STRENGTH OF MATERIALS
CHAPTER ONE
INTRODUCTION
CONCEPT OF STRESS
Prepared by : Eng. Ahmad Bani Yaseen
1
1.1 REVIEW OF THE METHODS OF STATICS
❑ The main objective of the mechanics of materials is to analyse and design a given structure
involving determination of stress and deformation.
2
❑ Review of Statics
• The structure is designed to
support a 30 kN load
• The structure consists of a
links and rod joined by pins at
the junctions and support.
• Perform a static analysis to
determine the internal force in
each structural member and
the reaction forces at the
supports.
3
❑ Structure Free-Body Diagram
• The structure is detached from supports and the
loads and reaction forces are indicated
• Conditions for static equilibrium:
• Ay and Cy can not be determined from these
equations
4
❑ Component Free-Body Diagram
• In addition to the complete structure, each
component must satisfy the conditions for
static equilibrium
• Consider a free-body diagram for the boom:
• substitute into the structure equilibrium
equation
• Results:
• Reaction forces are directed along link (AB) and
rod (CD).
5
❑ Method of Joints
• The link (AB) and rod (BC) are 2-force members,
i.e., the members are subjected to only two forces
which are applied at member ends
• For equilibrium, the forces must be parallel to an
axis between the force application points, equal in
magnitude, and in opposite directions
• Joints must satisfy the conditions for static
equilibrium which may be expressed in the form
of a force triangle:
6
1.2 STRESS IN THE MEMBERS OF A STRUCTURE
▪ Can the system withstand the force?
▪ Will the system break down?
▪ Its ability to withstand the depends on
1- Its material (Steel is stronger than Aluminum)
2- The cross-section of the rod
A
Stress is the intensity of the internal forces distributed over a given area
stress (N/m 2 = Pa)
P
=
A
Internal force (N)
Cross-section
area ( m2)
7
STRESS UNITS
8
❑ Analysis and design
• Assume rod BC is made of steel with
maximum allowable stress  all = 165 MPa
FBC = 50kN
=
FBC
50 kN
=
= 159 MPa
2
−3 2
 r  (10 10 )
   all
*Thus, rod BC can withstand the load
without breaking down
9
• Design of new structures requires selection of appropriate materials and component
dimensions to meet performance requirements
Example:
What will be the suitable diameter for rod BC without exceeding all = 100 MPa.
Solution:
P
 all =
A
A=
d2
A=
4
d=
4A

=
(
P
 all
=
4 500 10

50 103 N
100 106 Pa
−6
m
2
= 500 10− 6 m 2
) = 2.52 10−2 m = 25.2 mm
* The rod 26 mm or
more in diameter will be
suitable
10
❑ Axial loading, Normal stress
𝐏
The stress 𝜎 =
is the average stress on the cross−section.
𝐴
The stress at point 𝑄 is
Δ𝐅
𝜎 = lim
𝐴→0 Δ𝐴
𝑃 = න𝑑𝐹 = න 𝜎𝑑𝐴
𝐴
11
❑ The stress is considered uniformly distributed if:
I. The line of action of the concentrated load passes through the centroid of the
cross-section.
12
II. The cross-section is far from the edges where the load is applied.
13
SHEARING STRESS
❑ Transverse load is acting perpendicular to the rod (not in the normal direction).
❑ The load cause shear stress.
 ave
P F
= =
A A
14
EXAMPLE ON SHEAR FORCE
 ave
V F/2
= =
A
A
15
MORE EXAMPLES ON SHEAR
 ave
P F/4
= =
A
A
16
Example: given width w = 150 mm. Find the average shear stress along sections
a-a and b-b.
Solution:
 a−a
Va − a
3 103
=
=
= 200 kPa
Aa − a 0.1 0.15
 b −b
Vb −b
3 103
=
=
= 160 kPa
Ab −b 0.125  0.15
17
𝐄𝐱𝐚𝐦𝐩𝐥𝐞: Find 𝑑 and 𝑡 in order to support the 20kN, given
𝜎𝑎𝑙𝑙 = 60 MPa
𝜏𝑎𝑙𝑙 = 35 MPa
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
𝐏
𝐴=
𝜎𝑎𝑙𝑙
𝜋 2 20 × 103
𝑑 =
4
60 × 106
Then, we get =≫ 𝑑 = 20.6 mm
𝐕
𝐴=
𝜏𝑎𝑙𝑙
20 × 103
2𝜋 × 0.02 × 𝑡 =
35 × 106
𝑡 = 4.55 mm
18
𝐄𝐱𝐚𝐦𝐩𝐥𝐞:
𝑑 = 6 mm
𝑃 = 9 kN
Find
𝜏𝑎𝑣𝑒 in pins
𝜏𝑎𝑣𝑒 in the shadow planes (tear out)
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
9 × 103 /4
𝜏𝑎𝑣𝑒 (𝑝𝑖𝑛𝑠) = 𝜋
= 79.6 MPa
−3
2
(6 × 10 )
4
9 × 103 /4
𝜏𝑎𝑣𝑒 (𝑝𝑙𝑎𝑛𝑒𝑠) =
= 225 kPa
0.1 × 0.1
19
BEARING STRESS IN CONNECTIONS
• Bolts, rivets, and pins create stresses on the
points of contact or bearing surfaces of the
members they connect.
• The resultant of the force distribution on the
surface is equal and opposite to the force exerted
on the pin.
