LECTURE
Third Edition
RODS: AXIAL LOADING AND
DEFORMATION
• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering
3
Chapter
2.8
by
Dr. Ibrahim A. Assakkaf
SPRING 2003
ENES 220 – Mechanics of Materials
Department of Civil and Environmental Engineering
University of Maryland, College Park
Slide No. 1
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Uniform Member
– Consider a homogeneous rod as shown in
the figure of the next viewgraph.
– If the resultant axial stress σ = P/A dos not
exceed the proportional limit of the
material, then Hooke’s law can applied,
that is
σ = Eε
σ = axial stress, ε = axial strain
E = modulus of elasticity
1
Slide No. 2
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
ENES 220 ©Assakkaf
Deformations of Members under
Axial Loading
„
Uniform Member
L
δ
P
Slide No. 3
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Uniform Member
– From Hooke’s law, it follows that
σ = Eε (Hooke' s law)
P
= Eε
A
But ε =
δ
L
, therefore,
δ
PL
P
= E ⇒δ =
A
L
EA
2
Slide No. 4
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Uniform Member
– The deflection (deformation),δ, of the
uniform member subjected to axial loading
P is given by
PL
δ=
EA
(1)
Slide No. 5
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Multiple Loads/Sizes
– The expression for the deflection of the
previous equation may be used only if the
rod or the member is homogeneous
(constant E) and has a uniform cross
sectional area A, and is loaded at its ends.
– If the member is loaded at other points, or
if it consists of several portions of various
cross sections, and materials, then
3
Slide No. 6
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Multiple Loads/Sizes
– It needs to be divided into components
which satisfy individually the required
conditions for application of the formula.
– Denoting respectively by Pi, Li, Ai, and Ei,
the internal force, length, cross-sectional
area, and modulus of elasticity
corresponding to component i,then
n
n
PL
δ = ∑δ i = ∑ i i
i =1
i =1 Ei Ai
Slide No. 7
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Multiple Loads/Sizes
n
n
i =1
i =1
δ = ∑δ i = ∑
E1
E2
E3
L1
L2
L3
Pi Li
Ei Ai
4
Slide No. 8
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
ENES 220 ©Assakkaf
Deformations of Members under
Axial Loading
„
Multiple Loads/Sizes
– The deformation of of various parts of a rod
or uniform member can be given by
n
n
i =1
i =1
δ = ∑δi = ∑
Pi Li
Ei Ai
(2)
E1
E2
E3
L1
L2
L3
Slide No. 9
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Example 4
– Determine the deformation of the steel rod
shown under the given loads. Assume that
the modulus of elasticity for all parts is
– 29×106 psi
2
Area = 0.9 in
45 kips
30 kips
75 kips
Area = 0.3 in2
12 in
12 in
16 in
5
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Slide No. 10
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Example 4 (cont’d)
– Analysis of internal forces
1
2
45 kips
75 kips
P3
3
30 kips
30 kips
→ + ∑ Fx = 0; − P3 + 30 = 0
∴ P3 = 30 kips
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Slide No. 11
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Example 4 (cont’d)
– Analysis of internal forces
1
2
45 kips
75 kips
P2
3
45 kips
30 kips
30 kips
→ + ∑ Fx = 0; − P2 − 45 + 30 = 0
∴ P2 = −15 kips
6
Slide No. 12
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Example 4 (cont’d)
– Analysis of internal forces
2
1
3
45 kips
75 kips
30 kips
45 kips
75 kips
P1
30 kips
→ + ∑ Fx = 0; − P1 + 75 − 45 + 30 = 0
∴ P1 = 60 kips
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Slide No. 13
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Example 4 (cont’d)
– Deflection
• Input parameters
L1 = 12 in
L2 = 12 in
L3 = 16 in
A1 = 0.9 in 2
A2 = 0.9 in 2
A3 = 0.3 in 2
From analysis of internal forces,
P1 = 60 kips = 60,000 lb
P2 = −15 kips = −15,000 lb
P3 = 30 kips = 30,000 lb
7
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Slide No. 14
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Example 4 (cont’d)
– Carrying the values into the deformation
formula:
Pi Li 1  P1 L1 P2 L2 P3 L3 

