Post on 08-Oct-2015
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CIVL 222 STRENGTH OF
MATERIALS
Chapter 4-b
Axially Loaded Members
AXIAL LOADED MEMBERS
Todays Objectives:Students will be able to:
a) Determine the elastic deformation of axially loaded member
b) Apply the principle of superposition for total effect of different loading cases
c) Deal with compatibility conditions
d) Use force method of analysis.
In-class Activities:
Applications
Saint-Venants principle
Elastic deformation in axially
loaded member
Principle of superposition
Compatibility conditions
Force method of analysis
APPLICATIONS
Most concrete columns are reinforced with steel rods; and
these two materials work together in supporting the applied
load. Are both subjected to axial stress?
St. Venants Principle
The stresses and strains in a body at points that
are sufficiently remote from points of application
of load depend only on the static resultant of the
loads and not on the distribution of the loads.
ELASTIC DEFORMATION OF AN AXIALLY LOADED
MEMBER
and = P (x)
A (x) = d
dx
Provided these quantities do not exceed
the proportional limit, we can relate
them using Hookes Law, i.e. = E
P (x)
A (x)= E ( ) d
dx d =
P (x) dx
A (x) E
= L
0
P (x) dx
A (x) E
Fig. 4-2
Fig. 4-4
EXAMPLE 1
EXAMPLE 1 (continued)
PRINCIPLE OF SUPERPOSITION It can be used to simply problems having complicated loadings.
This is done by dividing the loading into components, then
algebraically adding the results.
It is applicable provided the material obeys Hookes Law and the
deformation is small.
Fig. 4-10
If P = P1 + P2And d d1 d2Then the deflection at location x is sum of two cases, ie
x = x1 + x2
COMPATIBILITY CONDITIONS When the force equilibrium condition alone cannot determine
the solution, the structural member is called statically
indeterminate.
In this case, compatibility conditions at the constraint locationsshall be used to obtain the solution. For example, the stresses
and elongations in the 3 steel wires are different, but their
displacement at the common joint A must be the same.
Fig. 4-51
COMPATIBILITY CONDITIONS
The distributed loading w= 46kN/m is supported by threesuspender bars. AB and EF aremade from aluminum and CD ismade from steel. If each bar has a
cross-sectional area of 450 mm2,determine the forces in each bar
when the distributed loading is
applied. Est =200GPa, Eal= 70Gpa.
EXAMPLE
FORCE METHOD OF ANALYSIS
It is also possible to solve statically indeterminate problem
by writing the compatibility equation using the superposition
of the forces acting on the free body diagram.
Fig. 4-16
EXAMPLE
EXAMPLE (continued)
THERMAL STRESS
A temperature change results in a change in length or
thermal strain. There is no stress associated with the thermal strain unless the elongation is restrained by
the supports.
coef.expansion thermal
AE
PLLT PT
Treat the additional support as redundant and apply
the principle of superposition.
0
0
AE
PLLT
PT
The thermal deformation and the deformation from
the redundant support must be compatible.
TEA
P
TAEPPT
0
Static Indeterminacy
Structures for which internal forces and reactions
cannot be determined from statics alone are said
to be statically indeterminate.
0 RL
Deformations due to actual loads and redundant
reactions are determined separately and then added
or superposed.
Redundant reactions are replaced with
unknown loads which along with the other
loads must produce compatible deformations.
A structure will be statically indeterminate
whenever it is held by more supports than are
required to maintain its equilibrium.
Determine the reactions at A and B for the steel bar and loading shown, assuming a close fit at
both supports before the loads are applied.
Solve for the reaction at A due to applied loads and the reaction found at B.
Require that the displacements due to the loads
and due to the redundant reaction be compatible,
i.e., require that their sum be zero.
Solve for the displacement at B due to the redundant reaction at B.
SOLUTION:
Consider the reaction at B as redundant, release the bar from that support, and solve for the
displacement at B due to the applied loads.
