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CIVE207,
Fall 2008,
Assignment
#1
(Due Sept. 16, Tuesday 12pm, Box 2 in Room MD475)
CEl
P1.34
A rectangular bar having width
w
=
6.00 in.
an~
thi~imes~
t
=
1.50 in. is subjected to a tension load
P.
The normal and shear
s.tresses on plane
AB
must not exceed 16 ksi and 8 ksi, respec
tIvely.
Det~rmi~e
the maximum load
P
that can be applied with
out exceedmg eIther stress limit.
FIGURE
P1.34/35
"'I'
'' .
.
.
,~r
P4.11
The simple pinconnected structure carries a concentrated
P3.23
Rigid bar
BCD
in Fig. P3.23 is supported by a pin at C and
load
P
as shown in Fig. P4.11. The rigid bar is supported by strut
by aluminum rod (I). A concentrated load
P
is applied to the lower
AB
and by a pin support at
C.
The steel strut
AB
has a cross
end of aluminum rod (2), which is attached to the rigid bar at D. The
sectional area of 0.75 in.
2
and a yield strength 0f60 ksi. The diameter
crosssectional area of each rod is
A
=
0.20 in? and the elastic mod
ulus of the aluminum material is
E
=
10,000 ksi. After the load
P
is
applied at
E,
the strain in rod (I) is measured as 900 JlE (tension).
of the steel pin at Cis 0.5 in., and the ultimate shear strength is 54
Determine:
ksi. If a factor of safety of 2.0 is required in both the strut and the
pin at C, determine the maximum load
P
that can be supported by
(a) the magnitude of load
P.
the structure.
(b)
the total deflection of point
E
relative to its initial position.
I'
'I
50 in.
~(l)l
Rigid bar
_._.m".,D
20 in.
Connection
detail
FIGURE P3.23
FIGURE P4.11
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View Full Document1.34
A rectangular bar having width
w =
6.00
in. and thickness
t
= 1.50 in. is subjected to a
tension load
P.
The normal and shear stresses
on plane
AB
must not exceed 16 ksi and 8 ksi,
respectively. Determine the maximum load
P
that can be applied without exceeding either
stress limit.
Fig. Pl.34
Solution
The general equations for normal and shear stresses on an inclined plane in terms
of the angle
B
are
P
an
=
2A
(l+cos2B)
(a)
and
Tnt
=
~sin2B
(b)
2A
The angle
B
for inclined plane
AB
is calculated from
3
tanB =  = 3
:.
B
= 71.5651 °
1
The crosssectional area
of the bar is
A
=
wxt
=
(6.00 in.)(1.50 in.)
=
9.0 in?
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 Fall '09
 Shao

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