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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 analyze 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.
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In-Class Activities
Reading Quiz
Applications
Normal Stress
Shear Stress
Normal Strain
Shear Strain
Cartesian Strain Components
Concept Quiz
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READING QUIZ
What is the normal stress in the bar if P=10 kN and 500 mm²?
0.02 kPa
20 Pa
20 kPa
200 N/mm²
20 MPa
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READING QUIZ (cont)
What is the average shear stress in the internal vertical surface AB (or CD), if F=20 kN, and AAB=ACD=1000 mm²?
20 N/mm²
10 N/mm²
10 kPa
200 kN/m²
20 MPa
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READING QUIZ (cont)
The center portion of the rubber balloon has a diameter of d = 100 mm. If the air pressure within it causes the balloon’s diameter to become d = 125 mm, determine the average normal strain in the rubber.
0.2
0.25 π
0.25
1.25
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READING QUIZ (cont)
4. What is the unit of strain? mm
mm/m
Micron
no unit
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APPLICATIONS
Where does the average normal stress occur?
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APPLICATIONS (cont)
Will the total shear force over the anchor length be equal to the total tensile force σtensile A in the bar?
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APPLICATIONS
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AVERAGE NORMAL STRESS
Will the total shear force over the anchor length be equal to the total tensile force σtensile A in the bar?
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NORMAL SHEAR STRESS
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EXAMPLE 1
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.
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EXAMPLE 1 (cont)
Solution
By inspection, different sections have different internal forces.
Graphically, the normal force diagram is as shown.
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EXAMPLE 1 (cont)
Solution
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
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16. DESIGN OF SIMPLE CONNECTION
For normal force requirement
For shear force requirement
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EXAMPLE 2
The rigid bar AB shown in Fig. 1–29a 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.
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EXAMPLE 2 (cont)
Solution
The allowable stresses are
There are three unknowns and we apply the equations of equilibrium,
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EXAMPLE 2 (cont)
Solution
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,
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EXAMPLE 2 (cont)
Solution
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,
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NORMAL STRAIN
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SHEAR STRAIN
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23. CARTESIAN STRAIN COMPONENTS
The approximate lengths of the sides of the parallelepiped are
The approximate angles between sides, again originally defined by the sides Δ x, Δ y and Δ z are
Notice that the normal strains cause a change in volume of rectangular element, whereas the shear strain cause a change in shape
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24. CARTESIAN STRAIN COMPONENTS (cont)
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EXAMPLE 3
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.
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EXAMPLE 3 (cont)
Solution
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
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EXAMPLE 3 (cont)
Solution
Part (b)
Assumes the rod has an original length of 200 mm and a change in length of 2.39 mm. Hence,
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EXAMPLE 4
Due to a loading, the plate is deformed into the dashed shape shown in Fig. 2–6a. 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.
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EXAMPLE 4 (cont)
Solution
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.
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EXAMPLE 4 (cont)
Solution
Part (b)
As noted, the once 90° angle BAC between the sides of the plate, referenced from the x, y axes, changes to θ’ due to the displacement of B to B’.
Since then is the angle shown in the figure.
Thus,
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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.
C > B > D
C > D > B
B > C > D
D > B > C
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CONCEPT QUIZ (cont)
2) The rectangular membrane has an un-stretched length L1 and width L2. If the sides are increased by small amounts ΔL1 and ΔL2, determine the normal strain along the diagonal AB.
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CONCEPT QUIZ (cont)
3) The rectangular plate is subjected to the deformation shown by the dashed line. Determine the average shear strain γxy of the plate.
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