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2597358439

# 2597358439 - 134 CHAPTER 2 Axially Loaded Members Problem...

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Problem 2.6-16 A prismatic bar is subjected to an axial force that produces a tensile stress 63 MPa and a shear stress 21 MPa on a certain inclined plane (see figure). Determine the stresses acting on all faces of a stress element oriented at 30° and show the stresses on a sketch of the element. Solution 2.6-16 Bar in uniaxial stress 134 CHAPTER 2 Axially Loaded Members 21 MPa 63 MPa 63 MPa 21 MPa I NCLINED PLANE AT ANGLE x cos 2 63 MPa x cos 2 (1) (2) Equate (1) and (2): or From (1) or (2): x 70.0 MPa (tension) tan u 21 63 1 3 u 18.43 63 MPa cos 2 u 21 MPa sin u cos u s x 21 MPa sin u cos u 21 MPa s x sin u cos u t u s x sin u cos u s x 63 MPa cos 2 u S TRESS ELEMENT AT 30 Plane at 30 90 120 N OTE : All stresses have units of MPa. 30.31 MPa t u ( 70 MPa)(sin 120 )(cos 120 ) s u (70 MPa)(cos 120 ) 2 17.5 MPa 30.31 MPa ( 70 MPa)(sin 30 )(cos 30 ) t u s x sin u cos u 52.5 MPa s u s x cos 2 u (70 MPa)(cos 30 ) 2 30 ° 17.5 30.31 52.5 30.31 52.5 y x 0

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Problem 2.6-17 The normal stress on plane pq of a prismatic bar in tension (see figure) is found to be 7500 psi. On plane rs , which makes an angle 30° with plane pq , the stress is found to be 2500 psi. Determine the maximum normal stress max and maximum shear stress max in the bar. Solution 2.6-17 Bar in tension SECTION 2.6 Stresses on Inclined Sections 135 q p r P P s Eq. (2-29a): x cos 2 30 P LANE pq : 1 x cos 2 1 1 7500 psi P LANE rs : 2 x cos 2 ( 1 ) 2 2500 psi Equate x from 1 and 2 : (Eq. 1) or (Eq. 2) cos 2 u 1 cos 2 ( u 1 b ) s 1 s 2 cos u 1 cos( u 1 b ) B s 1 s 2 s x s 1 cos 2 u 1 s 2 cos 2 ( u 1 b ) S UBSTITUTE NUMERICAL VALUES INTO E Q . (2): Solve by iteration or a computer program: 1 30 M AXIMUM NORMAL STRESS ( FROM E Q . 1) M AXIMUM SHEAR STRESS t max s x 2 5,000 psi 10,000 psi s max s x s 1 cos 2 u 1 7500 psi cos 2 30 cos u 1 cos( u 1 30 ) B 7500 psi 2500 psi 3 1.7321 q p r P P s
Problem 2.6-18 A tension member is to be constructed of two pieces of plastic glued along plane pq (see figure). For purposes of cutting and gluing, the angle must be between 25° and 45°. The allowable stresses on the glued joint in tension and shear are 5.0 MPa and 3.0 MPa, respectively. (a) Determine the angle so that the bar will carry the largest load P . (Assume that the strength of the glued joint controls the design.) (b) Determine the maximum allowable load P max if the cross-sectional area of the bar is 225 mm 2 . Solution 2.6-18 Bar in tension with glued joint 136 CHAPTER 2 Axially Loaded Members q p P P 25 45 A 225 mm 2 On glued joint: allow 5.0 MPa allow 3.0 MPa A LLOWABLE STRESS x IN TENSION (1) x sin cos Since the direction of is immaterial, we can write: | | x sin cos or (2) G RAPH OF E QS . (1) AND (2) s x t u sin u cos u 3.0 MPa sin u cos u s u s x cos 2 u s x s u cos 2 u 5.0 MPa cos 2 u (a) D ETERMINE ANGLE FOR LARGEST LOAD Point A gives the largest value of x and hence the largest load. To determine the angle corresponding to point A , we equate Eqs. (1) and (2).

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