VERTICAL ARM AB M B. MAXIMUM STRESSES occur on opposite sides of the vertical arm. MAXIMUM TENSILE STRESS. s t P A M(d 2 2)

Transcription

1 54 CHER 8 pplications of lane Stress Combined Loadings he problems for Section 8.5 are to be solved assuming that the structures behave linearl elasticall and that the stresses caused b two or more loads ma be superimposed to obtain the resultant stresses acting at a point. Consider both in-plane and out-of-plane shear stresses unless otherwise specified. roblem bracket CD having a hollow circular cross section consists of a vertical arm, a horizontal arm C parallel to the 0 ais, and a horizontal arm CD parallel to the z 0 ais (see figure). he arms C and CD have lengths b 1 3. ft and b.4 ft, respectivel. he outer and inner diameters of the bracket are d 8.0 in. and d in. vertical load 1500 lb acts at point D. Determine the maimum tensile, compressive, and shear stresses in the vertical arm. z 0 0 b b D 1 C 0 Solution racket CD b 1 b C D VERICL RM M b 1 3. ft b.4 ft 1500 lb CRSS SECIN d 8.0 in. d in. d 1 d 4 (d d 1 ) in. I 64 (d 4 d 1 4 ) in lb M (distance D ) b 1 b (1500 lb)(4.0 ft) 6,000 lb-ft 7,000 lb-in. MXIMUM SRESSES occur on opposite sides of the vertical arm. MXIMUM ENSILE SRESS s t M(d ) I 1500 lb (7,000 lb-in.)(4.0 in.) in in psi psi 3330 psi MXIMUM CMRESSIVE SRESS s c M(d ) 17.3 psi psi I 3590 psi MXIMUM SHER SRESS Uniaial stress. t ma s c 1790 psi

2 SECIN 8.5 Combined Loadings 55 roblem 8.5- gondola on a ski lift is supported b two bent arms, as shown in the figure. Each arm is offset b the distance b 180 mm from the line of action of the weight force W. he allowable stresses in the arms are 100 Ma in tension and 50 Ma in shear. If the loaded gondola weighs 1 kn, what is the mininum diameter d of the arms? W d b (a) W (b) Solution 8.5- Gondola on a ski lift W W d b M Wb W W 1 kn b 180 mm W 6 kn allow 100 Ma (tension) allow 50 Ma Find d min d 4 S d3 3 SUSIUE NUMERICL VLUES: s t 4W s allow 4W (100 Ma) 13, (6 kn) m 8b 1.44 m 13,090 d 3 d (d meters) Solve numericall: d m d min 48.4 mm MXIMUM ENSILE SRESS s t W M S 4W 3 Wb d d 3 or s t 4W d3 d 8b 0 MXIMUM SHER SRESS t ma s t (uniaial stress) Since allow is one-half of allow, the minimum diameter for shear is the same as for tension.

3 56 CHER 8 pplications of lane Stress roblem he hollow drill pipe for an oil well (see figure) is 6.0 in. in outer diameter and 0.75 in. in thickness. Just above the bit, the compressive force in the pipe (due to the weight of the pipe) is 60 k and the torque (due to drilling) is 170 k-in. Determine the maimum tensile, compressive, and shear stresses in the drill pipe. Solution Drill pipe for an oil wall s 60 k 4850 psi in. t (d ) 5864 psi 0 I compressive force torque 60 k 170 k-in. d outer diameter t thickness d 1 inner diameter d 6.0 in. t 0.75 in. d 1 d t 4.5 in. 4 (d d 1 ) in. I 3 (d 4 d 1 4 ) in. 4 SRESSES HE UER SURFCE RINCIL SRESSES s 1, s s t 45 psi (45) (5864) 45 psi 6345 psi psi 8770 psi MXIMUM ENSILE SRESS t 390 psi MXIMUM CMRESSIVE SRESS c 8770 psi MXIMUM IN-LNE SHER SRESS s t ma t 6350 psi NE: Since the principal stresses have opposite signs, the maimum in-plane shear stress is larger than the maimum out-of-plane shear stress.

