Stress Transformation Equations: u = +135 (Fig. a) s x = 80 MPa s y = 0 t xy = 45 MPa. we obtain, cos u + t xy sin 2u. s x = s x + s y.


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1 014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 9 7. Determine the normal stress and shear stress acting on the inclined plane AB. Solve the problem using the stress transformation equations. Show the result on the sectioned element. Stress Transformation Equations: MPa B 80 MPa u = +135 (Fig. a) s x = 80 MPa s y = 0 t xy = 45 MPa A we obtain, s x = s x + s y + s x  s y cos u + t xy sin u = cos sin 70 = 5 MPa s x  s y t x y =  sinu + t xy cos u = sin cos 70 = 40 MPa The negative sign indicates that s x is a compressive stress. These results are indicated on the triangular element shown in Fig. b. s x = 5 MPa, t x y = 40 MPa 84
2 014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently The state of stress at a point is shown on the element. Determine (a) the principal stress and (b) the maximum shear stress and average normal stress at the point. Specify the orientation of the element in each case. 60 MPa 30 MPa 45 MPa s x = 45 MPa s y = 60 MPa t xy = 30 MPa a) s 1, = s x + s y s x  s y ; A a b + txy = ; A a 45  (60) b + (30) s 1 = 53.0 MPa s = MPa Orientation of principal stress: tan u p = t xy (s x  s y )> = 30 (45  (60))> = u p = 14.87, Use Eq. 9 1 to determine the principal plane of and : s 1 s s x = s x + s y + s x  s y cos u + t xy sin u, where u = = 45 + (60) (60) cos sin 9.74 = 53.0 MPa Therefore u p1 = 14.9 and u p = b) t = maxin  plane A a s x  s y b (60) txy = A a b + 30 = 60.5 MPa s avg = s x + s y = 45 + (60) = MPa Orientation of maximum shear stress: tan u y = (s x  s y )> t xy = (45  (60))> 30 = u y = and u y = 59.9 By observation, in order to preserve equilibrium along AB, direction shown. t max has to act in the s 1 = 53.0 MPa, s = MPa, u p1 = 14.9 and u p = 75.1, s avg = MPa, = 60.5 MPa, u s = and 59.9 t max 850
3 014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently A point on a thin plate is subjected to the two successive states of stress shown. Determine the resultant state of stress represented on the element oriented as shown on the right MPa 60 MPa 30 s y t xy s x 85 MPa For element a: s x = s y = 85 MPa t xy = 0 u = 45 (s x ) a = s x + s y + s x  s y cos u + t xy sin u = cos (90 ) + 0 = 85 MPa (s y ) a = s x + s y + s x  s y cos u  t xy sin u = cos (90 )  0 = 85 MPa (t x y ) a =  s x  s y sin u + t xy cos u =  For element b: sin (90 ) + 0 = 0 s x = s y = 0 t xy = 60 MPa u = 60 (s x ) b = s x + s y + s x  s y cos u + t xy sin u = sin (10 ) = MPa (s y ) b = s x + s y  s x  s y cos u  t xy sin u = sin (10 ) = MPa (t x y ) b =  s x  s y =  sin u  t xy cos u sin (10 ) + 60 cos (10 ) = 30 MPa s x = (s x ) a + (s x ) b = 85 + (51.96) = 33.0 MPa s y = (s y ) a + (s y ) b = = 137 MPa t xy = (t x y ) a + (t x y ) b = 0 + (30) = 30 MPa s x = 33.0 MPa, s y = 137 MPa, t xy = 30 MPa 855
4 014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 9 5. The wooden block will fail if the shear stress acting along the grain is 550 psi. If the normal stress s x = 400 psi, determine the necessary compressive stress s y that will cause failure. s y 58 s x 400 psi t x y = a s x  s y b sin u + t xy cos u 550 = a s y b sin s y = 84 psi s y = 84 psi 863
5 014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 9 7. The bracket is subjected to the force of 3 kip. Determine the principal stress and maximum shear stress at point B on the cross section at section a a. Specify the orientation of this state of stress and show the results on elements. 3 kip a 3 in. 3 kip a A 0.5 in. Internal Loadings: Consider the equilibrium of the freebody diagram of the bracket s left cut segment, Fig. a. : + F x = 0; N  3 = 0 N = 3 kip M O = 0; 3(4)  M = 0 M = 1 kip # in 0.5 in. in. B 0.5 in. 1 in. Section a a Normal and Shear Stresses: The normal stress is the combination of axial and bending stress. Thus, The cross  sectional area and the moment of inertia about the z axis of the bracket s cross section is For point B, y = 1 in. Then s B = s = N A  My I A = 1() (1.5) = in I = 1 1 (1)A3 B (0.75)A1.53 B = in (1)(1) = .90 ksi Since no shear force is acting on the section, t B = 0 The state of stress at point A can be represented on the element shown in Fig. b. In  Plane Principal Stress: s x = .90 ksi, s y = 0, and t xy = 0. Since no shear stress acts on the element, s 1 = s y = 0 s = s x = .90 ksi The state of principal stresses can also be represented by the elements shown in Fig. b. Maximum In  Plane Shear Stress: t max = s x  s y C + t xy = a b + 0 = 11.5 ksi B Orientation of the Plane of Maximum In  Plane Shear Stress: tan u s =  As x  s y B> (.90)> =  = q t xy 0 u s = 45 and
