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

Equilibrium of a Deformable Body

ü External Loads

Surface & Body forces

ü Support Reactions

ü Equations of Equilibrium

ü Internal Resultant Loadings

External loads

Support Reactions

Equations of Equilibrium

Internal Resultant Loadings

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

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

APPLICATIONS (cont)

Will the total shear force over the anchor length be equal to the total tensile force σtensile A in the bar?

APPLICATIONS

AVERAGE NORMAL STRESS

Will the total shear force over the anchor length be equal to the total tensile force σtensile A in the bar?

AVERAGE SHEAR STRESS

→∆ 0limσ

→∆

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.

EXAMPLE 1 (cont)

• Graphically, the normal force diagram is as shown.

• By inspection, different sections have different internal forces.

Solutions

EXAMPLE 1 (cont)

• 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

PBCBCσ

DESIGN OF SIMPLE CONNECTION

• For normal force requirement

• For shear force requirement

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-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)

• The allowable stresses are

• There are three unknowns and we apply the equations of equilibrium,

Solutions

( )( )

MPa 4502

MPa 352

MPa 3402

failal

allowal

failst

allowst

( ) ( )

( ) ( ) (2) 075.02 ;0

(1) 0225.1 ;0

=−=+

∑∑

EXAMPLE 2 (cont)

• 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

28.106==P

( ) ( ) ( )[ ] kN 0.631018001035 66 === −BallowalB AF σ

( )( )kN 168

20.63==P

EXAMPLE 2 (cont)

• 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

25.114==P

(Ans) kN 168=P

EXAMPLE 3 (internal loadings)

EXAMPLE 3 (cont)

EXAMPLE 4 (double shear stress)

EXAMPLE 4 (cont)

EXERCISES

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

Normal & Shear Stress - Chapter 1

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Transcript of Normal & Shear Stress - Chapter 1

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