Chapter Objectives ü Understand the concepts of normal and shear stress ü Analyze and design of members subjected to axi
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Chapter Objectives ü Understand the concepts of normal and shear stress ü Analyze and design of members subjected to axial load or shear
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Equilibrium of a Deformable Body ü External Loads Surface & Body forces ü Support Reactions ü Equations of Equilibrium ü Internal Resultant Loadings
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External loads
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Support Reactions
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Equations of Equilibrium
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Internal Resultant Loadings
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Internal Resultant Loadings
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READING QUIZ 1.
What is the normal stress in the bar if P=10 kN and 500mm²? a)
0.02 kPa
b)
20 Pa
c)
20 kPa
d)
200 N/mm²
e)
20 MPa
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READING QUIZ (cont) 2.
What is the average shear stress in the internal vertical surface AB (or CD), if F=20kN, and AAB=ACD=1000mm²? a)
20 N/mm²
b)
10 N/mm²
c)
10 kPa
d)
200 kN/m²
e)
20 MPa
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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?
P σ= A
AVERAGE SHEAR STRESS ∆Fz σ z = lim ∆A→0 ∆A
∆Fx τ zx = lim ∆ A → 0 ∆A ∆Fy τ zy = lim ∆A → 0 ∆A
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EXAMPLE 1 (average normal stress) The bar in Fig. 1–16a 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) Solutions •
By inspection, different sections have different internal forces.
• Graphically, the normal force diagram is as shown.
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EXAMPLE 1 (cont) Solutions •
By inspection, the largest loading is in region BC,
PBC = 30 kN •
Since the cross-sectional area of the bar is constant, the largest average normal stress is
σ BC
( )
PBC 30 103 = = = 85.7 MPa (Ans) A (0.035)(0.01)
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DESIGN OF SIMPLE CONNECTION •
For normal force requirement A=
•
P
σ allow
For shear force requirement
A=
V
σ allow
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EXAMPLE 2 (single shear stress) 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 1800mm2. The 18-mmdiameter pins at A and C are subjected to single shear. If the failure stress for the steel and aluminum is (σ ) = 680 MPa and (σ ) = 70 MPa respectively, and the failure shear stress for each pin is τ = 900 MPa , determine the largest load P that can be applied to the bar. Apply a factor of safety of FS=2. st
al
fail
fail
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fail
EXAMPLE 2 (cont) Solutions •
The allowable stresses are
(σ st )allow =
(σ st ) fail
680 = 340 MPa 2 F .S . (σ al ) fail 70 = = 35 MPa (σ al )allow = 2 F .S . τ fail 900 = = 450 MPa τ allow = F .S . 2
•
=
There are three unknowns and we apply the equations of equilibrium, + ∑ M B = 0; + ∑ M A = 0;
P (1.25) − FAC (2 ) = 0
(1)
FB (2 ) − P(0.75) = 0
(2)
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EXAMPLE 2 (cont) Solutions •
We will now determine each value of P that creates the allowable stress in the rod, block, and pins, respectively.
[
]
•
For rod AC, FAC = (σ st )allow ( AAC ) = 340(10 6 ) π (0.01)2 = 106.8 kN
•
Using Eq. 1, P =
•
For block B, FB = (σ al )allow AB = 35(10 6 )[1800(10 −6 )] = 63.0 kN
•
Using Eq. 2, P =
(106.8)(2) = 171 kN 1.25
(63.0)(2) = 168 kN 0.75
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EXAMPLE 2 (cont) Solutions
[
]
•
For pin A or C, V = FAC = τ allow A = 450(106 ) π (0.009)2 = 114.5 kN
•
Using Eq. 1, P =
•
When P reaches its smallest value (168 kN), it develops the allowable normal stress in the aluminium block. Hence,
(114.5)(2) = 183 kN 1.25
P = 168 kN (Ans)
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EXAMPLE 3 (internal loadings)
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EXAMPLE 3 (cont)
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EXAMPLE 3 (cont)
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EXAMPLE 3 (cont)
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EXAMPLE 4 (double shear stress)
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EXAMPLE 4 (cont)
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EXAMPLE 4 (cont)
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EXAMPLE 4 (cont)
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EXERCISES
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EXERCISES
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EXERCISES
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EXERCISES
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EXERCISES
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EXERCISES
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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 BC and D. a)
C>B>D
b)
C>D>B
c)
B>C>D
d)
D>B>C
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