4 SECIN 8.5 Combined Loadings 57 roblem segment of a generator shaft is subjected to a torque and an aial force, as shown in the figure. he shaft is hollow (outer diameter d 80 mm and inner diameter d 1 30 mm) and delivers 1800 kw at 4.0 Hz. If the compressive force 55 kn, what are the maimum tensile, compressive, and shear stresses in the shaft? robs and Solution Generator shaft s 55 kn Ma m t (d ) (71,60 ˇm)(140 mm) Nˇ I m Ma 0 (Determine the stresses) 55 kn d 80 mm d 1 30 mm 0 power kw f frequenc f 4.0 Hz torque 0 f (Eq. 3-40) W 71,60 (4.0 Hz) Nˇ ˇm 4 (d d 1 ) m I 3 (d 4 d 1 4 ) m 4 SRESSES HE SURFCE F HE SHF RINCIL SRESSES s 1, s s s s t Ma 33.0 Ma Ma 46.3 Ma MXIMUM ENSILE SRESS t Ma MXIMUM CMRESSIVE SRESS c 46.3 Ma MXIMUM IN-LNE SHER SRESS t ma s 1 s s s t 33. Ma NE: Since the principal stresses have opposite signs, the maimum in-plane shear stress is larger than the maimum out-of-plane shear stress.

5 58 CHER 8 pplications of lane Stress roblem segment of a generator shaft of hollow circular cross section is subjected to a torque 0 k-in. (see figure). he outer and inner diameters of the shaft are 8.0 in. and 6.0 in., respectivel. What is the maimum permissible compressive load that can be applied to the shaft if the allowable in-plane shear stress is allow 6500 psi? Solution Generator shaft Units: pounds psi t (d ) (0 k-in.)(4.0 in.) I in. 4 3,01.9 psi (Determine the maimum allowable load ) 0 k-in. d 8.0 in. d in. allow 6500 psi (In-plane shear stress) 4 (d d 1 ) in. I 3 (d 4 d 1 4 ) in. 4 MXIMUM IN-LNE SHER SRESS s s t ma t (6500) ()(1.9911) R (3,01.9) 3,001,700; ,810 lb or ma 49 k s t allow t NE: he maimum in-plane shear stress is larger than the maimum out-of-plane shear stress (in this eample). SRESSES HE SURFCE F HE SHF 0 s in.

6 SECIN 8.5 Combined Loadings 59 roblem clindrical tank subjected to internal pressure p is simultaneousl compressed b an aial force F 7 kn (see figure). he clinder has diameter d 100 mm and wall thickness t 4 mm. Calculate the maimum allowable internal pressure p ma based upon an allowable shear stress in the wall of the tank of 60 Ma. F F Solution Clindrical tank with compressive force F p F 1 F 7 kn p internal pressure d 100 mm t 4 mm allow 60 Ma CIRCUMFERENIL SRESS (ENSIN) s 1 pr p(50 mm) t 4 mm 1.5 p Units: 1 Ma p Ma U-F-LNE SHER SRESSES LNGIUDINL SRESS (ENSIN) s pr t F pr t F rt 7,000 N 6.5p (50 mm)(4 mm) 6.5p Ma Units: Ma p Ma Case : t ma s p; 60 Ma 6.5p Solving, p 9.60 Ma Case 3: t ma s 3.15 p Ma 60 Ma 3.15 p Ma Solving, p Ma CSE, U-F-LNE SHER SRESS GVERNS p ma 9.60 Ma IXIL SRESS IN-LNE SHER SRESS (CSE 1) t ma s 1 s 3.15 p Ma 60 Ma 3.15 p Ma Solving, p Ma roblem clindrical tank having diameter d.5 in. is subjected to internal gas pressure p 600 psi and an eternal tensile load 1000 lb (see figure). Determine the minimum thickness t of the wall of the tank based upon an allowable shear stress of 3000 psi.

