Bài giảng Axial Load SBVL

Axial Load


Stress-Strain

Normal stress

Normal strain

Hooke’s Law

The two aluminum rods AB and AC have diameters of 10 mm and 8 mm,
respectively. Determine the largest vertical force P that can be
supported. The allowable tensile stress for the aluminum is σallow = 150
MPa.


Elastic Deformation
How to determine δ = displacement of one
point on the bar relative to the other point

N ( x)
Hooke’s Law
A ( x)
σ =E ( x) ϵ
dδδ
ϵ=
dδx
L

σ=

For the entire length L of the bar,

δ =∫
0

P = constant, AE = constant

General loads, AE = constant on Li
n

Li

SN
δ =∑
i=1 ( AE )i
Li

S N is area of N on Li
Sign Convention

dδδ=

N ( x) dδx
E ( x ) A( x )

N ( x) dδx
E ( x ) A( x )

Segments AB and CD of the assembly are solid circular rods, and

segment BC is a tube. If the assembly is made of 6061-T6 aluminum
(E=68.9 GPa), determine the displacement of end D with respect to
end A.


External Work and Strain Energy
Work of a
Force

Axial Load

The work done by dFz

Strain Energy
If no energy is lost: the
external work done by
the loads → converted converted
into internal work
called strain energy
which is caused by the
action of either normal
or shear stress.

Conservation
of Energy

Strain energy of normal stress

The strain energy

If A=100 mm2, E = 200 Gpa, determine the
horizontal displacement at point B.

N 1= P / √ 3
∂ N 1 / ∂ P=1/ √ 3
L 1=1 m

2

1
N L
Pδ=∑
2
2 AE

A1 = A
Castigliano’s
Theorem
n

δ j =∑ N i (
i =1

∂ N i Li
)
∂ P ( AE)i

N 3= P
N 2 =−2 P / √ 3
∂ N 2 /∂ P=−2 / √ 3 ∂ N 3 /∂ P=1

L 2 =2 m
L 3= √ 3 m
A2 = A
A3 = A
E 3 =E
E 2= E

E 1= E
Method 2: Castigliano’s Theorem
Method 1: Conservation
of Energy
1 P
1
2P 2
δ=
(
×
×1
+
× ×2 + P×1×√ 3)
1
1
2
2
2
AE
3
3
3
√

√
√
√3
Pδ=
(L N + L N + L N )
2
2 AE 1 1 2 2 3 3
(3+ √ 3) P
(3+ √ 3) P
δ=4.732 mm→
δ=
δ=4.732 mm→
δ=
AE
AE


Statically indeterminate axially loaded member
Principle of superposition

Compatibility conditions

Force Method for statically
indeterminate problem

11 X1   p 0

 Ni 0  fi  0, P 
Ni  fi  X 1 , P   
 Ni1  fi  1, 0 

N1

N i21
1
11 
Li 
L4
EA
EA




i 1
i
4
3

i=1

3

X1= N4
i=2

N2

N3
i=3

 p 
i 1

 X 1 

N i 0 N i1
L
 EA i i

 p

11

 Ni  fi  X1 , P 

The equations of equilibrium are not sufficient
to determine all the reactions on a member.
→ converted statically indeterminate problem
Compatibility
conditions

Determine the force
developed in each bar. Bars
AB and EF each have a crosssectional area of 50 mm2 , and
bar CD has a cross-sectional
area of 30 mm2 .


Review problems

The assembly shown in Fig.a 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. Take Est = 200 Gpa, Eal = 70 GPa.

Rigid beam AB rests on the two short posts shown in Fig. a .
AC is made of steel and has a diameter of 20 mm, and BD is
made of aluminum and has a diameter of 40 mm. Determine
the displacement of point F on AB if a vertical load of 90 kN
is applied over this point. Take Est = 200 GPa, Eal = 70 GPa.

The rigid bar is supported by the pinconnected rod CB that has a crosssectional area of 14 mm2. Determine the
vertical deflection of the bar at D when
the distributed load is applied


Review problems
Determine the support reactions at the rigid supports A
and C . The material has a modulus of elasticity of E.

Determine the vertical displacement at C