Post on 11-Oct-2015
description
Chapter Objectives
To show how to use the method of sections for determining the internal loadings in a member.
To introduce the concepts of normal and shear stress.
To analyse and design members subject to axial load and direct shear.
To define normal and shear strain, and show how they can be determined for various types of problems.
Copyright 2011 Pearson Education South Asia Pte Ltd
Reading Quiz
Applications
Normal Stress
Shear Stress
Normal Strain
Shear Strain
Cartesian Strain Components
Concept Quiz
In-Class Activities
Copyright 2011 Pearson Education South Asia Pte Ltd
READING QUIZ
1. What is the normal stress in the bar if P=10
kN and 500 mm?
a) 0.02 kPa
b) 20 Pa
c) 20 kPa
d) 200 N/mm
e) 20 MPa
Copyright 2011 Pearson Education South Asia Pte Ltd
READING QUIZ (cont)
2. What is the average shear stress in the
internal vertical surface AB (or CD), if F=20
kN, and AAB
=ACD
=1000 mm?
a) 20 N/mm
b) 10 N/mm
c) 10 kPa
d) 200 kN/m
e) 20 MPa
Copyright 2011 Pearson Education South Asia Pte Ltd
READING QUIZ (cont)
3. The center portion of the rubber balloon has
a diameter of d = 100 mm. If the air pressure
within it causes the balloons diameter to become d = 125 mm, determine the average
normal strain in the rubber.
a) 0.2
b) 0.25
c) 0.25
d) 1.25
Copyright 2011 Pearson Education South Asia Pte Ltd
READING QUIZ (cont)
4. What is the unit of strain?
a) mm
b) mm/m
c) Micron
d) no unit
Copyright 2011 Pearson Education South Asia Pte Ltd
APPLICATIONS
Copyright 2011 Pearson Education South Asia Pte Ltd
Where does the average normal stress occur?
Copyright 2011 Pearson Education South Asia Pte Ltd
APPLICATIONS (cont)
Copyright 2011 Pearson Education South Asia Pte Ltd
Will the total shear force over the anchor length be equal to
the total tensile force tensile A in the bar?
APPLICATIONS
Copyright 2011 Pearson Education South Asia Pte Ltd
AVERAGE NORMAL STRESS
Copyright 2011 Pearson Education South Asia Pte Ltd
Will the total shear force over the anchor length be equal to the total tensile force tensile A in the bar?
A
P
NORMAL SHEAR STRESS
Copyright 2011 Pearson Education South Asia Pte Ltd
A
FzA
z
0lim
A
F
A
F
y
Azy
x
Azx
0
0
lim
lim
EXAMPLE 1
Copyright 2011 Pearson Education South Asia Pte Ltd
The bar in Fig. 7-14a 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.
EXAMPLE 1 (cont)
Copyright 2011 Pearson Education South Asia Pte Ltd
Graphically, the normal force diagram is as shown.
By inspection, different sections have different internal forces.
Solution
EXAMPLE 1 (cont)
Copyright 2011 Pearson Education South Asia Pte Ltd
By inspection, the largest loading is in region BC,
Since the cross-sectional area of the bar is constant, the largest average normal stress is
Solution
kN 30BCP
(Ans) MPa 7.8501.0035.0
1030 3
A
PBCBC
DESIGN OF SIMPLE CONNECTION
Copyright 2011 Pearson Education South Asia Pte Ltd
For normal force requirement
For shear force requirement
allow
PA
allow
VA
EXAMPLE 2
Copyright 2011 Pearson Education South Asia Pte Ltd
The rigid bar AB shown in Fig. 129a is supported by a steel rod AC having a diameter of 20 mm and an aluminum block having a cross sectional area of 1800 mm2. The 18-mm-diameter pins at A and C are subjected to single shear. If the failure stress for the steel and aluminum is and , respectively, and the failure shear stress for each pin is , determine the largest load P that can be applied to the bar. Apply a factor of safety of FS=2.
MPa 680failst
MPa 70failal
MPa 900fail
EXAMPLE 2 (cont)
Copyright 2011 Pearson Education South Asia Pte Ltd
The allowable stresses are
There are three unknowns and we apply the equations of equilibrium,
Solution
MPa 4502
900
..
MPa 352
70
..
MPa 3402
680
..
SF
SF
SF
fail
allow
failal
allowal
failst
allowst
(2) 075.02 ;0
(1) 0225.1 ;0
PFM
FPM
BA
ACB
EXAMPLE 2 (cont)
Copyright 2011 Pearson Education South Asia Pte Ltd
We will now determine each value of P that creates the allowable stress in the rod, block, and pins, respectively.
