MEE 320 Problems Chapter 7, Spring 2018

SO OLUTION F    F    60 MPa 90 M 608 PRO For th the o based  0: A  9   180sin30 c   0: 90   A 2  90(cos 30  MPa OBLEM 7.2 he given state blique face of d on the equilib 90 sin30 cos A  cos30 60co   0 sin 30 sin 3  A 2  sin 30 )   2 e of stress, det f the shaded t brium of that e s30 90 co A   2 os 30 30 90 cos   A 60cos30 sin  termine the no triangular elem element, as wa s30 sin30    30 cos30    30 ormal and she ment shown. U as done in the 60 cos30 c A   60 cos30 sin  A earing stresses Use a method derivations of cos30   0 3   n 30   0 7   s exerted on d of analysis f Sec. 7.1A. 32.9 MPa  71.0 MPa  PROBLEM 7.6 For the given state of stress, determine (a) the principal planes, (b) the principal stresses. SOLUTION 28 MPa 140 MPa 42 MPa x y xy σ σ τ = = = − (a) 2 (2)( 42) tan 2 = 0.750 28 140 xy p x y τ θ σ σ − = = − − 2 36.87 θp = ° 18.43 ,108.43 θp = ° °  (b) 2 2 max,min 2 2 + 2 2 28 140 28 140 ( 42) 2 2 84 70 x y x y xy σ σ σ σ σ τ + −   = ±     + −   = ± + −     = ±    max 154 MPa σ =   min 14 MPa σ =  PROBLEM 7.1 For the given state of stress, determine (a) the orientation of the planes of maximum in-plane shearing stress, (b) the corresponding normal stress. SOLUTION 63 MPa 42 MPa 28 MPa x y xy σ σ τ = = − = (a) 63 42 tan 2 1.875 2 (2)(28) x y s xy σ σ θ τ − + = − = − = − 2 61.93 s θ = − ° 30.96 , 59.04 s θ = − ° °  (b) 2 2 max 2 x y xy σ σ τ τ −   = +      2 2 63 42 (28) 59.5 MPa 2 +   = + =      (c) ave 63 42 10.5 MPa 2 2 x y σ σ σ σ + − ′ = = = =  2 SO For All 50 m OLUTION r plane a-a,  lowable value P mm a PR Two a-a stres cent   65 .  0 s s x P P           of P is the sm a 25� OBLEM 7.2 o members of that forms an sses for the glu tric load P that 2 2 0, 0, cos (50 sin 65 ( )sin sin 65 cos65 xy x y x y A A                  maller one. 22 uniform cross n angle of 25 ued joint are  t can be applie 2 3 2 3 sin 2 10 )(80 1 sin 6 cos ( (50 10 )( y y xy xy P A                sin s section 50  5 with the h   800 kPa a ed. 3 3 2 2 sin cos 0 )(800 10 65 (cos sin (8          0 10 3 )(60 n 65 cos65     80 mm are g horizontal. Kn and 600 k   2 3 3 0 sin 65 ) 3.90 10 ) sin65 P A P A        3 0 10 ) 6.27   glued together nowing that th kPa, determin 5 0 N cos65 0     3 7 10 N  P  r along plane he allowable ne the largest  3.90 kN  PROBLEM 7.2 A mechanic uses a crowfoot wrench to loosen a bolt at E. Knowing that the mechanic applies a vertical 100 N force at A, determine the principal stresses and the maximum shearing stress at point H located as shown on top of the 18 mm diameter shaft. SOLUTION Equivalent force-couple system at center of shaft in section at point H. 100 N (100)(150) 15000 N mm (100)(250) 25000 N mm V M T = = = ⋅ = = ⋅ Shaft cross section: 4 4 4 1 18 mm, 9 mm 2 1 10306 mm 5153 mm 2 2 d c d J c I J π = = = = = = = Torsion: 2 (25000)(9) 21.8 N/mm 21.8 MPa 10306 Tc τ = J = = = Bending: 2 (15000)(9) 26.2 N/mm 26.2 MPa 5153 Mc I σ = = = = Transverse shear: At point H stress due to transverse shear is zero. Resultant stresses: ave 2 2 2 2 26.2 MPa, 0, 21.8 MPa 1 ( ) 13.1MPa 2 13.1 21.8 25.4 MPa 2 x y xy x y x y xy R σ σ τ σ σ σ σ σ τ = = = = + = −   = + = + =     ave 38.5 MPa a R σ = σ + =  ave 12.3 MPa b R σ = σ − = −   max 25.4 MPa R τ = =   5 SO Forc Tor 10 kN