7 530 CHER 8 pplications of lane Stress Solution Clindrical tank with tensile load 1000 lb p 600 psi d.5 in. p allow 3000 psi IN-LNE SHER SRESS (CSE 1) t ma s 1 s 3000 psi t Solving, t in t t CIRCUMFERENIL SRESS (ENSIN) s 1 pr (600 psi)(1.5 in) 750 t t t Units: 1 psi t inches psi LNGIUDINL SRESS (ENSIN) s pr t pr t rt lb t (1.5 in.)t 375 t t t U-F-LNE SHER SRESSES Case : t ma s ; t t Solving, t 0.15 in. Case 3: t ma s ; t t Solving, t in. CSE, U-F-LNE SHER SRESS GVERNS t min 0.15 in. IXIL SRESS 1 roblem he torsional pendulum shown in the figure consists of a horizontal circular disk of mass M 60 kg suspended b a vertical steel wire (G 80 Ga) of length L m and diameter d 4 mm. Calculate the maimum permissible angle of rotation ma of the disk (that is, the maimum amplitude of torsional vibrations) so that the stresses in the wire do not eceed 100 Ma in tension or 50 Ma in shear. L = m d = 4 mm ma M = 60 kg

8 SECIN 8.5 Combined Loadings 531 Solution orsional pendulum L d ma M L.0 m d 4.0 mm M 60 kg G 80 Ga allow 100 Ma allow 50 Ma d mm 4 W Mg (60 kg)(9.81 ms ) N RQUE: GI p f ma (EQ. 3-15) L SHER SRESS: t r (EQ. 3-11) GI p f ma t r Gr f ma ( a)f L I L ma 80 ma Units: Ma ma radius ENSILE SRESS W W I p s W Ma RINCIL SRESSES s 1, s s s 1, 3.40 (3.40) 6400 f ma(ma) Note that 1 is positive and is negative. herefore, the maimum in-plane shear stress is greater than the maimum out-of-plane shear stress. MXIMUM NGLE F RIN SED N ENSILE SRESS 1 maimum tensile stress f ma allow 100 Ma 100 Ma 3.40 (3.40) 6400 f ma ( ) (3.40) 6400 f ma ma rad 5.º MXIMUM NGLE F RIN SED N IN-LNE SHER SRESS s s t ma t (3.40) 6400 f ma allow 50 Ma (50) (3.40) 6400 f ma Solving, ma 0.55 rad 31.6º SHER SRESS GVERNS ma 0.55 rad 31.6º s s t 50 (3.40) 6400 f ma 0 t Ma 80 ma (Ma)

9 53 CHER 8 pplications of lane Stress roblem Determine the maimum tensile, compressive, and shear stresses at point on the biccle pedal crank shown in the figure. he pedal and crank are in a horizontal plane and point is located on the top of the crank. he load 160 lb acts in the vertical direction and the distances (in the horizontal plane) between the line of action of the load and point are b in. and b.5 in. ssume that the crank has a solid circular cross section with diameter d 0.6 in. b 1 = 5.0 in. = 160 lb Crank d = 0.6 in. b =.5 in. Solution edal crank 37,730 psi d b 1 b 9431psi 160 lb b in. d 0.6 in. b.5 in. SRESS RESULNS on cross section at point : orque: b 400 lb-in. Moment: M b lb-in. Shear force: V 160 lb SRESSES IN 16 t psi d s M S 3M 3 37,730 psi d (he shear force V produces no shear stresses at point.) 1, 18,860 psi 1,090 psi 1 39,950 psi 30 psi s s t ma t 1,090 psi MXIMUM ENSILE SRESS: t 39,950 psi MXIMUM CMRESSIVE SRESS: c,30 psi MXIMUM IN-LNE SHER SRESS: ma 1,090 psi RINCIL SRESSES ND MXIMUM SHER SRESS s 1, s s s s t 0 37,730 psi 9431 psi NE: Since the principal stresses have opposite signs, the maimum in-plane shear stress is larger than the maimum out-of-plane shear stress.