Example
SOLUTION:
Solve for the displacement at B due to the applied loads with the redundant constraint released,
EEA
LP
LLLL
AAAA
PPPP
i ii
ii9
L
4321
2643
2621
34
3321
10125.1
m 150.0
m10250m10400
N10900N106000
Solve for the displacement at B due to the redundant constraint,
i
B
ii
iiR
B
E
R
EA
LP
LL
AA
RPP
3
21
262
261
21
1095.1
m 300.0
m10250m10400
Example
Require that the displacements due to the loads and due to
the redundant reaction be compatible,
kN 577N10577
01095.110125.1
0
3
39
B
B
RL
R
E
R
E
Find the reaction at A due to the loads and the reaction at B
kN323
kN577kN600kN 3000
A
Ay
R
RF
kN577
kN323
B
A
R
R
Example
Question # 1: Coaxial tube and core
An aluminum tube (1) encases a
brass core (2). The two
components are bonded together
to comprise an axial member that
is subjected to a downward force
of 30 kN. Tube (1) has an outer
diameter of D= 30 mm and an
inner diameter of d= 22 mm. The
elastic modulus of the aluminum
is 70 GPa. The brass core (2) has
a diameter of D= 22 mm and an
elastic modulus of 105 GPa.
Compute the normal stresses in
tube (1) and core (2).
Question # 2: Coaxial tube and core
Question # 3: Compound bars with temperature change
The compound bar, composed of three segments shown; bronze
segment [A= 2000 mm2, E= 83 GPA: =19 x 10-6
mm/mm/C], aluminum segment [A= 1400 mm2, E= 70 GPA:=23 x 10-6 mm/mm/C], and steel segment [A= 800 mm2, E=200 GPA: =11.7 x 10-6 mm/mm/C], is initially stress-free.Compute the stress in each material if the temperature drops
25C . Assume that the walls do not yield.
Question # 4: Coaxial bars with temperature change
A rectangular bar 30-mm wide
and 24-mm thick made of
aluminum [E= 70 GPA: =23 x10-6 mm/mm/C] and tworectangular copper bars 30-mm
wide and 12-mm thick [E= 120
GPA: =16 x 10-6 mm/mm/C]are connected (in longitudinal
direction) by two smooth 11-mm
diameter pins. When the pins are
initially inserted into the bars,
both the copper and aluminum
bars are stress free. After the
temperature of the assembly has
increased by 65C.
(a) Determine the internal
axial force in the
aluminum bar.
(b) Determine the normal
strain in the copper
bars.
(c) Determine the shear
stress in the 11-mm
diameter pins.
READING QUIZ1) The stress distributions at different cross sections are different
(see figure below).
However, at locations far enough away from the support and the
applied load, the stress distribution becomes uniform. This is due to
A)Principle of superposition C) Poissons effect
B) Inelastic property D) Saint Venants Principle
Fig. 4-1
READING QUIZ
2) The principle of superposition is valid provided that
a) The loading is linearly related to the stress or displacement
b) The loading does not significantly change the original geometry
of the member
c) The Poissons ratio v 0.45
d) Youngs Modulus is small
A)a, b and c C) a and b only
B) a, b and d D) All
CONCEPT QUIZ1) The assembly consists of two posts made from material 1
having modulus of elasticity of E1 and a cross-sectional area
A1 and a material 2 having modulus of elasticity E2 and cross-
sectional area A2. If a central load P is applied to the rigid cap,
determine the force in each post. The support is also rigid.
Let r = E1A1E2A2
A) P1 = ( ) Pr
2r + 1
P2 = ( ) P1
2r + 1
B) P1 = ( ) P1
2r + 1
P2 = ( ) Pr
2r + 1
C) P1 = r P
P2 = (2r-1) P
D) P1 = r (r+1) P
P2 = (r+1) P
Prob. 4-62