LUTION ce-couple syst sion: At po 200 mm 6 m z N C tem at center o oint K, place l mm 1 T A y K H A B P Th th de at  4 1 2 2 4.1855 1 2 o o o d r J r I J        of tube in the p 10 kN 10 10 (10 1 2000 N (10 1500 x y z F M M            local x-axis in 51 ( 24.37 24.37 y o xy T M c r Tc J           150 mm 51 mm x D PROBLEM 7 he steel pipe A hickness. Kno etermine the p t point K.  4 6 4 6 102 51 mm 2 4.1855 5 10 m 2.0927 10 m ir        plane containi 3 3 3 3 0 N 0 )(200 10 N m 10 )(150 10 0 N m       negative glob 3 6 6 2000 N m 1 10 m (2000)(51 10 4.1855 10 10 Pa MPa       7.26 AB has a 102 wing that arm principal stres 6 4 4 5 10 mm m i o r r t     ing points H a 3 3 ) )  bal z-direction 3 6 0 )  2-mm outer di m CD is rigid sses and the m 45 mm  and K: n. ameter and a dly attached t maximum she 6-mm wall to the pipe, aring stress PROBLEM 7.26 (Continued) Transverse shear: Stress due to transverse shear x V  F is zero at point K. Bending: 3 6 6 | | (1500)(51 10 ) | | 36.56 10 Pa 36.56 MPa 2.0927 10 z y M c I           Point K lies on compression side of neutral axis. 36.56 MPa  y   Total stresses at point K: ave 2 2 0, 36.56 MPa, 24.37 MPa 1 ( ) 18.28 MPa 2 30.46 MPa 2 x y xy x y x y xy R                            max ave 18.28 30.46 R        max 12.18 MPa    min ave 18.28 30.46 R        min 48.7 MPa      max R    max 30.5 MPa    SO Plo (a) OLUTION otted points fo   25  90 MP 3   r Mohr’s circl : : : X Y C tan 2  p 2  p R  . 2 5   2       x      x y     y Pa 30 MPa 60 MPa PROBLEM Solve Prob. PROBLEM normal and s (a) 25 clock ave 60 MP 90 MPa, 30 MPa 2 x y xy x y             le: : ( 60 MPa, : (90 MPa, 30 : (15 MPa, 0)   30 0 75    FX FC 21.80    P FC2 FX   50 2 2 50 P      ave cos    R R sin ave cos    R M 7.36 7.14, using M M 7.13 throug shearing stres kwise, (b) 10 Pa, , 15 MPa  30 MPa) 0 MPa)  0.4 10.90  2 752 30 X   21.80 28     Mohr’s circle. gh 7.16 For th ses after the e counterclock 02 80.78 MP  8.20 he given state element shown kwise. Pa e of stress, de n has been rota    5 x     3 x y 8 y    etermine the ated through 56.2 MPa  38.2 MPa  86.2 MPa  (b)   10  . 2 20   2        x     x y R     y PROB 0 2 21.8     p ave cos    R sin R ave cos    R BLEM 7.36 ( 80 20 41     (Continued) .80 d)    45 x     53 x y    75 y 5.2 MPa  3.8 MPa  5.2 MPa  PROBLEM 7.44 Solve Prob. 7.22, using Mohr’s circle. PROBLEM 7.22 Two members of uniform cross section 50 80 mm are glued together along plane a-a that forms an angle of 25 with the horizontal. Knowing that the allowable stresses for the glued joint are   800 kPa and   600 kPa, determine the largest centric load P that can be applied. SOLUTION 3 3 3 2 0 0 / (50 10 )(80 10 ) 4 10 m (1 cos50 ) 2 2 1 cos50 x xy y P A A P A A P                       3 3 3 (2)(4 10 )(800 10 ) 1 cos50 3.90 10 N P P         3 3 3 2 (2)(4 10 )(600 10 ) sin50 6.27 10 N 2 sin50 sin50 P A P A              Choosing the smaller value, P  3.90 kN  P a 25� 50 mm a PROBLEM 7.55 Determine the principal planes and the principal stresses for the state of plane stress resulting from the superposition of the two states of stress shown. SOLUTION Consider the state of stress on the left. We shall express it in terms of horizontal and vertical components. Principal axes and principal stress: ave 1 (118.3 56.7) 