10 SECIN 8.5 Combined Loadings 533 roblem clindrical pressure vessel having radius r 300 mm and wall thickness t 15 mm is subjected to internal pressure p.5 Ma. In addition, a torque 10 kn m acts at each end of the clinder (see figure). (a) Determine the maimum tensile stress ma and the maimum in-plane shear stress ma in the wall of the clinder. (b) If the allowable in-plane shear stress is 30 Ma, what is the maimum allowable torque? Solution Clindrical pressure vessel 10 kn m r 300 mm t 15 mm.5 Ma SRESSES IN HE WLL F HE VESSEL s pr 5 Ma t t (Eq. 3-11) I p r 3 r t (Eq. 3-18) Ip t Ma r t (a) RINCIL SRESSES s pr 50 Ma t (b) MXIMUM LLWLE RQUE allow 30 Ma (in-plane shear stress) s s t ma t s pr 5 Ma t t r t Units: Ma N m Substitute into Eq. (1): ma allow 30 Ma (1.5 Ma) ( ) Square both sides, rearrange, and solve for : (30) (1.5) ( ) , , (Nˇ ˇm) N m ma 31 kn m s pr 50 Ma t (1) s 1, s s s s t Ma Ma 18.6 Ma ma 56.4 Ma Maimum IN-LNE SHER SRESS s s t ma t 18.9 Ma

11 534 CHER 8 pplications of lane Stress roblem n L-shaped bracket ling in a horizontal plane supports a load 150 lb (see figure). he bracket has a hollow rectangular cross section with thickness t 0.15 in. and outer dimensions b.0 in. and h 3.5 in. he centerline lengths of the arms are b 1 0 in. and b 30 in. Considering onl the load, calculate the maimum tensile stress t, maimum compressive stress c, and maimum shear stress ma at point, which is located on the top of the bracket at the support. b = 30 in. b 1 = 0 in. = 150 lb t = 0.15 in. b =.0 in. h = 3.5 in. Solution L-shaped bracket 150 lb b 1 0 in. b 30 in. t 0.15 in. h 3.5 in. b.0 in. FREE-DY DIGRM F RCKE b 1 V SRESS ELEMEN IN (his view is looking downward at the top of the bracket.) stress element at point op of racket b b SRESS RESULNS HE SUR orque: b (150 lb)(30 in.) 4500 lb-in. Moment: M b 1 (150 lb)(0 in.) 3000 lb-in. Shear force: V 150 lb RERIES F HE CRSS SECIN For torsion: m (b t)(h t) (1.875 in.)(3.375 in.) in. For bending: I 1 1 (bh3 ) 1 (b t)(h t) (.0 in.)(3.5 in.)3 1 (1.75 in.)(3.5 in.) in. 4 c h 1.75 in. SRESSES IN N HE F HE RCKE t 4500 lb-in. 844 psi t m (0.15 in.)(6.381 in. ) s Mc (3000 lb-in.)(1.75 in.) 454 psi I.1396 in. 4 (he shear force V produces no stresses at point.) psi 844 psi RINCIL SRESSES ND MXIMUM SHER SRESS s 1, s s 17 psi 3097 psi psi 1870 psi s s t ma t 3097 psi MXIMUM ENSILE SRESS: t 430 psi s s t