P
b =
td
20
APPLICATION TO THE ANALYSIS AND DESIGN OF SIMPLE
STRUCTURES
We will determine the normal stresses, shearing stresses, and
bearing stresses
Normal Stress in Link AB and Rod BC
1- Normal stress on rod (BC)
𝐅𝐁𝐂
50 × 103
𝜎𝐵𝐶_𝑚𝑖𝑑 = 𝜎𝑎𝑣𝑔 = 2 =
= 159 MPa
𝜋𝑟
314 × 10−6
𝐴BC_end = 𝐴BC_C = (20 )(40−25) × 10−6
= 300 × 10−6 m2
𝜎𝐵𝐶−𝐵
𝐅𝐁𝐂
50 × 103
=
=
= 167 MPa
𝐴𝐵𝐶−𝐵 300 × 10−6
𝜎𝐵𝐶−𝐶 = 𝜎𝐵𝐶−𝐵 = 167 MPa
21
2- Normal stress on link (AB)
𝜎𝐴𝐵
𝐅𝐀𝐁
−40 × 103
=
=
= −26.7 MPa
𝐴𝐴𝐵 30 × 50 × 10−6
❖ Note that there is no stress on the
joints A and B as the rod is under
compression.
Shearing Stress in Various Connections.
1- Shearing stress (Pin A)
A =  r 2 =  (12.5 10−3 ) 2 = 49110−6 m 2
FAB
P=
= 20 kN
2
P
 A = = 40.7 MPa
A
22
2- Shearing stress (Pin C)
𝐴 = 𝜋𝑟 2 = 𝜋(12.5 × 10−3 )2
= 491 × 10−6 m2
𝐏 = 𝐅𝐁𝐂 = 50 kN
𝐏
𝜏𝐶 = = 102 MPa
𝐴
23
3- Shearing stress (Pin B)
PG
25 103
B = 2 =
= 50.9 MPa
−6
r
49110
Bearing stress
1- Bearing stress at point A
1− on the link
𝐏
40 × 103
𝜎𝑏 =
=
= 53.3 MPa
𝑡𝑑 30 × 25 × 10−6
2 − on the brackets
𝐏
40 × 103
𝜎𝑏 =
=
= 32.0 MPa
𝑡𝑑 2 × 25 × 25 × 10−6
24
2- Bearing stress at point C
1− on the link
𝐏
50 × 103
𝜎𝑏 =
=
= 100.0 MPa
𝑡𝑑 20 × 25 × 10−6
2 − on the brackets
𝐏
50 × 103
𝜎𝑏 =
=
= 100.0 MPa
𝑡𝑑 20 × 25 × 10−6
25
1.3 STRESS ON AN OBLIQUE PLANE UNDER AXIAL LOADING
=
F
A
=
V
A
A0
A =
cos 
𝑃 cos 𝜃
𝜎=
𝐴0 / cos 𝜃
𝑃
𝜎=
cos 2 𝜃
𝐴0
𝑃 sin 𝜃
𝜏=
𝐴0 / cos 𝜃
𝑃
𝜏=
sin 𝜃 cos 𝜃
𝐴0
26
Problem
5.0 cm
5.0 cm
The 15 kN load P is supported by two wooden members of uniform cross
section that are joined by the simple glued scarf splice shown. Determine
the normal and shearing stresses in the glued splice.
Solution :
F
P
V
30
60 o
o
𝑃
𝜎=
cos 2 𝜃
𝐴0
𝑃
𝜏=
sin 𝜃 cos 𝜃
𝐴0
15 × 103
2 30 = 4.5 𝑀𝑃𝑎
𝜎=
cos
25 × 10−4
15 × 103
𝜏=
cos 30 sin 30 = 2.59 𝑀𝑃𝑎
25 × 10−4
27
1.5 DESIGN CONSIDERATION
❑ Allowable Load and Allowable Stress: Factor of Safety
Structural members or machines must be designed such that the
working stresses are less than the ultimate strength of the
material.
1- Determination of the ultimate strength of a material.
PU
U =
A
2- Allowable stress; factor of safety
Factor of safety = F.S =
Ultimate stress
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
28
COMMON SUPPORT REACTIONS
29
𝐄𝐱𝐚𝐦𝐩𝐥𝐞: Determine the required diameter of the bolts if the failure shear stress
is 𝜏𝐹𝑎𝑖𝑙 = 350 MPa. Use a factor of safety F. S = 2.5.
Solution :
A=
V
 all
20 103
 2
=
=
d
6
350 10 / 2.5 4
then, we get
d = 13.5 mm
30
𝐄𝐱𝐚𝐦𝐩𝐥𝐞: Given
(𝜎𝐴𝐵 )𝑎𝑙𝑙 = 175 MPa
(𝜎𝐵𝐶 )𝑎𝑙𝑙 = 150 MPa
Find 𝒅𝑨𝑩 and 𝒅𝑩𝑪
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
dAB
𝐅𝐀𝐁 = 70 kN, and 𝐅𝐁𝐂 = 30 kN
𝐅𝐀𝐁
(𝜎𝐴𝐵 )𝑎𝑙𝑙 = 𝜋
2
𝑑
𝐴𝐵
4
𝐅𝐁𝐂
(𝜎𝐵𝐶 )𝑎𝑙𝑙 = 𝜋
2
4 𝑑𝐵𝐶
dBC
𝑑𝐴𝐵 = 22.6 mm
𝑑𝐵𝐶 = 16 mm.
31
Problem 1.55
In the structure shown, an 8-mm-diameter pin is used at A, and 12-mm diameter pins are used at B and D.
Knowing that the ultimate shearing stress is 100 MPa at all connections and that the ultimate normal stress
is 250 MPa in each of the two links joining B and D, determine the allowable load P if an overall factor of
safety of 3.0 is desired.
32
33