1. Third Edition
LECTURE
RODS: AXIAL LOADING AND
DEFORMATION
• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering
3 by
Dr. Ibrahim A. Assakkaf
SPRING 2003
Chapter ENES 220 – Mechanics of Materials
2.8 Department of Civil and Environmental Engineering
University of Maryland, College Park
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 1
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
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

2. LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 2
ENES 220 ©Assakkaf
Deformations of Members under
Axial Loading
Uniform Member
L
δ
P
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 3
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
Uniform Member
– From Hooke’s law, it follows that
σ = Eε (Hooke' s law)
P
= Eε
A
δ
But ε = , therefore,
L
P δ PL
= E ⇒δ =
A L EA

3. LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 4
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
Uniform Member
– The deflection (deformation),δ, of the
uniform member subjected to axial loading
P is given by
PL
δ= (1)
EA
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 5
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
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

4. LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 6
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
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
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 7
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
Multiple Loads/Sizes n n
Pi Li
δ = ∑δ i = ∑
i =1 i =1 Ei Ai
E1 E2 E3
L1 L2 L3

5. LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 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
Pi Li
δ = ∑δi = ∑ (2)
i =1 i =1 Ei Ai
E1 E2 E3
L1 L2 L3
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 9
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
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

6. LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 10
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
Example 4 (cont’d)
– Analysis of internal forces
1 2 3
45 kips
30 kips
75 kips
30 kips
P3
→ + ∑ Fx = 0; − P3 + 30 = 0
∴ P3 = 30 kips
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 11
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
Example 4 (cont’d)
– Analysis of internal forces
1 2 3
45 kips
30 kips
75 kips
P2 45 kips 30 kips
→ + ∑ Fx = 0; − P2 − 45 + 30 = 0
∴ P2 = −15 kips

7. LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 12
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
Example 4 (cont’d)
– Analysis of internal forces
1 2 3
45 kips
30 kips
75 kips
45 kips
P1 30 kips
75 kips
→ + ∑ Fx = 0; − P + 75 − 45 + 30 = 0
1
∴ P = 60 kips
1
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 13
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
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,
P = 60 kips = 60,000 lb
1
P2 = −15 kips = −15,000 lb
P3 = 30 kips = 30,000 lb

8. LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 14
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
Example 4 (cont’d)
– Carrying the values into the deformation
formula:
3
Pi Li 1  P L1 P2 L2 P3 L3 
δ =∑ =  1 + + 
i =1 Ei Ai E  A1
 A2 A3 

1  60,000(12) − 15,000(12) 30,000(16) 
 + + 
29 × 106  0.9 0.9 0.3 
δ = 0.0759 in
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 15
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
Relative Deformation
A δA A
PL
δ B/ A = δ B − δ A =
AE
L
C D δB
B
P

9. LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 16
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
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.
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 17
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
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
PL (3)
δ B/ A = δ B − δ A =
EA

10. LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 18
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
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
ENES 220 ©Assakkaf
Axial Loading
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.

11. LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 20
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
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
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 21
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
Example 5 (cont’d)
Displacement of B:
PL
δB =
AE
(− 60×103 N)(0.3m)
=
(500×10-6 m2 )(70×109 Pa)
= −514×10−6 m
δ B = 0.514 mm↑

12. LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 22
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
Example 5 (cont’d)
Displacement of D:
PL
δD =
AE
(90 ×103 N)(0.4 m)
=
(600×10-6 m2 )(200×109 Pa )
= 300 ×10−6 m
δ D = 0.300 mm ↓
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 23
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
Example 5 (cont’d)
Displacement of D:
BB′ BH
=
DD′ HD
0.514 mm (200 mm ) − x
=
0.300 mm x
x = 73.7 mm
EE ′ HE
=
DD′ HD
δE
=
(400 + 73.7 )mm
0.300 mm 73.7 mm
δ E = 1.928 mm
δ E = 1.928 mm ↓

13. LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 24
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
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
dδ
ε=
dx
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 25
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
Nonuniform Deformation
dδ
ε= L
dx x
P

14. LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 26
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
Nonuniform Deformation
– Solving for dδ and substituting for ε
dδ = εdx
– But ε = σ / E, and σ = P/A. therefore
σ P
(4)
dδ = εdx = dx = dx
E EA
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 27
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
Nonuniform Deformation
– The total deformation δ of the rod or bar is
obtained by integrating Eq. 4 over the
length L as
L
Px
δ =∫ dx (5)
0
EAx

15. LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 28
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
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
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 29
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
Example 6 (cont’d)
– Consider a slice of thickness dy
– P = weight of above slice
– = ρg (volume above)
y h
1 2  δy r
P = ρg  πr y 
3 
1 
ρg  πr 2 y 
Pdy 3 ρg
dδ = =  2 = ydy
EA E (πr ) 3E

16. LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 30
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
Example 6 (cont’d)
ρg
dδ = ydy
3E
h
h
ρg ρg y 2
δ =∫ ydy =
0
3E 3E 2 0
ρgh 2
δ=
6E
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 31
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
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 h2 x
h(x)
L

17. LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 32
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
Normal Stresses in Tapered Bar
h1 h2 x
h(x)
L
– The following relationship gives the height
h of tapered bar as a function of the
location x
x
hx = h1 + (h2 − h1 ) (6)
L
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 33
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
Normal Stresses in Tapered Bar
h1 h2 x
h(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 )  (7)
 L

18. LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 34
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
Normal Stresses in Tapered Bar
h1 h2 x
h(x)
L
– The normal stress σx as a function of x is
given
P P
σx = =
Ax  x (8)
t h1 + (h2 − h1 ) 
 L
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 35
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
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 h(x) h2
– P = 5,000 lb
L

19. LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 36
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
Example 7 (cont’d)
– Applying Eq. 8, the normal stress as a
function of x is given by
P P
σx = =
Ax  x
t h1 + (h2 − h1 ) 
 L
5000 5000
σx = =
 x x
32 + (6 − 2) )  6 +
 36  3
15000
σx =
18 + x
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 37
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
Example 7 (cont’d) x (in) σ (psi)
– Max σ = 833.3 psi 0 833.3
3 714.3
• At x =0 6 625.0
9 555.6
– Min σ = 277.8 psi 12 500.0
• At x =36 in 15 454.5
18 416.7
21 384.6
24 357.1
h1 h2 x
h(x) 27 333.3
30 312.5
33 294.1
L 36 277.8

20. LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 38
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
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 h2 x
h(x)
L
LECTURE 3. RODS: AXIAL LOADING AND DEFORMATION (2.8) Slide No. 39
Deformations of Members under
ENES 220 ©Assakkaf
Axial Loading
Deflection of Tapered Bar
– Recall Eq. 5 L
Px
δ =∫ dx
0
EAx
– Substitute Eq. 7
– into Eq. 5, therefore
L
PL 1
δ= ∫ h1L + (h2 − h1 )x dx
Et 0
(9)

21. 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
PL  1 
δ=  h − h  ln[(h2 − h1 )L ] (10)

Et  2 1  