Giáo trình Strength of Materials - Chapter 3: Axial forces

CHAPTER 3: AXIAL FORCES 
3.1 Concept, Internal Forces 
3.2 Normal Strain and Normal Stress 
3.3 Stress – Strain diagram 
3.4 Hooke’s law: Modulus of Elasticity 
3.5 Deformation under axial loading 
3.6 Safety Factor, Allowable Stress 
3.7 Statically Indeterminate Systems 
3.1 CONCEPT, INTERNAL FORCES 
• Suitability of a structure or machine may depend on the deformations in 
the structure as well as the stresses induced under loading. Statics 
analyses alone are not sufficient. 
• Considering structures as deformable allows determination of member 
forces and reactions which are statically indeterminate. 
• Determination of the stress distribution within a member also requires 
consideration of deformations in the member. 
• Chapter 3 is concerned with deformation of a structural member under 
axial loading. Later chapters will deal with torsional and pure bending 
loads. 
3.1 CONCEPT, INTERNAL FORCES 
A structural member under axial loading means: 
Internal forces: only axial force N (along longitudinal axis of the bar) 
Method of Section: the same as discussed in previous chapter 
Imaginary cut is made through the body in the region where the internal 
loading is to be determined. 
The two parts are separated and a free body diagram of one of the parts is 
drawn. Only then applying equilibrium would enable us to relate the 
resultant internal force and moment to the external forces. 
3.1 CONCEPT, INTERNAL FORCES 
WHAT DOES IT LOOK LIKE? 
3.1 CONCEPT, INTERNAL FORCES 
WHAT DOES IT LOOK LIKE? 
3.1 CONCEPT, INTERNAL FORCES 
HOW CAN WE KNOW THAT IT IS A PURELY AXIAL LOADING 
PROBLEM? 
(1) Structure member is subjected to axial (longitudinal direction) forces only. For example: 
OR 
(2) Structure members are connected together by hinges (so-called nodes) and there is no 
load on the member, just have loads at nodes. For example: 
3.1 CONCEPT, INTERNAL FORCES 
EXAMPLES: DERTEMINE INTERNAL FORCES OF STRUCTURES 
BELOWS? 
(1) 
(2) 
3.2 NORMAL STRAIN AND NORMAL STRESS 
strain normal
stress


L
A
P



L
A
P
A
P





2
2
LL
A
P





2
2
a force P produces a deformation . In engineering, we usually transform 
this force into stress and the deformation into strain and we define these as 
follows: 
3.2 NORMAL STRAIN AND NORMAL STRESS 
Strain has no unit’s since it is a ratio of length to length. Most engineering 
materials do not stretch very mush before they become damages, so strain 
values are very small figures. It is quite normal to change small numbers in to 
the exponent for 10-6( micro strain). 
Sign convention of normal stresses: 
 x > 0 : tensile stress 
 x < 0 : compression stress 
3.2 NORMAL STRAIN AND NORMAL STRESS 
Normal stress: at a section far from the bar ends, the normal stress 
can be regarded uniform, or considered as average value 
Stress has units of force per unite area (kN/m2; N/m2) 
Note: Most of engineering fields used kPa, MPa, GPa. 
3.2 NORMAL STRAIN AND NORMAL STRESS 
3.3 STRESS – STRAIN DIAGRAM 
3.3 STRESS – STRAIN DIAGRAM 
Ductile Materials 
3.3 STRESS – STRAIN DIAGRAM 
Brittle Materials 
3.3 STRESS – STRAIN DIAGRAM 
Stress Strain Diagram 
3.3 STRESS – STRAIN DIAGRAM 
  Elastic domain: OA 
  linear stress–strain relationship 
  point A: proportional limit 
  Const slope: Modulus of Elasticity, E 
  Transition domain: AB  B: Elastic Limit (A  B) 
  Region of Plasticity: BC 
  Deformation develops under constant yield stress, 
  C  end of the region of plasticity 
  Strain-hardening & necking regions: 
  Nonlinear Stress-Strain Relationship 
  Point D:  the ultimate stress, 
  Point E:  Fracture of the specimen 
3.4 HOOKE’S LAW: MODULUS OF ELASTICITY 
• Below the yield stress 
Elasticity of Modulus 
or Modulus Youngs

E
E
• Strength is affected by alloying, 
heat treating, and manufacturing 
process but stiffness (Modulus of 
Elasticity) is not. 
3.4 HOOKE’S LAW: MODULUS OF ELASTICITY 
• For a slender bar subjected to axial loading: 
0 zy
x
x
E



• The elongation in the x-direction is 
accompanied by a contraction in the other 
directions. Assuming that the material is 
isotropic (no directional dependence), 
0 zy 
• Poisson’s ratio is defined as 
x
z
x
y




 
strain axial
strain lateral
POISSON RATIO 
- Poisson’s ratio, determined by experiment; values from 
0.2 – 0.5 
3.5 DEFORMATION UNDER AXIAL LOAD 
POISSON RATIO 
AE
P
E
E 


• 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

3.5 DEFORMATION UNDER AXIAL LOAD 
Determine the deformation of 
the steel rod shown under the 
given loads. 
in. 618.0 in. 07.1
psi1029 6

 
dD
E
SOLUTION: 
• 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. 
Example 3.01 
3.5 DEFORMATION UNDER AXIAL LOAD 
SOLUTION: 
• Divide the rod into three 
components: 
2
21
21
in 9.0
in. 12


AA
LL
2
3
3
in 3.0
in. 16


A
L
• Apply free-body analysis to each 
component to determine internal forces, 
lb1030
lb1015
lb1060
3
3
3
2
3
1



