Mechanics of Solids
Chapter 2
1/25
AXIALLY LOADED MEMBERS
2.1 INTRODUCTION
i.e., SOLID BARS (truss members, jet eng struts)
CABLES
We will consider structural elements that:
COIL SPRINGS
1. have straight longitudinal axes
2. carry only axial forces (Tension or Compression)
Will determine:
1. Maximum Stresses
2. Displacements
Structural Analysis = the calculation of stresses, strains, displacements and
load-carrying capacity.
Design = The determination of the geometric configuration and other characteristics of a
structure (i.e., material) in order to fulfill all the prescribed functions.
Designing a structure is usually a repetitive chore (iterative).
 Design-analyze, if it does not meet specifications or perform as intended,
 Redesign – analyze…….
Design is more difficult than analyzing.
2.2 DISPLACEMENTS OF AXIALLY LOADED MEMEBERS
L
Prismatic Bar:
δ
P
CONDITIONS
For Eqn ( 1 )
P
Prismatic Bar 
 
P
________________(a)
A



________________(b)
L
Homogeneous Material
  E _______________(c)
Linearly Elastic 
From (b) into (c):
 E
From (a) into (d):
Sign Convention:

L
____________________(d)
P E

A
L
+ = lengthening
- = shorthening
or

PL
_______________(1)
EA
(usually not needed)
Mechanics of Solids
Chapter 2
2/25
Analogous to an axially loaded spring:
Stiffness (spring constant): k 
Flexibility (compliance): f 
From (1):
P

__________________(e)
1 

_______________(f)
k P

PL
EA
P
EA
_________________(g)
L


FROM PHYSICS:
F = - k (Δx)
From (f) into (g):
k
EA
__________________(2)
L
f 
L
EA
(used a lot)
(seldom used)
2.3 MULTIPLE LOADS and / or MULTIPLE SECTIONS ( each one prismatic bar )
P
1
P
1
A
a
P
2
B
a
b
C
P
2
c
b
D
n
 
i 1
Pi Li
Ei Ai
_____________________(3)
Pi = axial force in section i of the bar
n = total No. of sections
section = (discontinuity) load change, geometry change, or material change
don’t cut section at a point where the discontinuity occurs because won’t know which
part to apply load or geometry or material to.
Mechanics of Solids
Chapter 2
3/25
CONTINUOUSLY VARYING AXIAL FORCE and / or CROSS-SECTIONAL AREA
Can not use above equation (eqn 3)
A
FBD
C
B
A
C
P(x)
dx
X
X
L
from eqn. (1):  
PL
EA
using a differential element:
d 
P( x) dx
E A( x)
Integrating:
L
L
0
0
   d  
P( x) dx
_________________________(4)
E A( x)
Based on the following assumption:
- stress distribution is uniform over the entire cross section ( based on σ = P/A ).
(not valid for tapered bars)
However, eqn (4) gives accurate results for small tapers.  < 20°
- Due to presence of E, material must be Linearly Elastic

Mechanics of Solids
Chapter 2
4/25
EXAMPLE No. 1
GIVEN:
The composite shown is made from 2 segments, AB and BD having cross-sectional areas
AAB = 1 in2 and ABD = 2 in2, respectively. It is subject to the loads shown.
Let E = 29 x 103 ksi.
15 kip
A
FIND:
Vertical displacement of end A.
4 kip
B
8 kip
8 kip
SOLn:
2 ft
4 kip
1.5 ft
C
D
1 ft
Mechanics of Solids
Chapter 2
5/25
EXAMPLE No. 2
GIVEN:
A member is made from a material that has a specific weight γ and modulus of elasticity
E. It is formed into the cone shown and suspended in the vertical position.
y
FIND:
r
0
The displacement of the end due to gravity
SOLn:
L
x
Mechanics of Solids
Chapter 2
6/25
2.4 STATICALLY INDETERMINATE STRUCTURES
Statically determinate = a structure for which the internal forces can be found using
equations of equilibrium.
Statically Indeterminate = a structure for which the equations of equilibrium are not
sufficient to find the internal forces.
A
A
P
P
B
B
STATICALLY DETERMINATE
STATICALLY INDETERMINATE
   Fy  RB  P  0
   Fy  RB  P  RA  0 ___( 1 )
1 eqn / 1 unkwn
1 eqn / 2 unkwns
2 methods for analyzing statically indeterminate members:
1. Flexibility Method
2. Stiffness Method
FLEXIBILITY METHOD (Force Method…..forces are unknown quantities)
RA
FBD:
A
P
a
P
RA + RB = P
L
C
b
B
RB
Let RA = statically redundant force – remove it.
From eqn ( 1 ):
(2)
Mechanics of Solids
Chapter 2
7/25
A
This is the RELEASED or PRIMARY structure.
P
Now, consider the effect of P on the displacement of A.
C
b
p 
B
Pb
EA
displacement is 
Next, consider the effect of RA
RA
A
R 
RA L
EA
displacement is 
B
Final displacement of A is:
δA = δP - δR
+  ( + is in direction of load)
We know: δA = 0, therefore;
δR = δP