= 
+
+
E
A
E
A
A
A
i =1
i i
2
3 
 1
1  60,000(12) − 15,000(12) 30,000(16) 
+
+


29 × 106 
0.9
0.9
0.3

δ = 0.0759 in
3
δ =∑
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Slide No. 15
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Relative Deformation
A
δA
A
δ B/ A = δ B − δ A =
L
C
B
D
PL
AE
δB
P
8
Slide No. 16
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Relative Deformation
– If the load P is applied at B, each of the
three bars will deform.
– Since the bars AC and AD are attached to
the fixed supports at C and D, their
common deformation is measured by the
displacement δA at point A.
Slide No. 17
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Relative Deformation
– On the other hand, since both ends of bars
AB move, the deformation of AB is
measured by the difference between the
displacements δA and δB of points A and B.
– That is by relative displacement of B with
respect to A, or
δ B/ A = δ B − δ A =
PL
EA
(3)
9
Slide No. 18
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Example 5
The rigid bar BDE is supported by two
links AB and CD.
Link AB is made of aluminum (E = 70
GPa) and has a cross-sectional area of 500
mm2. Link CD is made of steel (E = 200
GPa) and has a cross-sectional area of (600
mm2).
For the 30-kN force shown, determine the
deflection a) of B, b) of D, and c) of E.
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Slide No. 19
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Example 5 (cont’d)
SOLUTION:
• Apply a free-body analysis to the bar
BDE to find the forces exerted by
links AB and DC.
• Evaluate the deformation of links AB
and DC or the displacements of B
and D.
• Work out the geometry to find the
deflection at E given the deflections
at B and D.
10
Slide No. 20
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
SOLUTION:
Example 5 (cont’d)
Free body: Bar BDE
∑MB = 0
0 = −(30 kN × 0.6 m ) + FCD × 0.2 m
FCD = +90 kN tension
∑ MD = 0
0 = −(30 kN × 0.4 m ) − FAB × 0.2 m
FAB = −60 kN compression
Slide No. 21
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Example 5 (cont’d)
Displacement of B:
δB =
=
PL
AE
(− 60×103 N)(0.3m)
(500×10-6 m2 )(70×109 Pa)
= −514×10−6 m
δ B = 0.514 mm↑
11
Slide No. 22
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
„
ENES 220 ©Assakkaf
Example 5 (cont’d)
Displacement of D:
δD =
=
PL
AE
(90 ×103 N)(0.4 m)
(600×10-6 m2 )(200×109 Pa )
= 300 ×10−6 m
δ D = 0.300 mm ↓
Slide No. 23
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Example 5 (cont’d)
Displacement of D:
BB′ BH
=
DD′ HD
0.514 mm (200 mm ) − x
=
x
0.300 mm
x = 73.7 mm
EE ′ HE
=
DD′ HD
δE
0.300 mm
=
(400 + 73.7 )mm
δ E = 1.928 mm
73.7 mm
δ E = 1.928 mm ↓
12
Slide No. 24
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Nonuniform Deformation
– For cases in which the axial force or the
cross-sectional area varies continuously
along the length of the bar, then Eq. 1
– (PL / EA) is not valid.
– Recall that in the case of variable cross
section, the strain depends on the position
of point Q, where it is computed from
ε=
dδ
dx
Slide No. 25
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Nonuniform Deformation
ε=
dδ
dx
L
x
P
13
Slide No. 26
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Nonuniform Deformation
– Solving for dδ and substituting for ε
dδ = εdx
– But ε = σ / E, and σ = P/A. therefore
dδ = εdx =
σ
E
dx =
P
dx
EA
(4)
Slide No. 27
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Nonuniform Deformation
– The total deformation δ of the rod or bar is
obtained by integrating Eq. 4 over the
length L as
L
δ =∫
0
Px
dx
EAx
(5)
14
Slide No. 28
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Example 6
– Determine the deflection of point a of a
homogeneous circular cone of height h,
density ρ, and modulus of elasticity E due
to its own weight.
a
h
Slide No. 29
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Example 6 (cont’d)
– Consider a slice of thickness dy
– P = weight of above slice
–
= ρg (volume above)
y
1 2 
δy
P = ρg  πr y 
3


1
ρg  πr 2 y 
Pdy
ρg
3
dδ =
ydy
=  2 =
EA
E (πr )
3E
r
h
15
Slide No. 30
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Example 6 (cont’d)
dδ =
h
δ =∫
ρg
3E
ρg
3E
0
δ=
ρgh
ydy
ydy =
ρg y 2
3E 2
h
0
2
6E
Slide No. 31
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Normal Stresses in Tapered Bar
– Consider the following tapered bar with a
thickness t that is constant along the entire
length of the bar.
h1
h(x)
h2
x
L
16
Slide No. 32
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Normal Stresses in Tapered Bar
h1
h(x)
x
h2
L
– The following relationship gives the height
h of tapered bar as a function of the
location x
hx = h1 + (h2 − h1 )
x
L
(6)
Slide No. 33
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Normal Stresses in Tapered Bar
h1
h(x)
h2
x
L
– The area Ax at any location x along the
length of the bar is given by
x

Ax = thx = t h1 + (h2 − h1 ) 
L

(7)
17
Slide No. 34
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Normal Stresses in Tapered Bar
h1
h2
h(x)
x
L
– The normal stress σx as a function of x is
given
σx =
P
P
=
x
Ax

t h1 + (h2 − h1 ) 
L

(8)
Slide No. 35
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Example 7
– Determine the normal stress as a function of x
along the length of the tapered bar shown if
– h1 = 2 in
– h2 = 6 in
– t = 3 in, and
– L = 36 in
x
h1
h2
h(x)
– P = 5,000 lb
L
18
Slide No. 36
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Example 7 (cont’d)
– Applying Eq. 8, the normal stress as a
function of x is given by
P
P
=
x
Ax

t h1 + (h2 − h1 ) 
L

5000
5000
σx =
=
x
x

32 + (6 − 2) )  6 +
3
36 

σx =
σx =
15000
18 + x
Slide No. 37
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Example 7 (cont’d)
x (in)
– Max σ = 833.3 psi
• At x =0
– Min σ = 277.8 psi
• At x =36 in
h1
h(x)
L
h2
x
0
3
6
9
12
15
18
21
24
27
30
33
36
σ (psi)
833.3
714.3
625.0
555.6
500.0
454.5
416.7
384.6
357.1
333.3
312.5
294.1
277.8
19
Slide No. 38
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Deflection of Tapered Bar
– Consider the following tapered bar with a
thickness t that is constant along the entire
length of the bar.
h1
dx
x
h2
h(x)
L
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Slide No. 39
Deformations of Members under
Axial Loading
ENES 220 ©Assakkaf
„
Deflection of Tapered Bar
– Recall Eq. 5
L
δ =∫
0
– Substitute Eq. 7
Px
dx
EAx
– into Eq. 5, therefore
L
PL
1
δ=
dx
∫
Et 0 h1 L + (h2 − h1 )x
(9)
20
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8)
Slide No. 40
ENES 220 ©Assakkaf
Deformations of Members under
Axial Loading
„
Deflection of Tapered Bar
Integrating Eq. 9, the deflection of a
tapered bar is given by
δ=
PL  1 

 ln[(h2 − h1 )L ] (10)
Et  h2 − h1 
21