P
P
P
• Evaluate total deflection, 
     
in.109.75
3.0
161030
9.0
121015
9.0
121060
1029
1
1
3
333
6
3
33
2
22
1
11








 













A
LP
A
LP
A
LP
EEA
LP
i ii
ii
 in. 109.75 3
3.5 DEFORMATION UNDER AXIAL LOAD 
Example 3.02 
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. 
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. 
3.5 DEFORMATION UNDER AXIAL LOAD 
Displacement of B: 
  
  
m10514
Pa1070m10500
m3.0N1060
6
926-
3





AE
PL
B
 mm 514.0B
Displacement of D: 
  
  
m10300
Pa10200m10600
m4.0N1090
6
926-
3





AE
PL
D
 mm 300.0D
Free body: Bar BDE 
 
 
ncompressioF
F
tensionF
F
M
AB
AB
CD
CD
B
 kN60
m2.0m4.0kN300
0M
 kN90
m2.0m6.0kN300
0
D








SOLUTION: 
3.5 DEFORMATION UNDER AXIAL LOAD 
Displacement of D: 
 
mm 7.73
mm 200
mm 0.300
mm 514.0






x
x
x
HD
BH
DD
BB
 mm 928.1E
 
mm 928.1
mm 7.73
mm7.73400
mm 300.0






E
E
HD
HE
DD
EE


3.5 DEFORMATION UNDER AXIAL LOAD 
Example 3.03 
3.6 SAFETY FACTOR – ALLOWABLE STRESS 
ALLOWABLE LOAD / ALLOWABLE STRESS 
 Max load that a structural member/machine component will be allowed to 
carry under normal conditions of utilization is considerably smaller than the 
ultimate load 
 This smaller load = Allowable load / Working load / Design load 
 Only a fraction of ultimate load capacity of the member is utilised when 
allowable load is applied 
 The remaining portion of the load-carrying capacity of the member is kept in 
reserve to assure its safe performance 
 The ratio of the ultimate load/allowable load is used to define FACTOR OF 
SAFETY 
 FACTOR OF SAFETY = ULTIMATE LOAD/ALLOWABLE LOAD 
 FACTOR OF SAFETY = ULTIMATE STRESS/ALLOWABLE STRESS 
3.6 SAFETY FACTOR – ALLOWABLE STRESS 
SELECTION OF F.S. 
1. Variations that may occur in the properties of the member under 
considerations 
2. The number of loadings that may be expected during the life of the structure 
/machine 
3. Types of loadings that are planned for in the design, or that may occur in the 
future 
4. Types of failures that may occur 
5. Uncertainty due to the methods of analysis 
6. Deterioration that may occur in the future because of poor maintenance / 
because of unpreventable natural causes 
7. The importance of a given member to the integrity of the whole structure 
3.6 SAFETY FACTOR – ALLOWABLE STRESS 
• Factor of Safety: n 
• Allowable Stresses, 
 Ductile Materials: 
 Brittle Materials: 
actual strength
Factor of safety n
required strength
  1
 allow or ( )
y
allow
n

 

  ultallow
n
3.6 SAFETY FACTOR – ALLOWABLE STRESS 
Saint-Venant’s Principle 
• Loads transmitted through rigid 
plates result in uniform distribution 
of stress and strain. 
• Saint-Venant’s Principle: 
Stress distribution may be assumed 
independent of the mode of load 
application except in the immediate 
vicinity of load application points. 
• Stress and strain distributions 
become uniform at a relatively short 
distance from the load application 
points. 
• Concentrated loads result in large 
stresses in the vicinity of the load 
application point. 
3.7 STATICALLY INDETERMINACY 
• Structures for which internal forces and reactions 
cannot be determined from statics alone are said 
to be statically indeterminate. 
0 RL 
• Deformations due to actual loads and redundant 
reactions are determined separately and then added 
or superposed. 
• Redundant reactions are replaced with 
unknown loads which along with the other 
loads must produce compatible deformations. 
• A structure will be statically indeterminate 
whenever it is held by more supports than are 
required to maintain its equilibrium. 
3.7 STATICALLY INDETERMINACY 
Determine the reactions at A and B for the steel 
bar and loading shown, assuming a close fit at 
both supports before the loads are applied. 
• Solve for the reaction at A due to applied loads 
and the reaction found at B. 
• Require that the displacements due to the loads 
and due to the redundant reaction be compatible, 
i.e., require that their sum be zero. 
• Solve for the displacement at B due to the 
redundant reaction at B. 
SOLUTION: 
• Consider the reaction at B as redundant, release 
the bar from that support, and solve for the 
displacement at B due to the applied loads. 
Example 3.04 
3.7 STATICALLY INDETERMINACY 
SOLUTION: 
• Solve for the displacement at B due to the applied 
loads with the redundant constraint released, 
EEA
LP
LLLL
AAAA
PPPP
i ii
ii
9
L
4321
26
43
26
21
3
4
3
321
10125.1
m 150.0
m10250m10400
N10900N106000







• Solve for the displacement at B due to the redundant 
constraint, 
 







i
B
ii
ii
R
B
E
R
EA
LP
δ
LL
AA
RPP
3
21
26
2
26
1
21
1095.1
m 300.0
m10250m10400
3.7 STATICALLY INDETERMINACY 
• Require that the displacements due to the loads and due to 
the redundant reaction be compatible, 
 
kN 577N10577
0
1095.110125.1
0
3
39







B
B
RL
R
E
R
E


• Find the reaction at A due to the loads and the reaction at B 
kN323
kN577kN600kN 3000

 
A
Ay
R
RF
kN577
kN323


B
A
R
R
3.7 STATICALLY INDETERMINACY 
• MORE EXAMPLES - PRACTICES