Compatibility Eqn
R A L Pb

EA EA
RA 
Pb
__________ ( 3a )
L
Solving for RB in eqn ( 2 ):
RB = P – RA
Substituting from eqn ( 3 ):
RB  P 
Pb
L
RB 
PL  Pb
P ( L  b)

L
L
RB 
Pa
_____________ ( 3b )
L
NOTE: a  L  b
Mechanics of Solids
Chapter 2
8/25
FLEXIBILITY METHOD SUMMARY:
1. select 1 unknown reaction as redundant and release it from the structure.
2. calculate displacement due to the actual load.
3. calculate displacement due to the redundant force.
4. combine the 2 displacements into an equation of compatibility ( δR = δP ).
5. substitute expressions for displacements and solve for redundant force.
6. solve for the remaining unknown force using equilibrium eqn.
called Flexibility Method because eqn of compatibility contains:
f 
L
EA
Also called, FORCE METHOD….forces are unknown quantities.
RESTRICTIONS: material must behave in a linearly elastic manner.
Flexibility Method is based on:
1. eqns of static equilibrium
2. compatibility eqns…..will always have as many compatibility eqns. as redundants
 will always have sufficient no. of eqns to solve for all unknowns
STIFFNESS METHOD (Displacement Method…displacements are unknown quantities)
RA
A
P
a
P
L
C
δC is taken as the unknown.
C
b
B
RB
Assume C is moving downward:
 AC elongates
R a
C  A
EA
RA 
EA
 C ________ ( 4a )
a
 BC shortens
R b
C  B
EA
RB 
EA
 C _________ ( 4b )
b
Mechanics of Solids
Chapter 2
Isolate C as freebody:
9/25
+Fy = RA + RB – P = 0
RA
RA + RB = P
P
C
Substituting from eqns 4a & 4b:
RB
EA
EA
C 
C  P
a
b
Stiffness Terms k 
Thus,
C 
Pab
Pab

EA(a  b) EAL
NOTE: L = a + b
Substituting this back into eqns 4a & 4b:
RA 
EA  Pab 


a  EAL 
RB 
EA  Pab 


b  EAL 
RA 
Pb
L
RB 
Pa
L
STIFFNESS METHOD SUMMARY:
1. select suitable displacement as unknown quantity
2. express forces in terms of this displacement
3. substitute into the equilibrium eqn
4. solve for displacement
5. solve for forces
2.5 TEMPERATURE & PRE-STRAIN EFFECTS
εT =
UNIFORM THERMAL STRAIN
 = Coefficient of Thermal Expansion
- property of material
- values for various materials from Table H – 4 (text appendix H, pg 915)
 T   T  _________________ ( 5 )
units of  :
SI
1
(Kelvin) or
k
1
C

sign convention: + = expansion
– = contraction
US
1
F

EA
L
Mechanics of Solids
Chapter 2
10/25
Changes in dimensions, such as length:
 T   T L   T L ___________ ( 6 )
- usually only length of structural members are considered.
Transverse direction is usually neglected.
OMIT
statically determinant – no thermal stress because member is free to
expand or contract ( uniform ΔT )
READ TEXT
Pg 93 - 100
statically indeterminate – thermal stresses
non-uniform ΔT also produces thermal stresses.
Consider a statically indeterminate bar:
R
Remove upper support
FBD
δT
ΔT
L
 T   T L
R
Applying R
R 
R
RL
EA
δT – δR = 0
 T L 
RL
0
EA
R = (ΔT)EA
STRESS:

R
  T E
A
(compression)
read about PRE-STRAIN {important for pre-loading in bolts}
pg 101 – 104
analysis is essentially the same as that of temperature change.
Mechanics of Solids
Chapter 2
11/25
EXAMPLE No 1
GIVEN:
The aluminum post shown is reinforced with a brass core. It supports a resultant
compressive axial load of P = 9 kip. EAL = 10x103 ksi EBR = 15 x 103 ksi
9 kip
P
FBD
A
A
1.5’
FBR
r = 1”
r = 2”
FAL
y
FIND:
Average normal stress in the aluminum and brass via the Flexibility Method.
SOLn:
Mechanics of Solids
Chapter 2
EXAMPLE No. 2
GIVEN:
Same as example No. 1
FIND:
Average normal stress in the aluminum and brass via the Stiffness Method.
SOLn:
12/25
Mechanics of Solids
Chapter 2
13/25
OMIT
EXAMPLE No. 3
GIVEN:
The rigid bar shown below is fixed to the top of the 3 posts made of steel and aluminum.
The posts have a length of 250 mm when no load is applied to the bar and the
temperature is 20°C. The bar is subjected to a uniform distributed load of 150 kN/m and
the temperature is raised to 80°C.
600 mm
STEEL: Est = 200(109) Pa
ALUM: Eal = 70(109) Pa
αst = 12(10-6)/ °C
αal = 23(10-6)/ °C
FIND:
The force supported by each post.
60 mm
40 mm
SOLn:
150 kN/m
st
st
al
40 mm
250 mm
Mechanics of Solids
Chapter 2
14/25
2.6 STRESSES ON INCLINED SURFACES
Axially loaded prismatic bar
y
m
P
P
x
O
n
z
y
m
P
O
x 
x
P
A
( 1 ) normal stress on m n
y
n
z
x
x
x
In 2-D:
y
z
O
m
P
x
x 
O
n
P
A
There are other internal forces acting on the bar.
y
p