87.5 2      1 (118.3 56.7) 30.8 2 2       y x 2 2 (30.8) (75) 81.08    R 75 tan 2 2 67.67 30.8      p p 33.8    p , and 123.8  ave max 87.5 81.08       R max 168.6 MPa     in ave m 87.5 81.08       R min 6.42 MPa     100 MPa 50 MPa 50 MPa 75 MPa + 308 50cos30 43.30 43.30 50sin30 25.0 x y xy            PROBLEM 7.68 For the state of stress shown, determine the maximum shearing stress when (a)  y  40 MPa, (b)  y 120 MPa. (Hint: Consider both in-plane and out-of-plane shearing stresses.) SOLUTION (a) ave 140 MPa, 40 MPa, 80 MPa 1 ( ) 90 MPa 2 x y xy x y             2 2 2 2 50 80 94.34 MPa 2 x y xy R                ave ave 184.34 MPa (max) 4.34 MPa (min) 0 a b c R R              max(in-plane) 1 ( ) 94.34 MPa 2 a b R        max max min 1 1 ( ) ( ) 94.3 MPa 2 2 a b           max 94.3 MPa    (b) ave 140 MPa, 120 MPa, 80 MPa 1 ( ) 130 MPa 2 x y xy x y             2 2 2 2 10 80 80.62 MPa 2 x y xy R                ave ave 210.62 MPa (max) 49.38 MPa 0 (min) a b c R R             max min 210.62 MPa 0 a c         max(in-plane) 86.62 MPa R    max max min 1 ( ) 105.3 MPa 2       max 105.3 MPa    80 MPa y z x 140 MPa σy PROBLEM 7.71 For the state of stress shown, determine the maximum shearing stress when (a) z  0, (b) z  60 MPa, (c) z  60 MPa. SOLUTION ave 1 (170 100) 135 2     1 1 ( ) (170 100) 35 2 2       x y 2 2 (35) (84) 91    R 135 91 226 MPa     A 135 91 44 MPa     B (a) 0.  z  Point Z corresponding to the z axis is located at O, outside the circle drawn through A and B. The largest of the 3 Mohr’s circles is the circle through O and A. We have max 1 1 1 ( ) (226) 2 2 2     A  OA max 113.0 MPa    (b) 60 MPa.  z   Point Z is located between B and A. The largest of the 3 circles is the one drawn through A and B. max 91.0 MPa   R   (c) 60 MPa.  z   Point Z is located outside the circle drawn through A and B. The largest of the 8 Mohr’s circles is the circle through Z and A. We have max 1 1 ( ) (60 226) 2 2     ZA max 143.0 MPa    z x 84 MPa y 170 MPa 100 MPa z PROBLEM 7.7 For the state of stress shown, determine the value of τxy for which the maximum shearing stress is (a) 63 MPa, (b) 84 MPa. SOLUTION ave 105 MPa 42 MPa 1 ( ) 73.5 MPa 2 31.5 MPa 2 x y x y x y U σ σ σ σ σ σ σ = = = + = − = = τ (MPa) (a) For max 63 MPa τ = Center of Mohr’s circle lies at point C. Lines marked (a) show the limits on max. τ Limit on σmax is max 2 max 126 MPa. σ = τ = For the Mohr’s circle max σa = σ corresponds to point Aa. ave 2 2 2 2 2 2 126 73.5 52.5 MPa 52.5 31.5 42 MPa a xy xy R R U R U σ σ τ τ = − = − = = + = ± − = ± − = ±  (b) For max 84 MPa τ = Center of Mohr’s circle lies at point C. R = 84 MPa 2 2 78.7 MPa xy R U τ = ± − = ±  Checking 73.5 84 157.5 MPa 73.5 84 10.5 MPa 0 a b c σ σ σ = + = = − = − = max max min 1 ( ) 84 MPa O.K. 2 τ σ σ = − =  5 PROBLEM 7.79 For the state of stress shown, determine two values of y  for which the maximum shearing stress is 80 MPa. SOLUTION 90 MPa, 0, 60 MPa x z xz       Mohr’s circle of stresses in zx plane: ave 2 2 2 2 1 ( ) 45 MPa 2 45 60 75 MPa 2 x z x y zx R                      ave ave 120 MPa, 30 MPa a b R R            Assume max 120 MPa. a     min max max 2 120 (2)(80) y          40.0 MPa  y    Assume min 30 MPa. b      max min max 2 30 (2)(80) y           130.0 MPa  y   z y x σy 90 MPa 60 MPa