12 SECIN 8.5 Combined Loadings 535 MXIMUM CMRESSIVE SRESS: c 1870 psi MXIMUM SHER SRESS: ma 3100 psi NE: Since the principal stresses have opposite signs, the maimum in-plane shear stress is larger than the maimum out-of-plane shear stress. roblem semicircular bar ling in a horizontal plane is supported at (see figure). he bar has centerline radius R and weight q per unit of length (total weight of the bar equals qr). he cross section of the bar is circular with diameter d. btain formulas for the maimum tensile stress t, maimum compressive stress c, and maimum in-plane shear stress ma at the top of the bar at the support due to the weight of the bar. R d Solution Semicircular bar d diameter of bar R radius of bar q weight of bar per unit length W weight of bar qr Weight of bar acts at the center of gravit From Case 3, ppendi D, with, we get c R ending moment at : M W C qr orque at : WR qr (Shear force at produces no shear stress at the top of the bar.) SRESSES HE F HE R s M (d) I t (d) I R W (qr )(d) d qR d 3 (qr )(d) 16qR d 4 3 d 3 SRESS ELEMEN HE F HE R c d 0 RINCIL SRESSES: s 1, s s s s t 3qR d 3 16qR d 3 ( 4 ) MXIMUM ENSILE SRESS s t s 1 16qR d 3 ( 4 ) 9.15 qr d 3 s s t 3qR d 3 16qR d 3 MXIMUM CMRESSIVE SRESS s c s 16qR d 3 ( 4 ) 8.78 qr d 3 MXIMUM IN-LNE SHER SRESS (EQ. 7-6) t ma 1 (s 1 s ) 16qR 4 d qr d 3

13 536 CHER 8 pplications of lane Stress roblem n arm C ling in a horizontal plane and supported at (see figure) is made of two identical solid steel bars and C welded together at a right angle. Each bar is 0 in. long. Knowing that the maimum tensile stress (principal stress) at the top of the bar at support due solel to the weights of the bars is 93 psi, determine the diameter d of the bars. z C Solution q q Horizontal arm C z SRESSES HE F HE R normal stress due to M M(d ) M(d ) 3 M 1 gl s I d d d shear stress due to torque (d ) t I p (d ) d gl 3 d d (5) (6) SRESS ELEMEN N F HE R C L length of and C d diameter of and C cross-sectional area d 4 weight densit of steel q weight per unit length of bars d 4 (1) RESULN FRCES CING N weight of and C ql Ld 4 () torque due to weight of C L ql (ql) (3) gl d 8 M bending moment at M L L 3L 3L d 8 (4) 1 principal tensile stress (maimum tensile stress) s 1 s s s s t (7) 0 (8) Substitute (8) into (7): s 1 s (9) s t Substitute from (5) and (6) and simplif: s 1 gl d gl (6 40) (3 10) d (10)

14 ˇ SECIN 8.5 Combined Loadings 537 SLVE FR d d gl (3 10) s 1 (11) SUSIUE NUMERICL VLUES IN EQ. (11): 490 lbft lbin. 3 L 0 in psi d 1.50 in. roblem pressurized clindrical tank with flat ends is loaded b torques and tensile forces (see figure). he tank has radius r 50 mm and wall thickness t 3 mm. he internal pressure p 3.5 Ma and the torque 450 N m. What is the maimum permissible value of the forces if the allowable tensile stress in the wall of the clinder is 7 Ma? Solution Clindrical tank r 50 mm t 3.0 mm p 3.5 Ma 450 N m allow 7 Ma CRSS SECIN rt (50 mm)(3.0 mm) mm I r 3 t (50 mm) 3 (3.0 mm) mm 4 SRESSES IN HE WLL F HE NK MXIMUM ENSILE SRESS s ma s allow 7 Ma s s s s t ( ) [14.583ˇ ˇ( ) ] ˇ ˇ(9.5493) ( ) Square both sides and simplif: SLVE FR 34,080 N R ma 34.1 kn s pr t (3.5 Ma)(50 mm) (3.0 mm) Ma Units: Ma, newtons s pr Ma t t r (450 ˇm)(50 mm) Nˇ I mm Ma mm

