Homework #5 Solution
ME 311Winter 2016
Due Date 02/19/2016
1
4.5. The C
200
×
20
channel section of Figure P4.5 has second moments of area
I
x
=
15
×
10
6
mm
4
,
I
y
=
0
.
637
×
10
6
mm
4
. It is loaded by the bending moments
M
x
=
2200
Nm and
M
y
=

350
Nm. Find the magnitude and location of the
maximum tensile stress in the beam.
O
x
y
350 Nm
15.6
2200 Nm
203
59.5
9.9
Figure P4.5
We first calculate
E
R
x
=
M
x
I
x
=
2200
×
10
3
15
×
10
6
=
0
.
147 N/mm
3
E
R
y
=
M
y
I
y
=

350
×
10
3
0
.
637
×
10
6
=

0
.
549 N/mm
3
,
from equations (4.16, 4.17) and hence
σ
zz
=
Ey
R
x

Ex
R
y
=
0
.
147
y
+
0
.
549
x
in MPa if
x
,
y
are in mm.
The maximum tensile stress will occur when both
x
and
y
are at their most posi
tive — i.e. at the top right hand corner whose coordinates are
x
C
=
59
.
5

15
.
6
=
43
.
9
;
y
C
=
203
2
=
101
.
5
.
We therefore have
σ
max
zz
=
0
.
147
×
101
.
5
+
0
.
549
×
43
.
9
=
39 MPa
.
1
4.8. The Zsection of Figure P4.8 has second moments of area
I
x
=
560
,
000
mm
4
,
I
y
=
290
,
000
mm
4
,
I
xy
=
300
,
000
mm
4
. It is used for a beam of length
2 m which is simply supported at its ends and loaded by a uniformly distributed
vertical load of 1000 N/m. Find the location and magnitude of the maximum
tensile stress.
x
y
10
10
10
O
40
40
50
all dimensions in mm
Figure P4.8
Figure P4.8.1(a) shows a freebody diagram of the beam, from which, using sym
metry, we deduce that the reactions at each end must be
R
1
=
R
2
=
1000
×
2
2
=
1000 N
.
1000 N/m
2 m
R
2
R
1
1000 N/m
1000 N
y
V
1 m
M
x
(a)
(b)
Figure P4.8.1
The maximum bending moment will occur at the midpoint and can be deter
mined from the freebody diagram of the beam segment shown in Figure P4.8.1(b).
We find
M
max
x
=

1000
×
1
+
1000
×
0
.
5
=

500 Nm
.