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Strength Of Materials
Chapter One Simple stresses
Strength of materials extends the study of forces that was begun in Engineering
Mechanics, but there is a sharp distinction between the two subjects.
Fundamentally, the field of mechanics covers the relation between forces acting
on rigid bodies; in statics, the bodies are in equilibrium , whereas in dynamics ,
they are accelerated but can be but in equilibrium by applying correctly inertia
forces.
In contrast to mechanics, strength of materials deals with the relation between
externally applied loads and their internal effects on bodies. Moreover, the bodies
are no longer assumed to be ideally rigid; the deformations, however small, are of
major interest. The properties of the material of which a structure or machine is
made affects both its choice and dimensions that will satisfy the requirements of
strength and rigidity.
The difference between mechanics and strength of materials can be further
emphasized by the following example:-
∑ 𝑀𝐴 = 0 (in statics)
We can find the load P
Member AB assumed to be rigid enough and strong enough to permit the desired
action.
In strength of materials we must investigate the bar itself to be sure that it will
neither break nor be so flexible that it bends without lifting the load.
SI Units (System International Units)
A. Selected SI Units
W
P
A
B
Quantity Name SI Symbol
Energy Joule J (1J=1 N.m )
Force Newtons N (1N = 1Kg.m/s )
Length meter m
Mass Kilogram Kg
Moment (torque) Newton meter N.m
Plane angle Radian
dgree
Rad
Rotational frequency Revolution per second r/s
Stress (pressure) pascal Pa (1 Pa =1 N/m
2
Temperature Degree celsius ̊C
Time second s
Power watt W (1 W = 1 J/s)
B.Commonly used SI Prefixes
Multiple Factor Prefix SI Symbol
9
giga G
6
mega M
3
kilo K
milli m
micro μ
nano n
Units
British Metric S.I.(System
International)
Force
(1 N = 4.448 lb)
lb(lebra),kip,Ton
1 kip = 1000 lb
1 ton = 2240 lb
g (gram), kg
1 kg = 1000 g
1 Ton = 1000 kg
N(Newton),kN
1 KN =1000 N
1 kg = 10 N
Length
(1 in = 2.54 cm )
In (inch), ft
1 ft = 12 in
mm, cm,m
1 cm = 10 mm
1 m =100 cm
1 m = 1000 mm
Mm,cm,m
1 cm =10 mm
1m =100 cm
1 m = 1000 mm
Stress
(force/area)
kPa = 6.894)
psi (ib/in^2)
ksi (kip/in ^2)
Papascal, kPa,Mpa,GPa
MPa[Mega Pascal] =10^6 Pa (N/mm^2)
GPa[Giga Pascal] = 10 ^9 Pa (kN/mm^2)
No.of unknowns =
No.of equations =
Indeterminate to the 4
th
degree.
Indeterminate to the 2
nd
degree.
The hinge will now add another equation (equation of condition) and the beam is
now determinate.
No. of unkowns =
No.of equations =3+
3 static equilibrium equations
2 equations of condition (2-hinges)
So the beam is indeterminate to the 1
st
degree.
Example :- Find the reactions.
b
2 kN/m
Hinge
8 kN
6 kN
a a
x
c
R
c
5 m
6 m 3 m 3 m 10 m
a
y
Ma
P 2 P
P1 P
P1 hinge P
P
hinge
member bc as F.B.D
∑ 𝑀𝑏 = 0 Type equation here.
20*5 =10 Rc
Rc = 10 KN
The whole frame as F.B.D
∑ 𝐹𝑥 = 0 , ∑ 𝐹𝑦 = 0 Type equation here.
ax =
ay - 8 - 6 - 20+10 =
ay = 24 kN
Member ab as F.B.D
Ma +86 +63 - 24*12 = 0
Ma = 222 kN.m
Example :- For the frame shown , find the reactions.
Member cd as F.B.D
∑ 𝑀𝑐 = 0 Type equation here.