6 014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 9 7. Continued Substituting u = 45 into t x y =  s x  s y = sin u + t xy cos u sin = 11.5 ksi = t max This indicates that is directed in the positive sense of the y axes on the t max element defined by u s = 45. Average Normal Stress: s avg = s x + s y = = ksi The state of maximum in  plane shear stress is represented by the element shown in Fig. c. s 1 = 0, s = .90 ksi, t max = 11.5 ksi, u s = 45 and
7 014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently The clamp bears down on the smooth surface at E by tightening the bolt. If the tensile force in the bolt is 40 kn, determine the principal stress at points A and B and show the results on elements located at each of these points. The crosssectional area at A and B is shown in the adjacent figure. 300 mm 50 mm Support Reactions: As shown on FBD(a). E B 100 mm A 100 mm 50 mm 30 mm B A 5 mm Internal Forces and Moment: As shown on FBD(b). Section Properties: I = 1 1 (0.03) A0.053 B = A106 B m 4 Q A = 0 Q B = y A =0.015(0.05)(0.03) = 9.375A106 B m 3 Normal Stress: Applying the flexure formula s =  My. I s A = .40(103 )(0.05) 0.315(106 ) = 19 MPa s B = .40(103 )(0) 0.315(106 ) = 0 Shear Stress: Applying the shear formula t A = 4.0(10 3 )(0) 0.315(106 )(0.03) = 0 t B = 4.0(103 )C9.375(106 )D 0.315(106 )(0.03) t = VQ It = 4.0 MPa InPlane Principal Stresses: s x = 0, s y = 19 MPa, and t xy = 0 for point A. Since no shear stress acts on the element. s 1 = s x = 0 s = s y = 19 MPa s x = s y = 0 and t xy = 4.0 MPa for point B. Applying Eq. 95 s 1, = s x + s y ; a s x  s y b + t C xy = 0 ; 0 + (4.0) = 0 ; 4.0 s 1 = 4.0 MPa s = 4.0 MPa 875
8 014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently Continued Orientation of Principal Plane: Applying Eq. 94 for point B. tan u p = t xy As x  s y B> = 4.0 = q 0 u p = and 45.0 Substituting the results into Eq. 91 with u = yields s x = s x + s y + s x  s y cos u + t xy sin u = [4.0 sin (90.0 )] = 4.0 MPa = s 1 Hence, u p1 = u p = 45.0 Point A: s 1 = 0, s = 19 MPa, u p1 = 0, u p = 90, Point B: s 1 = 4.0 MPa, s = 4.0 MPa, u p1 = 45.0, u p =
9 014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 9 4. The drill pipe has an outer diameter of 3 in., a wall thickness of 0.5 in., and a weight of 50 lb>ft. If it is subjected to a torque and axial load as shown, determine (a) the principal stresses and (b) the maximum shear stress at a point on its surface at section a. 800 lb ft 1500 lb 0 ft a Internal Forces and Torque: As shown on FBD(a). 0 ft Section Properties: A = p 4 A3 .5 B = p in J = p A B = in 4 Normal Stress: s = N A = 500 = psi p Shear Stress: Applying the torsion formula. t = T c J = 800(1)(1.5) = psi a) InPlane Principal Stresses: s x = 0, s y = psi and t xy = psi for any point on the shaft s surface. Applying Eq. 95, s 1, = s x + s y ; a s x  s y b + t C xy = 0 + ( ) ; a 0  ( ) b + (3497.5) C = ; s 1 = 966 psi =.97 ksi s = 414 psi = 4.1 ksi b) Maximum InPlane Shear Stress: Applying Eq. 9 7, t max = a s x  s y b + t C xy = 0  ( ) + (3497.5) C = 3545 psi = 3.55 ksi s 1 =.97 ksi, s = 4.1 ksi, t max = 3.55 ksi 886
10 014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently Determine the equivalent state of stress if an element is oriented 5 counterclockwise from the element shown. 550 MPa A(0, 550) B(0, 550) C(0, 0) R = CA = CB = 550 s x = 550 sin 50 = 41 MPa t x y = 550 cos 50 = 354 MPa s y = 550 sin 50 = 41 MPa s x = 41 MPa, t x y = 354 MPa, s y = 41 MPa 90
11 014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently Draw Mohr s circle that describes each of the following states of stress. 5 MPa 0 ksi 5 MPa 0 ksi (a) (b) 18 MPa (c) 905
12 014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently The thinwalled pipe has an inner diameter of 0.5 in. and a thickness of 0.05 in. If it is subjected to an internal pressure of 500 psi and the axial tension and torsional loadings shown, determine the principal stress at a point on the surface of the pipe. 00 lb 0 lb ft 0 lb ft 00 lb Section Properties: A = pa B = p in J = p A B =.84768A103 B in 4 r Normal Stress: Since, thin wall analysis is valid. t = = 10 s long = N A + pr t = p + 500(0.5) = ksi (0.05) s hoop = pr t = 500(0.5) 0.05 = 5.00 ksi Shear Stress: Applying the torsion formula, t = Tc J = 0(1)(0.75).84768(103 = 3.18 ksi ) Construction of the Circle: In accordance with the sign convention s x = ksi, s y = 5.00 ksi, and t xy = ksi. Hence, s avg = s x + s y = = ksi The coordinates for reference points A and C are A(7.350, ) C(6.175, 0) The radius of the circle is R = ( ) = ksi InPlane Principal Stress: The coordinates of point B and D represent and, respectively. s 1 s s 1 = = 9.4 ksi s = = ksi s 1 = 9.4 ksi, s = ksi 910