P
P
x 
x
P
_________ ( 1 )
A
q
IMPORTANT:  is w.r.t. direction of load
And outward normal !!
//
┴
N

P
P sin 

P cos 
V
N = Normal Force (force normal to surface)
V = Shearing Force (force parallel to surface)
Mechanics of Solids
Chapter 2
+
15/25
F┴ = 0
F// = 0
+
P sin  – V = 0 __________ ( 2b )
N – P cos  = 0 ________ ( 2a )
These forces produce stresses:
//
P
A
σ = normal stress ( ┴ to surface)
┴
σ
τ = shearing stress ( // to cut surface)
τ
A = cross sectional area of cut surface
SIGN CONVENTION:
σ = + when in tension
τ = + when it produces a CCW rotation of material (matches moment sign)
A
 

A = A cos 
A
A 
N
A
N    A __________ ( 4a )
A
________________( 3 )
cos 
  
V
A
V    A __________ ( 4b )
FROM eqns ( 2 ) & ( 4 ):
  A  P cos  0
P sin     A  0
  A  P cos 
  A   P sin 
FROM eqn ( 3 ):

A
 P cos 
cos 
 
P
cos 2 
A
FROM eqn ( 1 ):
    x cos 2  ________ ( 5a )

A
  P sin 
cos 
P
sin  cos 
A
    x sin  cos 
  
ASIDE:
sin  cos   12 sin( 2 )
    12  x sin( 2 ) _________ ( 5b )
Mechanics of Solids
Chapter 2
16/25
Eqns 5 gives the normal and shear stress at an angle  w.r.t. the axial load
in terms of x which is the normal stress in the direction of the axial load.
These eqns (5a & b) are only good for axial loading. However, they were derived based
only on the equilibrium eqns so they are valid for any material whether it behaves
linearly or nonlinearly, elastically or nonelastically.
FROM eqn ( 5 ): cos  = max when  = 0°
sin (2) = max when 2 = 90°
 = 45°
 max   x @  0
THEREFORE:
 max  
x
2
@  45
 Maximum Normal Stress
 Maximum Shear Stress
STRESS ELEMENT
45°
P
y
σx
σx
P
σx
σx
2
 = 45°
 = 0°
σx
2
σ
τmax = x
2
τmax =
σx
x
σx
2
σx
2
y
x
τmax = σx
2
τmax = σx
2
σx
2
FROM eqn ( 5 ):
(cos 0°)2 = 1  σ = σx = σmax
sin ( 2 x 0° ) = 0
 τ = 0
(cos 45° )2 = ½
sin (2 x 45°) = 1
 σ = ½ σ x
 τ = - ½ σx = τmax
Mechanics of Solids
Chapter 2
17/25
For   0° or   45°
{ for angles of  other than 0° and 45° }
σ2
σ1
τ
y
x
τ
τ
τ
σ2
σ1
Eqns ( 5a ) and ( 5b ):
1
    x cos 2 1
    12  x sin( 21 )
1
For angles other than 0° and 45° must consider the complimentary angle, 2 = 90° + 1
to obtain:
 2   x cos 2  2 _________________ ( 5c )
ALTERNATTE STRESS ELEMENT (found in some texts)
τ
x



- advantage: gives original stresses , τ
- disadvantage: doesn’t give stress on other planes
τxy
τyx
y
- homework should include stress element.
- stress element is just as important as a FBD
> gives stress at a point in the structure
Once you know the magnitude and direction of 1 shear stress, you have it for the others
on the element because they are equal but opposite…..keeps element in equilibrium. This
is not true for the normal stress.
Mechanics of Solids
Chapter 2
18/25
2.7 STRAIN ENERGY
OMIT
READ: pg 116 - 122
P is a static load. This Load is applied slowly
(gradually) from P = 0 to P = Pmax
δ
L
Thus, δ = 0 to δ = δmax
P
LOAD DISPLACEMENT DIAGRAM
P
Work = area under curve
dP1

W   P1 d 1
0
Pmax
P1
δ
δ1
dδ1
δ
Strain Energy (internal work) = the energy absorbed by the bar during the
loading process.