For rod AC,
Using Eq. 1,
For block B,
Using Eq. 2,
Solution
kN 8.10601.010340 26 ACallowstAC AF
kN 171
25.1
28.106P
kN 0.631018001035 66 BallowalB AF
kN 168
75.0
20.63P
EXAMPLE 2 (cont)
Copyright 2011 Pearson Education South Asia Pte Ltd
For pin A or C,
Using Eq. 1,
When P reaches its smallest value (168 kN), it develops the allowable normal stress in the aluminium block. Hence,
Solution
kN 5.114009.010450 26 AFV allowAC
kN 183
25.1
25.114P
(Ans) kN 168P
NORMAL STRAIN
Copyright 2011 Pearson Education South Asia Pte Ltd
s
ssavg
'
ss 1'
s
ss
nAB
'lim
along
SHEAR STRAIN
Copyright 2011 Pearson Education South Asia Pte Ltd
'lim2
along along
tACnAB
nt
CARTESIAN STRAIN COMPONENTS
Copyright 2011 Pearson Education South Asia Pte Ltd
The approximate lengths of the sides of the parallelepiped are
The approximate angles between sides, again originally defined by the sides x, y and z are
Notice that the normal strains cause a change in volume of rectangular element, whereas the shear strain cause a
change in shape
zyx zyx 1 1 1
xzyzxy
2
2
2
CARTESIAN STRAIN COMPONENTS (cont)
Copyright 2011 Pearson Education South Asia Pte Ltd
EXAMPLE 3
Copyright 2011 Pearson Education South Asia Pte Ltd
The slender rod creates a normal strain in the rod of
where z is in meters. Determine (a) displacement of end B
due to the temperature increase, and (b) the average normal
strain in the rod.
2/131040 zz
EXAMPLE 3 (cont)
Copyright 2011 Pearson Education South Asia Pte Ltd
Part (a)
Since the normal strain is reported at each point along the rod, it has a deformed length of
The sum along the axis yields the deformed length of the rod is
The displacement of the end of the rod is therefore
Solution
dzzdz 2/1310401'
m 20239.010401'2.0
0
2/13 dzzz
(Ans) mm39.2m00239.02.020239.0 B
EXAMPLE 3 (cont)
Copyright 2011 Pearson Education South Asia Pte Ltd
Part (b)
Assumes the rod has an original length of 200 mm and a change in length of 2.39 mm. Hence,
Solution
(Ans) mm/mm 0119.0200
39.2'
s
ssavg
EXAMPLE 4
Copyright 2011 Pearson Education South Asia Pte Ltd
Due to a loading, the plate is deformed into the dashed shape
shown in Fig. 26a. Determine (a) the average normal strain along the side AB, and (b) the average shear strain in the plate
at A relative to the x and y axes.
EXAMPLE 4 (cont)
Copyright 2011 Pearson Education South Asia Pte Ltd
Part (a)
Line AB, coincident with the y axis, becomes line AB after deformation, thus the length of this line is
The average normal strain for AB is therefore
The negative sign indicates the strain causes a contraction of AB.
Solution
mm 018.24832250' 22 AB
(Ans) mm/mm 1093.7240
250018.248' 3
AB
ABABavgAB
EXAMPLE 4 (cont)
Copyright 2011 Pearson Education South Asia Pte Ltd
Part (b)
As noted, the once 90 angle BAC between the sides of the plate, referenced from the x, y axes, changes to due to the displacement of B to B.
Since then is the angle shown in the figure.
Thus,
Solution
'2
xy xy
(Ans) rad 121.02250
3tan 1
xy
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 B, C 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
CONCEPT QUIZ (cont)
2) The rectangular membrane has an
unstretched length L1 and width L2. If the
sides are increased by small amounts L1 and L2, determine the normal strain along the diagonal AB.
Copyright 2011 Pearson Education South Asia Pte Ltd
2221
2
2
2
1
21
21
2
2
2
1
2
2
2
1
2
2
1
1
D) C)
B) A)
LL
LL
LL
LL
LL
LL
L
L
L
L
CONCEPT QUIZ (cont)
3) The rectangular plate is subjected to the
deformation shown by the dashed line.
Determine the average shear strain xy
of the
plate.
Copyright 2011 Pearson Education South Asia Pte Ltd
22 2001503
200
150
200
3
150
3
D) C)
B) A)