15 538 CHER 8 pplications of lane Stress roblem post having a hollow circular cross section supports a horizontal load 50 lb acting at the end of an arm that is 4 ft long (see figure on the net page). he height of the post is 5 ft, and its section modulus is S 10 in. 3 (a) Calculate the maimum tensile stress ma and maimum in-plane shear stress ma at point due to the load. oint is located on the front of the post, that is, at the point where the tensile stress due to bending alone is a maimum. (b) If the maimum tensile stress and maimum in-plane shear stress at point are limited to 16,000 psi and 6,000 psi, respectivel, what is the largest permissible value of the load? 5 ft 4 ft = 50 lb Solution ost with horizontal load 50 lb b length of arm 4.0 ft 48 in. h height of post 5 ft 300 in. S section modulus 10 in. 3 RECINS HE SUR M h 75,000 lb-in. b 1,000 lb-in. V 50 lb SRESSES IN 0 s M 75,000 lb-in. S 10 in psi t r I r outer radius of post 1,000 lb-in. S I t 600 psi r I r S (10 in. 3 ) (he shear force V produces no stresses at point.) (a) MXIMUM ENSILE SRESS ND MXIMUM SHER SRESS s ma s s 3750 psi (3750 psi) (600 psi) 3750 psi 3798 psi 7550 psi s s t ma t 3800 psi (b) LLWLE LD allow 16,000 psi allow 6,000 psi he stresses at point are proportional to the load. ased on tensile stress: allow s allow s ma 530 lb ased on shear stress: allow 6,000 psi allow (50 lb) t allow t 3,800 psi ma Shear stress governs: allow 395 lb s s 16,000 psi allow (50 lb) 7,550 psi 395 lb t

16 SECIN 8.5 Combined Loadings 539 roblem sign is supported b a pipe (see figure) having outer diameter 100 mm and inner diameter 80 mm. he dimensions of the sign are.0 m 0.75 m, and its lower edge is 3.0 m above the base. Note that the center of gravit of the sign is 1.05 m from the ais of the pipe. he wind pressure against the sign is 1.5 ka. Determine the maimum in-plane shear stresses due to the wind pressure on the sign at points,, and C, located on the outer surface at the base of the pipe. ipe X Rose s Editing Co. X C.0 m 100 mm 0.75 m 3.0 m C Section X-X Solution Sign supported b a pipe IE: d 100 mm d 1 80 mm t 10 mm SIGN:.0 m 0.75 m 1.50 m h height from the base to the center of gravit of the sign h 3.0 m 1 (0.75 m) m b horizontal distance from the center of gravit of the sign to the ais of the pipe b 1 (.0 m) 1 (100 mm) 1.05 m WIND RESSURE: SRESS RESULNS HE SE p 1.5 ka horizontal wind force on the sign p (1.5 ka)(1.50 m ) 50 N M h (50 N)(3.375 m) N m b (50 N)(1.05 m) 36.5 N m V 50 N t r d (36.5 ˇm)(0.1 m) Nˇ I I ( mm 4 ) Ma s s t ma t ( Ma) (0.380 Ma) Ma 68.6 Ma RERIES F HE UULR CRSS SECIN I 64 (d 4 d 1 4 ) mm 4 SRESSES IN I I mm 4 Q 3 (r 3 r 1 3 ) 1 1 (d 3 d 1 3 ) mm 3 (From Eq. 5-43b, Chapter 5) SRESSES IN 0 s Mc Md I I ( ˇm)(0.1 m) Nˇ ( mm 4 ) Ma

17 540 CHER 8 pplications of lane Stress 0 (Moment M produces no stresses at points and C) 0 t r VQ I Ib r I Ma VQ Ib (50 N)( mm 3 ) ( mm 4 )(0 mm) Ma Ma Ma Ma ure shear Ma SRESSES IN C 0 0 t r VQ I Ib Ma Ma 1.96 Ma ure shear. C.0 Ma C roblem sign is supported b a pole of hollow circular cross section, as shown in the figure. he outer and inner diameters of the pole are 10.0 in. and 8.0 in., respectivel. he pole is 40 ft high and weighs 3.8 k. he sign has dimensions 6 ft 3 ft and weighs 400 lb. Note that its center of gravit is 41 in. from the ais of the pole. he wind pressure against the sign is 30 lb/ft. (a) Determine the stresses acting on a stress element at point, which is on the outer surface of the pole at the front of the pole, that is, the part of the pole nearest to the viewer. (b) Determine the maimum tensile, compressive, and shear stresses at point. 40 ft X X 6 ft Julie s ffice 3 ft 10 in. 8 in. Section X-X Sign supported b a pole LE: d 10 in. d 1 8 in. W 1 weight of pole 3800 lb SIGN: 6 ft 3 ft, or 7 in. 36 in. 18 ft 59 in. W weight of sign 400 lb h height from the base to the center of gravit of the sign h 40 ft 1.5 ft 38.5 ft 46 in. b horizontal distance from the center of gravit of the sign to the ais of the pole b 1 (6 ft) 1 (d ) 41 in. WIND RESSURE: p 30 lb/ft psi horizontal wind force on the sign p ( psi) (59 in. ) 540 lb SRESS RESULNS HE SE ial force: N w 1 w 400 lb (compression) ending moment from wind pressure: M h (540 lb)(46 in.) 49,480 lb-in. (his moment causes tension at point.) ending moment from weight of sign: (his moment causes zero stress at point.) orque from wind pressure: b (540 lb)(41 in.),140 lb-in. Shear force from wind pressure: (his force causes zero shear stress at point.)