U = strain energy
U  W   P1 d 1 _____________ ( 1 )
0
P1 = any value of load between 0 and Pmax
UNITS: SI
1J = 1Nm
US ft lbs
Mechanics of Solids
Chapter 2
19/25
Now, we unload the bar:
P
Elastic limit (pt A) exceeded
B
A
Permanent deformation OD remains
LOADING
UNLOADING
δ
O
D
C
Elastic strain energy – recoverable energy
Inelastic strain energy – lost energy
Now, assume we stay below the elastic limit load
P
A
Material Conditions from O to A:
1. Elastic
2. Follows Hooke’s Law
LOADING
P
U
FROM eqn ( 1 ):
UNLOADING
B
O
δ
U  12 P _________ ( 2 )
δ
RECALL: for a prismatic bar:
FROM eqn ( 2 ):  

PL
__________ ( a )
EA
2U
_______________ ( b )
P
FROM eqns ( a ) & ( b ):
2U PL

P
EA
P2L
U
____________ ( 3a ) in terms of load
2 EA
Mechanics of Solids
Chapter 2
20/25
FROM eqn ( a ):
P
EA
________________ ( c )
L
FROM ( c ) & ( 3 ):
 EA 

 L
L 

U
2 EA
2
U
EA 2
_______________ ( 3b ) in terms of
2L
k
EA
______________ ( d )
L
deformation
LET: k = spring stiffness
FROM ( d ) & ( 3a ):
FROM ( d ) & ( 3b ):
U
2
P
2k
U
k 2
2
For a bar with changes in cross-section:
Pa
Pb
Pi = axial force in part (section) i of the bar.
n = number of parts (sections)
based on eqn ( 3a ):
n
U 
i 1
Pi 2 Li
2 Ei Ai
Mechanics of Solids
Chapter 2
21/25
Nonprismatic bar with varying axial force:
based on eqn ( 3a ):
P(x)
2

Px 
U 
dx
2 E Ax 
0
L
x
dx
L
Strain energy density = strain energy per unit volume
u
volume: V = AL
RECALL:
 
P
 E
A

U
U

V
AL
(strain energy density)

   L
L
FROM eqn ( 3a ):
u
 P2  1
P2L
P2
2





2 EA( AL) 2EA2  A 2  2 E 2E

u

u
2
2E
FROM eqn ( 3b ):
u
EA 2
E 2 E (L) 2 E 2



2L( AL) 2L2
2
2 L2
E 2
2
Mechanics of Solids
Chapter 2
22/25
10 mm
EXAMPLE No. 1
600 N
GIVEN: An axial force of 600 N acts on the steel bar.
150 mm
600 N
3200 mm
mm
FIND: The stresses on all faces of a stress element for  = - 30°.
SOLn:
Mechanics of Solids
Chapter 2
EXAMPLE No. 2
23/25
OMIT
Pass out SOLn
GIVEN: A conical bar of diameter d at the support and length L hangs vertically under
its own weight. LET: γ = specific weight
d
E = modulus of elasticity
DERIVE: Formula for the strain energy, U, of the bar.
SOLn:
L
Mechanics of Solids
Chapter 2
24/25
2.10 STRESS CONCENTRATIONS
Changes in geometry cause stress concentrations:  
concentrations.
P
P
changes in AREA cause stress
A
P
Which can sustain greater load?
VS
For a prismatic bar, we know that:  
P
A
But what about the following?
c
P
/2
P
d
c
/2
STRESS DISTRIBUTION:
σmax
P
Stress Concentrations = highly localized stresses caused by changes in geometry. The
more abrupt the change, the higher the stress concentration.
Stress Raisers = the geometric features that cause stress concentrations
Stress Concentration Factor = K
K
 max
 nom
 nom 
P
P

A
c
2 t 
2

 nom 
for flat bar, above
K, stress concentration factor is equivalent to
an additional safety factor.
P
ct
(average stress)
t = thickness
ct = net area
RECALL: Factor of Safety
n
y
 allow
Text pg 141 & 142 gives the stress concentration factor, K, for common geometries.
Mark’s Engineering Hndbk and other engineering references books have stress
concentration factors
Mechanics of Solids
Chapter 2
25/25
EXAMPLE No. 1
GIVEN:
The steel strap shown below is subject to an axial load of 80kN.
σy = 700 x 106 Pa
E = 200 x 109 Pa
20 mm
80 kN
80 kN
40 mm
6 mm
300 mm
800 mm
300 mm
FIND:
a.) Maximum normal stress developed in the strap
b.) Displacement of one end of the strap w.r.t. the other end.
SOLn:
10 mm