13 014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently If the box wrench is subjected to the 50 lb force, determine the principal stress and maximum shear stress at point B on the cross section of the wrench at section a a. Specify the orientation of these states of stress and indicate the results on elements at the point. in. 1 in. 0.5 in. 50 lb a a A B Section a a Internal Loadings: Considering the equilibrium of the freebody diagram of the wrench s cut segment, Fig. a, F y = 0; Mx = 0; M z = 0; V y + 50 = 0 T + 50(1) = 0 Mz  50() = 0 V y = 50 lb T = 600 lb # in M z = 100 lb # in Section Properties: The moment of inertia about the z axis and the polar moment of inertia of the wrench s cross section are I z = p 4 (0.54 ) = p in 4 J = p (0.54 ) = p in 4 Referring to Fig. b, (Q y ) B = 0 Normal and Shear Stress: The normal stress is caused by the bending stress due to M z. (s x ) B =  M zy B =  100(0.5) = ksi I z p The shear stress at point B along the y axis is (t xy ) B = 0 since (Q y ) B. However, the shear stress along the z axis is caused by torsion. (t xz ) B = Tc J = 600(0.5) = ksi p The state of stress at point B is represented by the twodimensional element shown in Fig. c. 96
14 014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently Continued Construction of the Circle: s x = ksi, s z = 0, and t xz = ksi. Thus, s avg = s x + s y = = ksi The coordinates of reference point A and the center C of the circle are A(1.019, ) C( , 0) Thus, the radius of the circle is R = CA = [ ( )] + (3.056) = ksi Using these results, the circle is shown in Fig. d. InPlane Principal Stress: The coordinates of reference points B and D represent and, respectively. s 1 s s 1 = =.59 ksi s = = ksi Maximum InPlane Shear Stress: The coordinates of point E represent the maximum stress, Fig. a. t max = R = 3.10 ksi s 1 =.59 ksi, s = ksi, u p1 = 40.3, u p = 49.7, t = 3.10 ksi,u s = 4.73 max 97
15 014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently The solid propeller shaft on a ship extends outward from the hull. During operation it turns at v = 15 rad>s when the engine develops 900 kw of power. This causes a thrust of F = 1.3 MN on the shaft. If the shaft has a diameter of 50 mm, determine the maximum shear stress at any point located on the surface of the shaft m A T F Power Transmission: Using the formula developed in Chapter 5, P = 900 kw = 0.900A10 6 B N # m>s T 0 = P v = 0.900(106 ) 15 = 60.0A10 3 B N # m Internal Torque and Force: As shown on FBD. Section Properties: A = p 4 A0.5 B = p m J = p A0.154 B = A103 B m 4 Normal Stress: s = N A = 1.3(106 ) p Shear Stress: Applying the torsion formula. = MPa t = Tc J = 60.0(103 ) (0.15) (103 = MPa ) Maximum InPlane Principal Shear Stress: s x = MPa, s y = 0, and t xy = MPa for any point on the shaft s surface. Applying Eq. 97, t max = a s x  s y b + t C xy = a b + (19.56) C = 3. MPa t max = 3. MPa 94
16 014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently Determine the equivalent state of stress if an element is oriented 40 clockwise from the element shown. Use Mohr s circle. 10 ksi 6 ksi A(6, 0) B(10, 0) C(, 0) R = CA = CB = 8 s x = cos 80 = ksi t x y = 8 sin 80 = 7.88 ksi s y = cos 80 = ksi s x = ksi, t x y = 7.88 ksi,s y = ksi 945
17 014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently The crane is used to support the 350lb load. Determine the principal stresses acting in the boom at points A and B. The cross section is rectangular and has a width of 6 in. and a thickness of 3 in. Use Mohr s circle. 5 ft 3 in. A B 5 ft A = 6(3) = 18 in I = 1 1 (3)(63 ) = 54 in 4 Q B = (1.5)(3)(3) = 13.5 in 3 Q A = 0 For point A: s A =  P A  My I = (1)(3) 54 = 100 psi t A = 0 s 1 = 0 s = 100 psi = 1.0 ksi For point B: s B =  P A = = psi t B = VQ B It = 47.49(13.5) 54(3) = 0.6 psi A(33.19, 0.6) B(0, 0.6) C( , 0) R = = 6.47 s 1 = = 9.88 psi s = = psi Point A: s 1 = 0, s = 1.0 ksi, Point B: s 1 = 9.88 psi,s = psi 946
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