3*4 = 12 dx
dx = 1 kN
member bc as F.B.D
5 kN 4 kN
3 kN
4.5 kN
a
d
b c
0.75 kN/m
6m
6m
4 m 4 m 4 m 4 m
8kN
6 kN
Ma
ax
ay
a b
20 kN
Rc
The whole frame as F.B.D
8 *Ra = 18 * Rdy
Rdy = 11.85 KN
Rx = 26 .67 KN
Rcy = 11.85 KN
Analysis of Internal Forces
In engineering mechanics we would start by determining the
resultant of the applied forces to determine weather or not the
body remains at rest.
F 1
F
F
F
In strength of materials , we would make additional
investigation of the internal forces.
The internal forces reduce to a force and couple which resolved
into component normal and tangent to the section.
The plane a-a is normal to x-axis so it’s known as x-surface or x
face.
Pxx Axial Force : This component measure the pulling ( or pushing) action over
the section. It is often denoted by P.
Pxy , Pxz Shear forcec : These components of the total resistance to sliding the
portion to one side of the explorary section past the other. it often denoted by V.
Mxx Torque : This component measure the resistance to twisting the member
and is commonly given the symbol T.
Mxy ,Mxz Bending Moments : These components measure the resistance to
bending member about the Y or Z axes and are often denoted by My , Mz.
Units of stress
Pascal (Pa) = N/mm
2
MPa = MN/m
2
or equal to N/mm
2
GPa = GN/m
2
or equal to kN/m
2
Example : Which one of these two bars is stronger?
500 𝑁
10 × 10
− 6
= 50 × 10
6
2
5000 𝑁
1000 × 10
− 6
×𝑚^ 2
= 5 × 10
6
2
Type equation here.
The material of the bar 1 is ten times as stronger as material 2.
Normal Stress
The resisting area is perpendicular to the applied force, thus normal. There are
two types of normal stresses; tensile stress and compressive stress. Tensile stress
applied to bar tends the bar to elongate while compressive stress tend to shorten
the bar. Where P is the applied normal load in Newton and A is the area in mm2.
The maximum stress in tension or compression occurs over a section normal to
the load
Shearing Stress
Forces parallel to the area resisting the force cause shearing stress. It differs to
tensile and compressive stresses, which are caused by forces perpendicular to the
area on which they act. Shearing stress is also known as tangential stress.
Bar 1
Bar 2
500N
5000N
Where V is the resultant shearing force which passes through the centroid of the
area A being sheared. Type equation here.
Example
What force is required to punch a 20-mm-diameter hole in a plate that is 25 mm
thick? The shear strength is 350 MN/m2.
Bearing Stress
Bearing stress is a compressive stress but it is differs from the normal compressive
stress in that the latter is an internal stress caused by internal compressive force
whereas the former is a contact pressure between two separate bodies. Some
examples of bearing stress are soil pressure beneath the pier and the forces on
bearing plates. We now consider the contact pressure between a rivet or bolt and
the contact surface of the plate against which it pushes.
P
Example( popov): For the structure shown in the figure, calculate the size of the
bolt and the area of the bearing plates required if the allowable stresses are (
MPa) in tension and 3.44 MPa in bearing. Neglect the weight of the beams.
Solution:
Whole structure as F.B.D
Member cb as F.B.D , gives the tensile force in the bolt
- 73 × 10
3
124 × 10
6
= 5. 38 × 10
− 4
2
𝜋
4
2
, 5. 38 × 10
− 4
𝜋
4
2
𝑑 = 26 𝑚𝑚 (Diameter of the bolt)
2
2
− 5. 38 × 10
− 4
A 12-inches square steel bearing plate lies between an 8-inches diameter
wooden post and a concrete footing as shown in Fig. P-110. Determine
the maximum value of the load P if the stress in wood is limited to 1800
psi and that in concrete to 650 psi.
A
C B
40.03 kN
0.915m 1.83m 0.915 1.83m
One bolt
L
L
- The end chord of timber truss framed into the bottom chord as shown in the
figure. neglect friction (a)compute dimension b if the allowable shearing stress is
900 kPa ; and (b) determine dimension c so that the bearing stress does not
exceed 7 MPa.
