 4 in., determine the shearing stress at
(a) point a, (b) point b.
SOLUTION
3 24
1
1(4)(0.5) (4)(0.5)(3.75) 28.167 in
12
I=+=
34
2
1(5)(4) 106.67 in
3
I= =
4
12
4 2 326 inIII=+=
(a)
11 2 2
2
a
Q Ay Ay
= +
3
(2)(4)(0.5)(3.75) (5)(4)(2) 55 in= +=
5 in.
a
t=
(1200)(55) 40.5 psi
(326)(5)
a
aa
VQ
It
t
= = =
(b)
4
11
(4)(0.5)(3.75) 7.5 in
b
Q Ay= = =
0.5 in.
b
t=
(1200)(7.5) 55.2 psi
(326)(0.5)
b
bb
VQ
It
t
= = =
consent of McGraw–Hill Education.
PROBLEM 13.36
The vertical shear is 1200 lb in a beam having the cross section
shown. Determine (a) the distance d for which
t
a =
t
b, (b) the
corresponding shearing stress at points a and b.
SOLUTION
2
1
0.5 in ,Ad=
1
3.75 in.,y=
0.5 in.
b
t=
2
2
(5)(4) 20 in ,A= =
2
2 in.,y=
5 in.
a
t=
3
11
1.875 in
b
Q Ay d= =
1.875 3.75
0.5
b
bb
VQ V d Vd
It I I
t
= = =
22
2 (20)(2) (2)(1.875 )
ab
Q Ay Q d= += +
40 3.75d= +
5 in.
a
t=
(a)
(40 3.75 ) 8 0.75 3.75
(5)
a
ab
a
VQ V d V Vd Vd
It I I I I
tt
+
== =+==
8 0.75 3.75dd+=
82.6667 in.
3
d= =
2.67 ind=
.
(b)
3 24
1
1(2.6667)(0.5) (2.6667)(0.5)(3.75) 18.780 in
12
I=+=
34
2
1(0.5)(4) 106.667 in
3
I= =
4
12
4 2 288.46 inIII=+=
(3.75)(1200)(2.6667)
3.75 288.46
ab
Vd
I
tt
= = =
41.6 psi
a
t
=
consent of McGraw–Hill Education.
PROBLEM 13.37
A beam consists of three planks connected as shown by steel bolts with a
longitudinal spacing of 225 mm. Knowing that the shear in the beam is
vertical and equal to 6 kN and that the allowable average shearing stress in
each bolt is 60 MPa, determine the smallest permissible bolt diameter that
can be used.
SOLUTION
2
26 4
64
PROBLEM 13.38
Four L102 × 102 × 9.5 steel angle shapes and a 12 × 400–mm 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
PROBLEM 13.39
Three planks are connected as shown by bolts of
3
8
-in.
diameter spaced every
6 in. along the longitudinal axis of the beam. For a vertical shear of 2.5 kips,
determine the average shearing stress in the bolts.
SOLUTION
Locate neutral axis.
2
(2)(2)(10) (10)(4) 80 inAΣ= + =
3
(2)(2)(10)(5) (10)(4)(8) 520 inAyΣ= + =
520 6.5 in.
80
Ay
YA
Σ
= = =
Σ
32
3 24
1
2 (2)(10) (2)(10)(1.5)
12
1(10)(4) (10)(4)(1.5) 566.67 in
12
I
= +
++ =
3
(2)(10)(1.5) 30 inQ= =
(2.5)(30)(6) 0.79411kips
566.67
VQs
F qs I
= = = =
2
22
bolt bolt
30.110447 in
4 48
Ad
ππ
= = =
bolt bolt
0.79411 7.19 ksi
0.110447
F
A
t
= = =
consent of McGraw–Hill Education.