strength of material chapter 4
TRANSCRIPT
Strength of Materials PPB 25403
Lecture 4: Axial Load
Saint-Venant’s Principle
Saint-Venant’s principle states that both localized
deformation and stress tend to “even out” at a
distance sufficiently removed from these regions.
Elastic Deformation of an Axially Loaded Member
Using Hooke’s law and the definitions of stress and
strain, we are able to develop the elastic deformation
of a member subjected to axial loads.
Suppose an element subjected to loads,
dx
dδε
xA
xP and
L
ExA
dxxP
0
= small displacement
L = original length
P(x) = internal axial force
A(x) = cross-sectional area
E = modulus of elasticity
Elastic Deformation of an Axially Loaded Member
Constant Load and Cross-Sectional Area
When a constant external force is applied at each
end of the member,
Sign Convention
Force and displacement is positive when tension and
elongation and negative will be compression and
contraction.
AE
PL
2- 5
Deformations Under Axial Loading
AE
P
EE
From Hooke’s Law:
From the definition of strain:
L
Equating and solving for the deformation,
AE
PL
With variations in loading, cross-section or
material properties,
i ii
ii
EA
LP
Example 4.1The assembly consists of an aluminum tube AB having a cross-sectional area of
400 mm2. A steel rod having a diameter of 10 mm is attached to a rigid collar and
passes through the tube. If a tensile load of 80 kN is applied to the rod, determine
the displacement of the end C of the rod. (Est = 200 GPa, Eal = 70 GPa )
Solution:Find the displacement of end C with respect to end B.
m 001143.0001143.0107010400
4.0108096
3
AE
PLB
m 003056.010200005.0
6.010809
3
/AE
PLBC
Displacement of end B with respect to the fixed end A,
Since both displacements are to the right, mm 20.4m 0042.0/ BCCC
(Ans)
Principle of Superposition
Principle of superposition is to simplify stress and
displacement problems by subdividing the loading
into components and adding the results.
A member is statically indeterminate when equations
of equilibrium are not sufficient to determine the
reactions on a member.
Statically Indeterminate Axially Loaded Member
Example 4.2The steel rod has a diameter of 5 mm. It is attached to the fixed wall at A, and
before it is loaded, there is a gap between the wall at and B’ and the rod of 1 mm.
Find the reactions at A and B’ if the rod is subjected to an axial force of P = 20 kN.
Neglect the size of the collar at C. (Est = 200 GPa)
Solution:Equilibrium of the rod requires
(1) 01020 ;0 3
BAx FFF
(2) mN 0.39278.04.0
001.0/
BA
CBBACAAB
FF
AE
LF
AE
LF
The compatibility condition for the rod is .m 001.0/ AB
By using the load–displacement relationship,
Solving Eqs. 1 and 2 yields FA = 16.6 kN and FB = 3.39 kN. (Ans)
Example 4.3The A-36 steel rod shown has a diameter of 5 mm. It is attached to the fixed wall at
A, and before it is loaded there is a gap between the wall at and the rod of 1 mm.
Determine the reactions at A and B’.
Solution:
(1) 0.001 Bp
BBABB
B
ACP
FF
AE
LF
AE
PL
6
9
9
3
103056.0102000025.0
2.1
m 002037.0102000025.0
4.01020
Consider the support at B’ as redundant and using principle of superposition,
Thus,
By substituting into Eq. 1,
Solution:
(Ans) kN 39.31039.3
103056.0002037.0001.0
3
6
B
B
F
F
(Ans) kN 6.16
039.320 ;0
A
Ax
F
FF
From the free-body diagram,
Thermal Stress
Change in temperature cause a material to change
its dimensions.
Since the material is homogeneous and isotropic,
TLT
= linear coefficient of thermal expansion, property of the material
= algebraic change in temperature of the member
= original length of the member
= algebraic change in length of the member
TT
T
Example 4.12The rigid bar is fixed to the top of the three posts made of A-36 steel and 2014-T6
aluminum. The posts each have a length of 250 mm when no load is applied to the
bar, and the temperature is T1 = 20°C. Determine the force supported by each post
if the bar is subjected to a uniform distributed load of 150 kN/m and the temperature
is raised to T2 = 20°C.
Solution:
(2) alst
(1) 010902 ;0 3
alsty FFF
From free-body diagram we have
The top of each post is displaced by an equal amount and hence,
The final position of the top of each post is equal to its displacement caused by the
temperature increase and internal axial compressive force.
Solution:
FalTstFstTst
FalTalal
FstTstst
Applying Eq. 2 gives
With reference from the material properties, we have
(3) 109.165216.1
101.7303.0
25.025.020801023
1020002.0
25.025.020801012
3
92
6
92
6
alst
alst
FF
FF
Solving Eqs. 1 and 3 simultaneously yields (Ans) kN 123 and kN 4.16 alst FF
© 2008 Pearson Education South Asia Pte Ltd
Chapter 4: Axial Load
Mechanics of Material 7th Edition
Exercise 1
allow
fail
allow
fail
allow
fail
SF
SF
F
FSF
.
.
.
Ffail=406kN
Answer: 12.5mm, 5.8
© 2008 Pearson Education South Asia Pte Ltd
Chapter 4: Axial Load
Mechanics of Material 7th Edition
Solution
2- 16
Exercise 2
Determine the deformation
of the steel rod shown under
the given loads.
Steps:
Divide the rod into components
at the load application points.
Apply a free-body analysis on
each component to determine
the internal force
Evaluate the total of the
component deflections.
mm 73.1Answer
2- 17
SOLUTION:
Divide the rod into three
components:
26-
21
21
m10581
m3.0
AA
LL
26-
3
3
m10194
m4.0
A
L
Apply free-body analysis to each
component to determine internal forces,
N10120
N1060
N10240
3
3
3
2
3
1
P
P
P
Evaluate total deflection,
m1073.1
10194
4.010120
10581
3.01060
10581
3.010240
10200
1
1
3
6
3
6
3
6
3
9
3
33
2
22
1
11
A
LP
A
LP
A
LP
EEA
LP
i ii
ii
mm 73.1
Exercise 3
Answer Pmax= 186N
Answer