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fisrt law of mechanical
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LECTURE 2
NOOR MAZNI ISMAIL
FACULTY OF MANUFACTURING ENGINEERING
Example 02: Determine the internal loads
Q2: Determine the resultant internal loadings acting on the cross section at B of the
pipe. The pipe has a mass of 2 kg/m and is subjected to both a vertical force of 50 N and
a couple moment of 70 N·m at its end A. It is fixed to the wall at C.
Solution:Free-Body Diagram
( )( )( )( )( )( ) N 525.2481.925.12
N 81.981.95.02
==
==
AD
BD
W
W
Calculating the weight of each segment of pipe,
Applying the six scalar equations of equilibrium,
( )( )( )( ) (Ans) N 3.84
050525.2481.9 ;0
(Ans) 0 ;0
(Ans) 0 ;0
=
=−−−=
==
==
∑∑∑
zB
zBz
yBy
xBx
F
FF
FF
FF
( ) ( ) ( ) ( ) ( )( )
( ) ( ) ( ) ( )( )
( ) ( ) (Ans) 0 ;0
(Ans) mN8.77
025.150625.0525.24 ;0
(Ans) mN3.30
025.081.95.0525.245.05070 ;0
==
⋅−=
=++=
⋅−=
=−−−+=
∑
∑
∑
zBzB
yB
yByB
xB
xBxB
MM
M
MM
M
MM
Stress1. NORMAL STRESS
2. SHEAR STRESS
3. ALLOWABLE STRESS
Stress� Distribution of internal loading is important in mechanics of
materials.
� The intensity of internal force at a point is called
stress, also known as force per unit area.
� SI Unit: N/m2
Types of StressNormal Stress, σ (sigma)
� Force per unit area acting normal to ΔA. Since ΔFz is normal to
the area, then
Shear Stress, τ (tau)
� Force per unit area acting tangent to ΔA. The shaer stress
components,
A
Fz
Az ∆
∆=
→∆ 0limσ
A
F
A
F
y
Azy
x
Azx
∆
∆=
∆
∆=
→∆
→∆
0
0
lim
lim
τ
τ
Normal force acting “pull” on ΔA, tensile stress.
If acting “pushes” on ΔA, compressive stress.
Note:
the subscript notation z specifies the orientation of the area ΔA, and x and y indicate the axes along which shear stress acts.
1.0: Average Normal Stress in an Axially Loaded Bar
� When a cross-sectional area bar is subjected to axial
force through the centroid, it is only subjected to normal
stress.
� Stress is assumed to be averaged over the area.
Average Normal Stress Distribution
� When a bar is subjected to a
constant deformation,
Normal Stress Equilibrium
� 2 normal stress components
that are equal in magnitude
but opposite in direction.
A
P
AP
dAdFA
=
=
= ∫∫
σ
σ
σ
σ = average normal stressP = resultant normal forceA = cross sectional area of bar
Example 01The 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.
Solution:By inspection, different sections have different internal forces.
Graphically, the normal force diagram is as shown.
Solution:
By inspection, the largest loading is in region BC,
kN 30=BCP
Since the cross-sectional area of the bar is constant, the largest average normal stress is
( )( )( )
(Ans) MPa 7.8501.0035.0
10303
===A
PBCBCσ
1 N/m2 = 1 Pa
3kN/m 80=stγ
Example 02The casting is made of steel that has a specific weight of
. Determine the average compressive stress acting at points A and B.
Solution:By drawing a free-body diagram of the top segment,the internal axial force P at the section is
( )( ) ( )kN 042.8
02.08.080
0 ;0
2
=
=−
=−=↑+ ∑
P
P
WPF stz
π
The average compressive stress becomes
( )(Ans) kN/m 0.64
2.0
042.8 2
2===
πσ
A
P
2.0: Average Shear Stress� The average shear stress distributed over each sectioned
area that develops a shear force.
� 2 different types of shear:
A
Vavg =τ
τ = average shear stressV = internal resultant shear forceA = area at that section
a) Single Shear b) Double Shear
� If two parts are joined together, the applied load may
cause shearing of of the material, it is generally assumed
that an average shear stress acts over the cross-sectional
area.
� Shear Stress Equalibrium
All shear stresses must have equal magnitude and be directed
either toward or away from each other at opposite edges of
the element.
Example 01The inclined member is subjected to a compressive force of 3000 N. Determine the
average compressive stress along the smooth areas of contact defined by AB and BC,
and the average shear stress along the horizontal plane defined by EDB.
The shear force acting on the sectioned horizontal plane EDB is
Solution:
N 1800 ;0 ==→+ ∑ VFx
Average compressive stresses along the AB and BC planes are
( )( )
( )( )(Ans) N/mm 20.1
4050
2400
(Ans) N/mm 80.14025
1800
2
2
==
==
BC
AB
σ
σ
( )( )(Ans) N/mm 60.0
4075
1800 2==avgτ
Average shear stress acting on the BD plane is
The compressive forces acting on the areas of contact (inclined member) are
( )( ) N 240003000 ;0
N 180003000 ;0
5
4
5
3
=⇒=−=↑+
=⇒=−=→+
∑∑
BCBCy
ABABx
FFF
FFF
AVERAGE STRESS:
INTERNAL LOADING:
3.0: Allowable Stress & Factor of Safety
� Proper design is critical to ensure the
structure is safe.
� Many unknown factors that influence
the actual stress in a member.
� A factor of safety is needed to
obtained allowable load. Restrict the
applied load.
� The factor of safety (F.S.) is a ratio of
the failure load divided by the
allowable loadallow
fail
allow
fail
allow
fail
SF
SF
F
FSF
τ
τ
σ
σ
=
=
=
.
.
.
Example 01
The control arm is subjected to the loading. Determine to the nearest 5 mm the required diameter of the steel pin at C if the allowable shear stress for the steel is . Note in the figure that the pin is subjected to double shear.
Solution:For equilibrium we have
MPa 55=allowableτ
( ) ( ) ( )( )( )( ) kN 3002515 ;0
kN 502515 ;0
kN 150125.025075.0152.0 ;0
5
3
5
4
5
3
=⇒=−−=+↑
=⇒=+−−=+→
=⇒=−==+
∑∑∑
yyy
xxx
ABABC
CCF
CCF
FFM
Solution:
The pin at C resists the resultant force at C. Therefore,
( ) ( ) kN 41.3030522 =−=CF
mm 8.18
mm 45.2462
m 1045.2761055
205.15
2
26
3
2
=
=
×=×
== −
d
d
VA
allowable
π
τ
The pin is subjected to double shear, a shear force of 15.205 kN acts over its cross-sectional area between the arm and each supporting leaf for the pin.
The required area is
Use a pin with a diameter of d = 20 mm. (Ans)
25 October 2011 19
END OF LECTURE 02Thank you!
25 October 2011 20
ASSIGNMENT 1 !