Stress
Definition
Stress is force per unit area, or the distribution of
a force over the area in which it acts.
Stress = Force / area
Units of Stress
Stress = Force /Aera
International System of units (SI)
Stress is expressed in (N/m2). This unit is called Pascal
(1Pa = 1 N/m2), is small, therefore stress is normally
expressed in kilopascals (kPa) or megapascals (MPa)
Note: 1 MPa = 106 Pa = 106 N/m2 or
1 MPa = 1 N/mm2
1 GPa = 109 Pa = 109 N/m2 or
1GPa = 1000 MPa
U.S.Customary System of units(USCS)
Stress is expressed in pounds per square inch
(psi) or in kilopounds per square inch (ksi)
Note: 1 psi = 6895 Pa = 6.895 kPa
Type of stress
• Normal stress
• Shear stress
Normal Stress
Average Normal stress = Normal force / area on which the force acts
The symbol used for normal force is the lower case Greek letter sigma (σ)
σavg = F /A
READING QUIZ
1.
What is the normal stress in the bar if P=10
kN and 500mm²?
a)
0.02 kPa
b)
20 Pa
c)
20 kPa
d)
200 N/mm²
e)
20 MPa
Copyright © 2011 Pearson Education South Asia Pte Ltd
EXAMPLE 1
The bar has a constant width of 35 mm and a thickness of 10
mm. Determine the maximum average normal stress in the
bar when it is subjected to the loading shown.
Copyright © 2011 Pearson Education South Asia Pte Ltd
EXAMPLE 1 (cont)
Solutions
•
By inspection, different sections have different internal forces.
• Graphically, the normal force diagram is as shown.
Copyright © 2011 Pearson Education South Asia Pte Ltd
EXAMPLE 1 (cont)
Solutions
•
By inspection, the largest loading is in region BC,
PBC  30 kN
•
Since the cross-sectional area of the bar is constant, the largest
average normal stress is
 BC
 
PBC
30 103


 85.7 MPa (Ans)
A 0.0350.01
Copyright © 2011 Pearson Education South Asia Pte Ltd
CONCEPT QUIZ
1) The thrust bearing is subjected to the loads
as shown. Determine the order of average
normal stress developed on cross section
through BC and D.
a)
C>B>D
b)
C>D>B
c)
B>C>D
d)
D>B>C
Copyright © 2011 Pearson Education South Asia Pte Ltd
Example 2
Two steel wires, AB and BC, support a lamp weighing 18 lb. Both wires
have diameter 30 mils (one mil equals 0.001 in)
Determine the tensile stresses in each wireβ = 48º and α= 34º
Solution of Example 2
Factor of Safety
Structural members or machines
must be designed such that the
working stresses are less than the
ultimate strength of the material.
FS  Factor of safety
FS 
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u
ultimate stress

 all allowable stress
Factor of safety considerations:
uncertainty in material properties
uncertainty of loadings
uncertainty of analyses
number of loading cycles
types of failure
maintenance requirements and
deterioration effects
importance of member to integrity of
whole structure
risk to life and property
influence on machine function
Example 3
R.C. Hibbler
Solution Example 3
Shear Stress
Average Shear stress = Shear force/Area on which shear
force acts
Units of Stress
Stress = Force /Aera
International System of units (SI)
Stress is expressed in (N/m2). This unit is called Pascal
(1Pa = 1 N/m2), is small, therefore stress is normally
expressed in kilopascals (kPa) or megapascals (MPa)
Note: 1 MPa = 106 Pa = 106 N/m2 or
1 MPa = 1 N/mm2
1 GPa = 109 Pa = 109 N/m2 or
1GPa = 1000 MPa
U.S.Customary System of units(USCS)
Stress is expressed in pounds per square inch
(psi) or in kilopounds per square inch (ksi)
Note: 1 psi = 6895 Pa = 6.895 kPa
READING QUIZ (cont)
2.
What is the average shear stress in the
internal vertical surface AB (or CD), if
F=20kN, and AAB=ACD=1000mm²?
a)
20 N/mm²
b)
10 N/mm²
c)
10 kPa
d)
200 kN/m²
e)
20 MPa
Copyright © 2011 Pearson Education South Asia Pte Ltd
Stress on an Oblique Plane
Pass a section through the member forming an
angle q with the normal plane.
From equilibrium conditions, the distributed
forces (stresses) on the plane must be
equivalent to the force P.
Resolve P into components normal and
tangential to the oblique section,
F  P cos q
V  P sin q
The average normal and shear stresses on
the oblique plane are

(a) Section plane made
Oblique section through a two-force member.
at an angle q to the member normal plane, (b) Free-body diagram of
left section with internal resultant force P. (c) Free-body diagram of
resultant force resolved into components F and V along the section
plane’s normal and tangential directions, respectively. (d) Free-body
diagram with equivalent as normal stress, , and shearing stress, .
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
F
P cos q
P


cos 2 q
Aq A0
A0
cos q
V
P sin q
P


sin q cos q
Aq A0
A0
cos q
Maximum Stresses
Normal and shearing stresses on an oblique
plane

P
P
cos 2 q   sin q cos q
A0
A0
The maximum normal stress occurs when the
reference plane is perpendicular to the member
axis,
m 
P
A0
  0
The maximum shear stress occurs for a plane at +
45o with respect to the axis,
m 
Selected stress results for axial loading.
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P
P
sin 45 cos 45 

A0
2 A0
Example 4
Hibller
Stress Under General Loadings
A member subjected to a general
combination of loads is cut into
two segments by a plane passing
through Q
Multiple loads on a general body.
The distribution of internal stress
components may be defined as,
F x
 x  lim
A0 A
 xy  lim
A0
(a) Resultant shear and normal forces, Vx and Fx , acting
on small area A at point Q. (b) Forces on A resolved into
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force in coordinate directions.
V yx
A
Vzx
 xz  lim
A0 A
For equilibrium, an equal and
opposite internal force and stress
distribution must be exerted on
the other segment of the member.
State of Stress
Stress components are defined for the planes
cut parallel to the x, y and z axes. For
equilibrium, equal and opposite stresses are
exerted on the hidden planes.
Positive resultant forces on a small element at point Q
resulting from a state of general stress.
The combination of forces generated by the
stresses must satisfy the conditions for
equilibrium:
 Fx   Fy   Fz  0
Mx  My  Mz  0
Consider the moments about the z axis:
 M z  0   xy Aa   yx Aa
 xy   yx
similarly,  yz   zy
Free-body diagram of small element at Q viewed on
projected plane perpendicular to z’-axis. Resultant
forces on positive and negative z’ faces (not shown) act
through the z’-axis, thus do not contribute to the
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moment about that axis.
and  yz   zy
Only six components of stress are required to
define the complete state of stress
Example 5
Solution Example 5
Solution Example 5