18 SECIN 8.5 Combined Loadings 541 RERIES F HE UULR CRSS SECIN 4 (d d 1) 8.74 in. I 64 (d4 d 4 1) in. 4 I p I in. 4 c d 5.0 in. (a) SRESSES IN (b) MXIMUM SRESSES IN s 1, s s s s t 078 psi 087 psi psi 9 psi s s t ma t 087 psi Ma. tensile stress: t 4165 psi Ma. compressive stress: c 9 psi Ma. shear stress: ma 087 psi NE: Since the principal stresses have opposite signs, the maimum in-plane shear stress is larger than the maimum out-of-plane shear stress. s 0 s N Mc I t d Ip 400 lb (49,480 lb-in.)(5.0 in.) s 8.74 in in psi 4,304. psi 4156 psi (,140 lb-in.)(10 in.) t 191 psi (579.6 in. 4 ) roblem horizontal bracket C (see figure on the net page) consists of two perpendicular arms and C, the latter having a length of 0.4 m. rm has a solid circular cross section with diameter equal to 60 mm. t point C a load 1.0 kn acts verticall and a load 3.07 kn acts horizontall and parallel to arm. Considering onl the forces 1 and, calculate the maimum tensile stress t, the maimum compressive stress c, and the maimum in-plane shear stress ma at point p, which is located at support on the side of the bracket at midheight. 0 p 0 0 z m C p mm Cross section at

19 54 CHER 8 pplications of lane Stress Solution Horizontal bracket 1 vertical force.0 kn horizontal force 3.07 kn b length of arm C 0.4 m d diameter of solid bar 60 mm SRESSES IN p N HE SIDE F HE RCKE 0 p z 0 RERIES F HE CRSS SECIN d 87.4 mm 4 I d4 636,170 mm4 64 I p I mm 4 c d 30 mm r d 30 mm SRESS RESULNS SUR N 3070 N (compression) M b 18 N m M ma be omitted because it produces no stresses at point p. 1 b 808 N m V 1 00 N s N M c I 0 t r 4V I p 3 s 3070 N (18 ˇm)(30 mm) Nˇ 87.4 mm 636,170 mm Ma Ma Ma t (808 ˇm)(30 mm) 4(00 N) Nˇ mm 4 3(87.4 mm ) Ma Ma Ma MXIMUM SRESSES IN s 1, s s s s t Ma Ma Ma 65.1 Ma s s t ma t 35.6 Ma Ma. tensile stress: t 6.1 Ma Ma. compressive stress: c 65.1 Ma Ma. in-plane shear stress: ma 35.6 Ma

