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csueb mpa

csueb mpa

Description


Where does the average normal stress occur?




What is the average shear stress in the internal vertical surface AB (or CD), if F=20kN, and AAB=ACD=1000mm²?




What is the normal stress in the bar if P=10 kN and 500mm²?



Chapter Objectives ✔ Understand the concepts of normal and shear stress ✔ Analyze and design of members subjected on axial load  or shear Copyright © 2011 Pearson Education South Asia Pte LtdIn-class Activities 1. Check homework, if any 2. ReadingDon't forget about the age old question of sfsu geography
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Quiz 3. Applications 4. Normal Stress 5. Shear Stress 6. Concept Quiz Copyright © 2011 Pearson Education South Asia Pte LtdREADING 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 LtdREADING 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 LtdAPPLICATIONS  Where does the average normal stress occur?Copyright © 2011 Pearson Education South Asia Pte Ltd Copyright © 2011 Pearson Education South Asia Pte LtdAPPLICATIONS (cont)  Will the total shear force over the anchor length be equal to  the total tensile force σtensile A in the bar?Copyright © 2011 Pearson Education South Asia Pte Ltd APPLICATIONS  Copyright © 2011 Pearson Education South Asia Pte LtdAVERAGE NORMAL STRESS Will the total shear force over the anchor length be equal to  the total tensile force σtensile A in the bar? σ =Copyright © 2011 Pearson Education South Asia Pte Ltd P A NORMAL SHEAR STRESS Δ Fz τ= lim Δ F x σ lim =Δ →0 zΔ zx Δ → A 0 Δ A A A τ Δ = lim F y zy Δ → A 0 Δ A Copyright © 2011 Pearson Education South Asia Pte Ltd EXAMPLE 1  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.Copyright © 2011 Pearson Education South Asia Pte Ltd EXAMPLE 1 (cont) Solutions• By inspection, different sections have different internal forces. • Graphically, the normal force diagram is as shown. Copyright © 2011 Pearson Education South Asia Pte Ltd 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 ( ) PBC σ BC30 103 = = = ( )( )85.7 MPa(Ans) A 0.035 0.01 Copyright © 2011 Pearson Education South Asia Pte Ltd DESIGN OF SIMPLE CONNECTION • For normal force requirement Aσ = P allow • For shear force requirement Aσ =V allow Copyright © 2011 Pearson Education South Asia Pte Ltd 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 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  ( ) = 680 MPa σ st fail ( ) = 70 MPa σ al fail respectively, and the failure shear stress for each  pin is , determine the largest load P that can be  = 900 MPa fail τapplied to the bar. Apply a factor of safety of FS=2. Copyright © 2011 Pearson Education South Asia Pte Ltd EXAMPLE 2 (cont) Solutions • The allowable stresses are ( )( ) σ σ s t fail 680 s t allow = = = F S 340 MPa . . ( )( ) 2 σ σ a l fail 70 a l allow = = = F S 35 MPa τ τ fail . . 900 2 allow = = = F S 450 MPa . . 2 • There are three unknowns and we apply the equations of equilibrium, ∑ ( ) ( ) + = − = M P F 0; 1.25 2 0 (1) B AC∑ 0; (2) (0.75) 0 (2) + = − = M F P A B Copyright © 2011 Pearson Education South Asia Pte Ltd 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, • Using Eq. 1, • For block B, ( ) ( ) 340(10 )[ (0.01) ] 106.8 kN 6 2 = σ AC= π = FAC st allow A ( )( )171kN 106.8 2 P = = 1.25 ( ) 35(10 )[1800(10 )] 63.0 kN 6 6 − FBσ a l allowAB • Using Eq. 2, = = = ( )( )168 kN 63.0 2 P = =0.75 Copyright © 2011 Pearson Education South Asia Pte Ltd EXAMPLE 2 (cont) Solutions • For pin A or C, 450(10 )[ (0.009) ] 114.5 kN 6 2 V = FAC=τ allowA = π = ( )( )183 kN 114.5 2 • Using Eq. 1, P = = 1.25 • When P reaches its smallest value (168 kN), it develops the allowable  normal stress in the aluminium block. Hence, P =168 kN (Ans)Copyright © 2011 Pearson Education South Asia Pte Ltd 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

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