Solution :
𝜏 = 𝑉/𝐴 ; 900 × 10
3
50 𝑐𝑜𝑠 30 × 10
3
- 15 ×𝑏
7 × 10
6
50 × 10
3
×𝑐𝑜𝑠 30
- 15 ×𝑐
P= 50 kN
30 °
c
b
𝜎𝑑𝑓 = 188 𝑀𝑃𝑎 C ; 𝜎𝑐𝑒 = 113 𝑀𝑃𝑎 T ; 𝜎𝑏𝑑 = 80. 1 𝑀𝑃𝑎
108. Determine the outside diameter of hollow steel tube that will carry a tensile
load of 500 kN stress of 140 MN/m
2
.Assume that wall thickness to be one tenth
of the outside diameter.
Solution: Assume the outside diameter=D
𝑃
𝐴
500 × 10
− 3
140 × 10
− 6
; 𝐴 = 3. 57 × 10
− 3
2
0. 1 𝐷×𝜋𝐷 = 3. 57 × 10
− 3
120. Two blocks of wood, 50 mm wide and 20 mm thick, are glued together as
shown in figure. (a) Using the free body diagram concept illustrated before,
determine the shear load and from it the shearing stress on the glued joint if
P=6000 N. ( b)Generalize the procedure of part (a) to show that the shearing
stress on a plane inclined at any angle 𝜃 to a transverse section of area A is 𝜏 =
Solution: (a)
50
𝑙
𝐴 = 20 × 57. 735 = 1154. 7 𝑚𝑚
2
𝑉
𝐴
D
0.1D
60 °
50mm
P
P
3000
- 7
(b)
ℎ
𝑙
′
= 𝑏×𝑙 ; 𝐴
′
= 𝑏× ℎ ⁄𝑠𝑖𝑛𝜃
𝑉
𝐴
′
𝑃𝑐𝑜𝑠𝜃
𝑏ℎ 𝑠𝑖𝑛𝜃
⁄
𝑃𝑠𝑖𝑛 2 𝜃
2 𝐴
109. Part of landing gear for a light plane is shown in figure. Determine the
compressive stress in the strut AB caused by a landing reaction R=20 kN. Strut AB
is inclined at 53.1° with BC .Neglect weight of the members.
Solution:
20 × 650 − 𝐹𝑎𝑏×𝑠𝑖𝑛 53. 1 × 450 = 0
- 125 × 10
3
𝜋
4
( 40
2
− 30
2
)
Bolted and Riveted Connections
P
P
P P
P
P
P
P
ℎ𝑜𝑙𝑙𝑜𝑤 𝑠𝑡𝑟𝑢𝑡
OD=40 mm
ID=30 mm
R
A
B
C
200mm 450mm
V
P
N
60 °
l
h
P
P
l
b
𝑃 = 6 [ 0. 019 × 0. 019 × 372. 9 × 10
6
]
For each of the outer plates
= 6 [ 0. 019 × 0. 0125 × 372. 9 × 10
6
]
- Tensile Stresses
In this case you should check the critical section (1-1),(2-2) and (3-3).
Middle Plate ,
Sec(1-1)
× 0. 019 × 165. 5 × 10
6
Sec (2-2)
𝑃 = [ 0. 25 − 2 × 0. 019 ]× 0. 019 × 165. 5 × 10
6
Sec (3-3)
0. 25 − 3 × 0. 019
× 0. 019 × 165. 5 × 10
6
Cover plate
Sec (3-3)
= ( 0. 25 − 3 × 0. 019 )× 0. 0125 × 165. 5 × 10
6
Sec (2-2)
P
P
P/
P
P
P/
1/6(P/2)
2/6(P/2)
3/6(P/2)
1
2 3
− 3 ×
0. 25 − 2 × 0. 019
× 0. 0125 × 165. 5 × 10
6
Sec (3-3)
0. 25 − 1 × 0. 019
× 0. 0125 × 165. 5 × 10
6
∴ 𝑃 = 351. 8 𝑘𝑁 Is the safe load for this joint.
Thin-Walled Pressure Vessels
A tank or pipe carrying a fluid or gas under a pressure is subjected to
tensile forces, which resist bursting, developed across longitudinal and
transverse sections.
TANGENTIAL STRESS 𝝈𝒕
(Circumferential Stress)
Consider the tank shown being subjected to an internal pressure p. The
length of the tank is L and the wall thickness is t. Isolating the right half
of the tank:
