sm7 149 151

________________________________________________________________________
PROBLEMS (*7.149 through 7.151) Compute the shape factor f for an elastoplastic beam of
the cross sections shown in Figs. P7.149 through P7.151.
SOLUTION (*7.149)
For fully plastic case, N.A. divides the entire area into two equal parts:
5
22.4 6
h
bh==
y
12t
AAAA
3
=
++
2
1(2 2.4) 2.4 .
2in
=×=
2
111.2 .
2t
AA in==
Thus,
115
()1.2
226
bh h h==
From which
1.697 . 1.414 .hinb==in
Then
11 22 23
Z
Ay Ay Ay=+ +
1.414 1.697 1.697 1.414 0.703 0.703 2 0.703 (0.703)2
23 2323
×××
=+ +
3
1(4.072 0.699 1.976) 1.125 .
6in
=+ +=
32
1
36 2 3 24
I
BH BH
S
cH
== =
23
2(2.4) 0.48 .
24 in
==
Thus,
1.125 2.34
0.48
Z
fS == =
z’
A3
h
A1
A2
H
=2.4 in. b
0. .703 in
B
=2 in.
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.
Continued on next slide

SOLUTION (7.150)
= axis of elastic bending, = axis of plastic bending
z'z
z’
100 mm
h1
h22
A
20 mm
20 mm
160 m
m
y
N
.A.
z
c1
c2
Elastic bending
220(160)80 100(20)170
160(20) 100(20)
ii
i
Ay
cA
+
== +
∑
∑114.62 mm
=
1180 114.62 65.38cm=− = m
22
20 (160) 20(160)(34.62)
12
z
I=+ 32
100 (20) 20(100)(55.38)
12
++
64
16.86 10 mm=×
643
2
16.86 10 14.71 10
114.62
I
Sm
c
×
== = × m
Plastic bending
32
160 20 100 20 5.2 10Amm=×+×=×
3
22
5.2 10
(20) , 130
22
A
hh
×
== = mm
m
1180 130 50hm=−=
1100 20(40) 30 (30 2) 34.23
100 20 30 20
y
mm
×+×
==
×+×
2130 65
2
y
mm==
3
125.2 10
() (6534.23) 43
25.8 10 mm=×
22
A
Zyy ×
=+= +
Thus,
25.8 1.75
14.71
Z
fS
== =
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.
Continued on next slide

SOLUTION (7.151)
Elastic bending
33
1(100 120 60 80 )
12
z
I=×−×
64
11.84 10 mm=×
6
11.84 10
60
I
Sc
×
==
43
19.73 10 mm=×
Plastic bending
11 2 2
2( )
Z
Ay Ay=+
43
2(20 100 50 40 40 20) 26.4 10 mm=××+××=×
Thus,
26.4 1.34
19.73
Z
fS
== =
40 mm
20 mm
y
2 mm 0
80 mm
0 m
10
m
A
1
A2
z
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.