Normal & Shear Stress - Chapter 1
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Transcript of Normal & Shear Stress - Chapter 1
Chapter Objectives
ü Understand the concepts of normal and shear stress
ü Analyze and design of members subjected to axial load or shear
Copyright © 2011 Pearson Education South Asia Pte Ltd
Equilibrium of a Deformable Body
ü External Loads
Surface & Body forces
ü Support Reactions
ü Equations of Equilibrium
ü Internal Resultant Loadings
Copyright © 2011 Pearson Education South Asia Pte Ltd
External loads
Copyright © 2011 Pearson Education South Asia Pte Ltd
Support Reactions
Copyright © 2011 Pearson Education South Asia Pte Ltd
Equations of Equilibrium
Copyright © 2011 Pearson Education South Asia Pte Ltd
Internal Resultant Loadings
Copyright © 2011 Pearson Education South Asia Pte Ltd
Internal Resultant Loadings
Copyright © 2011 Pearson Education South Asia Pte Ltd
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
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=20kN, and AAB=ACD=1000mm²?
a) 20 N/mm²
b) 10 N/mm²
c) 10 kPa
d) 200 kN/m²
e) 20 MPa
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 σtensile A in the bar?
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?
A
P=σ
AVERAGE SHEAR STRESS
Copyright © 2011 Pearson Education South Asia Pte Ltd
A
Fz
Az
∆
∆=
→∆ 0limσ
A
F
A
F
y
Azy
x
Azx
∆
∆=
∆
∆=
→∆
→∆
0
0
lim
lim
τ
τ
EXAMPLE 1 (average normal stress)
Copyright © 2011 Pearson Education South Asia Pte Ltd
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.
EXAMPLE 1 (cont)
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• Graphically, the normal force diagram is as shown.
• By inspection, different sections have different internal forces.
Solutions
EXAMPLE 1 (cont)
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• 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
Solutions
kN 30=BCP
( )( )( )
(Ans) MPa 7.8501.0035.0
1030 3
===A
PBCBCσ
DESIGN OF SIMPLE CONNECTION
Copyright © 2011 Pearson Education South Asia Pte Ltd
• For normal force requirement
• For shear force requirement
allow
PA
σ=
allow
VA
σ=
EXAMPLE 2 (single shear stress)
Copyright © 2011 Pearson Education South Asia Pte Ltd
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-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 680=failstσ
( ) MPa 70=failalσ
MPa 900=failτ
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,
Solutions
( )( )
( )( )
MPa 4502
900
..
MPa 352
70
..
MPa 3402
680
..
===
===
===
SF
SF
SF
fail
allow
failal
allowal
failst
allowst
ττ
σσ
σσ
( ) ( )
( ) ( ) (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,
Solutions
( ) ( ) ( ) ( )[ ] kN 8.10601.01034026 === πσ ACallowstAC AF
( )( )kN 171
25.1
28.106==P
( ) ( ) ( )[ ] kN 0.631018001035 66 === −BallowalB AF σ
( )( )kN 168
75.0
20.63==P
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,
Solutions
( ) ( )[ ] kN 5.114009.01045026 ==== πτ AFV allowAC
( )( )kN 183
25.1
25.114==P
(Ans) kN 168=P
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
Copyright © 2011 Pearson Education South Asia Pte Ltd
EXERCISES
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EXERCISES
Copyright © 2011 Pearson Education South Asia Pte Ltd
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
Copyright © 2011 Pearson Education South Asia Pte Ltd