Solution for Homework Problem on stress concentration (H6) - F22.pdf - with grade points

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1 Prof. K.A. Narh ME-315-101: Fall 2022 Solution of Problems on Stress Concentration (H6) (1) (a) Stress at section AB: (i) ( )( ) 4300 9556 psi 3 0.15 9556 psi is below proportional limit, 55,000 psi, therefore, the stress is in elastic zone AB AB AB P BC P A = = = = = (ii) Stress at section BC: ( )( ) 4300 28,667 psi 1 0.15 28,667 psi < ; 55,000 psi. Therefore, the stress is in elastic zone BC BC BC P P BC P A = = = = = (b) Stress concentration factor K = 2.9 (given) ( ) ( )( ) ( ) max max max where 28,667 psi i 2.9 28,667 83,133 psi ii Since 83,133 psi > 55,000 psi,the material in the region of the fillet becomes plastic. ave ave BC ave K K = = = = = = = 55000 psi 1 (2) 1 1 1 (2) 1 1 (2) (25) Final grade
2 Problem 2 For this problem, we use the graph below, also given at the end of the notes on stress concentration. Solution To use the above graph, the two ratios w/d and r/d must first be computed; w is the width of the bar, and r is the radius of the groove. The net width, d , equals the width minus twice the groove radius. So ( ) 2.6 2 0.3 2. d = = Therefore, 2.6 0.3 1.3 and 0.15 2 2 w r d d = = = = To determine K from the graph we find the curve corresponding to w/d = 1.3, and find r/d = 0.15 on the horizontal axis. However, since there is no curve for w/d = 1.3, we use interpolation procedure as follows: Given: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 1 2 1 2 3 1 1 2 3 1 1 3 1 3 1 3 1 3 1 To find and : x = ; y = or 1.3 1.2 2.15 2.18 1.5 1.2 2.25 2.15 + + = = x y y y x x y y x x x x x y y y y y x x y y x x K K D/d K t x 1 (1.2) y 1 (2.15) x 2 (1.3) y 2 (K ) x 3 (1.5) y 3 (2.25) (4)
3 (3) ( )( ) ( )( ) 2 max 1.5 0.375 4 Implies that =2.27 from the graph 1 in Fig. 3, 0.04 0.015 0.015 0.000375 m 5000 13.333 MPa. 0.000375 Then, =K 2.27 13.333 MPa Maximum stress at large hole : = = = = = = = = = = ave n n ave ave d w K P A A w d t ( )( ) ( ) 2 30.27 MPa 2 / 3 0.2 (10 / 3) Implies that =2.48 from the graph 1 in Fig. 3. where = net area 10 2 0.015 0.0004 m 3 3 5000 N 5000 12.5 MPa. 0.0004 Maximum stress at small hole : = = = = = = = = = ave n n n ave d w K P A A A w d t P ( )( ) a t m x Then, =K 2.48 12.5 MPa 31 MPa 5000 5000 =10 MPa 10 / 3 0.015 0.000 Stress concentration factor K from Figure 4) r = 1/3 cm d = 10/3 cm 5 ( D Maximum stress at the fillet : = = = = = ave ave P d t t = 4 cm To determine K from the graph we find the curve corresponding to D/d = 1.2, and find r/d = on the horizontal axis. However, we don't have a curve for D/d = 1.2, s 4 (10 / 3) 1.2 0.1 0.1 = = = D d r d ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 1 2 1 2 3 1 1 2 3 1 1 3 1 3 1 3 1 3 1 To find and by interpolation we use one of the following Equations: x = ; y = or Using the o we use an interpolation procedue as follows: + + = x y y y x x y y x x x x x y y y y y x x y y x x ( ) ( ) ( ) ( ) 6 max 0 e above Equation: 1.2 1.05 1.5 0.099 1.5 1.5 0.2475 1.7 i 475 1.75 1 Max str ss d 1 ue at f lleted .45 1.05 2.16 1.5 0.4 (1.75)(10 0 ) 17.50 M area: P The maximum stress occurs at the side of = = = = = + = = = t t t t t K K K K K the small hole. 5000 N 5000 N 4 cm 10/3 cm 2 2 (4) 2 2 (4) 2 2 (4) (3)
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