20 SECIN 8.5 Combined Loadings 543 roblem clindrical pressure vessel with flat ends is subjected to a torque and a bending moment M (see figure). he outer radius is 1.0 in. and the wall thickness is 1.0 in. he loads are as follows: 800 k-in., M 1000 k-in., and the internal pressure p 900 psi. Determine the maimum tensile stress t, maimum compressive stress c, and maimum shear stress ma in the wall of the clinder. M 0 M 0 z 0 Solution Clindrical pressure vessal Internal pressure: p 900 psi ending moment: M 1000 k-in. orque: 800 k-in. uter radius: r 1 in. Wall thickness: t 1.0 in. Mean radius: r r t 11.5 in. uter diameter: d 4 in. Inner diameter: d 1 in. MMEN F INERI I 64 (d 4 d 4 1) in. 4 I p I in. 4 NE: Since the stresses due to and p are the same everwhere in the clinder, the maimum stresses occur at the top and bottom of the clinder where the bending stresses are the largest. RINCIL SRESSES s 1, s s s s t psi psi 1 10,479 psi 540 psi MXIMUM SHER SRESSES In-plane: 3970 psi ut-of-plane: t s 1 ors t s psi ma 540 psi MXIMUM SRESSES FR HE F HE CYLINDER t 10,480 psi ma 540 psi c 0 (No compressive stresses) R (a). F HE CYLINDER z 0 0 R (b). M F HE CYLINDER Stress element on the bottom of the clinder as seen from below. s pr t Mr psi psi I psi s pr 10,350 psi t t r psi I p Stress element on the top of the clinder as seen from above. s pr t Mr psi psi I 668. psi s pr 10,350 psi t z 0 0 t r psi I p

21 544 CHER 8 pplications of lane Stress RINCIL SRESSES s 1, s s psi psi 1 10,685 psi 7347 psi MXIMUM SHER SRESSES In-plane: 1669 psi ut-of-plane: s s t t s 1 ors t s psi ma 5340 psi MXIMUM SRESSES FR HE M F HE CYLINDER t 10,680 psi ma 5340 psi c 0 (No compressive stresses) R (c). ENIRE CYLINDER he largest stresses are at the bottom of the clinder. t 10,680 psi c 0 (No compressive stresses) ma 5340 psi roblem For purposes of analsis, a segment of the crankshaft in a vehicle is represented as shown in the figure. he load equals 1.0 kn, and the dimensions are b 1 80 mm, b 10 mm, and b 3 40 mm. he diameter of the upper shaft is d 0 mm. (a) Determine the maimum tensile, compressive, and shear stresses at point, which is located on the surface of the upper shaft at the z 0 ais. (b) Determine the maimum tensile, compressive, and shear stresses at point, which is located on the surface of the shaft at the 0 ais. 0 b 1 = 80 mm z 0 0 b = 10 mm b 3 = 40 mm = 1.0 kn Solution art of a crankshaft 0 M D 1.0 kn d 0 mm b 1 80 mm b 10 mm b 3 40 mm z 0 V 0 RECINS HE SUR M moment about the 0 ais (M produces compression at point and no stress at point ) M (b 1 b 3 ) 10 N m torque about the 0 ais ( produces shear stresses at points and ) b 10 N m V force directed along the z 0 ais (V produces shear stress at point and no stress at point ) V 1000 N

22 SECIN 8.5 Combined Loadings 545 MMENS F INERI ND CRSS-SECINL RE (b) SRESSES IN I d4 7,854.0 mm4 64 I p I 15,708.0 mm 4 d mm 4 0 (a) SRESSES IN z t d 4V Ma 4.44 Ma I p Ma MXIMUM SRESSES IN s Md Ma I 0 t d Ma I p MXIMUM SRESSES IN s 1, s s s s t Ma Ma Ma Ma Element is in URE SHER Ma 7. Ma ma 7. Ma Ma. tensile stress: t 7. Ma Ma. compressive stress: c 7. Ma Ma. shear stress: ma 7. Ma NE: Since the principal stresses have opposite signs, the maimum in-plane shear stress is larger than the maimum out-of-plane shear stress. s s t ma t Ma Ma. tensile stress: t 3 Ma Ma. compressive stress: c 184 Ma Ma. shear stress: ma 108 Ma NE: Since the principal stresses have opposite signs, the maimum in-plane shear stress is larger than the maimum out-of-plane shear stress.