________________________________________________________________________
PROBLEMS (7.49 through *7.51) A vertically symmetric beam of t =10 mm uniform wall
thickness as shown in Figs. P7.49 through P7.51, is acted upon by shear force V = 20 kN. Dimensions are
in millimeters. Determine:
(a) The shear stress at the sections indicated.
(b) The maximum shearing stress in the beam.
SOLUTION (7.49)
6.52
y
z
c
15
b
z
’
80
60
c
c
A
1
A
2
A
3
A
3
C
41.52
b1
b1
a
a
38.48
36.52
(c)
(
b
)
(
a
)
We have: mmtkNV 1020
=
=
11 2 2 33
123
Ay Ay Ay
yAAA
++
=++
60 10 75 2(60 10 40) 2(25 10 5) 41.52
60 10 2 60 10 2 25 10 mm
×× + ×× + ××
==
×+××+××
33
22
60(10) 10(60)
60 10(33.48) 2[ 10 60(1.52) ]
12 12
I=+× + +×
326
25(10)
2[ 25 10(36.52) ] 1.71 10
12 mm++× =×
4
(a) Section a-a
Q
33
10478.552.361015 mm
a×=××=
36
63
20 10 (5.478 10 ) 6.41
1.71 10 (10 10 )
aVQ
M
Pa
Ib
τ
−
−−
××
== =
××
Section b-b (Fig.b)
33
10694.652.11080 mmQQ ab ×=××+=
36
63
20 10 (6.694 10 ) 7.83
1.71 10 (10 10 )
bVQ
M
Pa
Ib
τ
−
−−
××
== =
××
Continued on next slide
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
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Section c-c (Fig.c)
33
10042.1052.61070 mmQQ ac ×=××+=
36
63
20 10 (10.042 10 ) 11.75
1.71 10 (10 10 )
cVQ
M
Pa
Ib
τ
−
−−
××
== =
××
(b) Refer to Fig. a:
33
max 10098.1476.201052.41 mmQQ a×=××+=
36
max
max 63
20 10 (14.098 10 ) 16.49
1.71 10 (10 10 )
VQ
M
Pa
Ib
τ
−
−−
××
== =
××
SOLUTION(7.50)
We have: mmtkNV 1020
=
=
11 2 2 33
123
Ay Ay Ay
yAAA
++
=++
70 10 45 2(15 10 17.5) 60 10 5 24.84
70 10 2 15 10 60 10 mm
×× + ×× + ××
==
×+××+×
33
22
10(70) 10(15)
70 10(20.16) 2[ 10 15(7.34) ]
12 12
I=+× + +×
323
60(10) 60 10(19.84) 833.3 10
12 mm++× =×
4
(a) Section a-a (Fig. a)
33
10048.1216.401030 mmQa×=××=
36
93
20 10 (12.048 10 ) 28.92
833.3 10 (10 10 )
aVQ
M
Pa
Ib
τ
−
−−
××
== =
××
7.34
y
z
a
25 b
z
’
55.16
60 c
c
A
1
A
3
A2
A
2
c
19.84
b
b
aa
40.16
24.84=
y
30
(a)
(
b
)
(
c
)
Dimensions are
in millim
e
t
e
r
s
Continued on next slide
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.
Section b-b (Fig.b)
33
101.134.71015 mmQb×=××=
36
93
20 10 (1.1 10 ) 2.64
833.3 10 (10 10 )
bVQ
M
Pa
Ib
τ
−
−−
××
== =
××
Section c-c (Fig.c)
33
1006.684.191025 mmQQ bc ×=××+=
36
93
20 10 (6.06 10 ) 14.54
833.3 10 (10 10 )
cVQ
M
Pa
Ib
τ
−
−−
××
== =
××
(b) 33
max 55.16
10 55.16( ) 15.213 10
2
Qm=× = × m
36
max
max 93
20 10 (15.213 10 ) 36.51
833.3 10 (10 10 )
VQ
M
Pa
Ib
τ
−
−−
××
== =
××
SOLUTION (*7.51)
We have: V mmtkN 1020
=
=
33 64
60(80) 15(60)
2[ ] 2.02 10
12 12
I
mm=− =×
(a) Section a-a
At this section, we have 0
=
a
τ
Section b-b (Fig.a)
33
104201020 mmQb×=××=
36
63
20 10 (4 10 ) 3.96
2.02 10 (10 10 )
bVQ
M
Pa
Ib
τ
−
−−
××
== =
××
20
y
z
b
d
c
d
d
c
15
(a)
(
b
)
(
c
)
b
aa
b
c
c
d
40
(d)
30
50
40
Dimensions are
in millim
e
t
e
r
s
Section c-c (Fig.b) – cut 2 surfaces:
33
105.7151050 mmQc×=××=
Continued on next slide
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.
36
63
20 10 (7.5 10 ) 3.71
2.02 10 (10 10 )2
cVQ
M
Pa
Ib
τ
−
−−
××
== =
××
Section d-d (Fig. c) – cut 2 surfaces:
33
105.21351050101040 mmQd×=××+××=
36
63
20 10 (21.5 10 ) 10.65
2.02 10 (10 10 )2
dVQ
M
Pa
Ib
τ
−
−−
××
== =
××
(b) Refer to Fig. d– cut 2 surfaces: 33
max 103451010)153015(2204060 mmQ ×=××−××−××=
36
max
max 63
20 10 (34 10 ) 16.83
2.02 10 (10 10 )2
VQ
M
Pa
Ib
τ
−
−−
××
== =
××
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.
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