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 8.5-1 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 1 7.0 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 8.5-1 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 1 7.0 in. d 1 d 4 (d d 1 ) 11.781 in. I 64 (d 4 d 1 4 ) 83.03 in. 4 1500 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.) 11.781 in. 83.03 in. 4 17.3 psi 3461.4 psi 3330 psi MXIMUM CMRESSIVE SRESS s c M(d ) 17.3 psi 3461.4 psi I 3590 psi MXIMUM SHER SRESS Uniaial stress. t ma s c 1790 psi
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) 13,089.97 1 4(6 kn) m 8b 1.44 m 13,090 d 3 d 1.44 0 (d meters) Solve numericall: d 0.04845 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.
56 CHER 8 pplications of lane Stress roblem 8.5-3 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 8.5-3 Drill pipe for an oil wall s 60 k 4850 psi 1.370 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 ) 1.370 in. I 3 (d 4 d 1 4 ) 86.977 in. 4 SRESSES HE UER SURFCE RINCIL SRESSES s 1, s s t 45 psi (45) (5864) 45 psi 6345 psi 1 390 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.
SECIN 8.5 Combined Loadings 57 roblem 8.5-4 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. 8.5-4 and 8.5-5 Solution 8.5-4 Generator shaft s 55 kn 0.08 10 3 6.1 Ma m t (d ) (71,60 ˇm)(140 mm) Nˇ I 38.70 10 6 m 4 30.50 Ma 0 (Determine the stresses) 55 kn d 80 mm d 1 30 mm 0 power 0 1800 kw f frequenc f 4.0 Hz torque 0 f (Eq. 3-40) 1800 103 W 71,60 (4.0 Hz) Nˇ ˇm 4 (d d 1 ) 0.08 10 3 m I 3 (d 4 d 1 4 ) 38.70 10 6 m 4 SRESSES HE SURFCE F HE SHF RINCIL SRESSES s 1, s s s s t 13.11 Ma 33.0 Ma 1 0.1 Ma 46.3 Ma MXIMUM ENSILE SRESS t 1 0.1 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.
58 CHER 8 pplications of lane Stress roblem 8.5-5 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 8.5-5 Generator shaft Units: pounds psi t (d ) (0 k-in.)(4.0 in.) I 74.889 in. 4 3,01.9 psi (Determine the maimum allowable load ) 0 k-in. d 8.0 in. d 1 6.0 in. allow 6500 psi (In-plane shear stress) 4 (d d 1 ) 1.9911 in. I 3 (d 4 d 1 4 ) 74.889 in. 4 MXIMUM IN-LNE SHER SRESS s s t ma t (6500) ()(1.9911) R (3,01.9) 3,001,700; 1934.43 48,810 lb or ma 49 k s t allow t 61.905 10 9 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 1.9911 in.
SECIN 8.5 Combined Loadings 59 roblem 8.5-6 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 8.5-6 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 57.96 Ma Units: Ma p Ma Case : t ma s 1 6.5 p; 60 Ma 6.5p Solving, p 9.60 Ma Case 3: t ma s 3.15 p 8.648 Ma 60 Ma 3.15 p 8.648 Ma Solving, p 3 8.37 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 8.648 Ma 60 Ma 3.15 p 8.648 Ma Solving, p 1 10.03 Ma roblem 8.5-7 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.
530 CHER 8 pplications of lane Stress Solution 8.5-7 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 13.84 t Solving, t 1 0.0413 in. 47.68 13.84 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 375 1000 lb t (1.5 in.)t 375 t 17.3 50.3 t t U-F-LNE SHER SRESSES Case : t ma s 1 375 ; 3000 375 t t Solving, t 0.15 in. Case 3: t ma s 51.16 ; 3000 51.16 t t Solving, t 3 0.0837 in. CSE, U-F-LNE SHER SRESS GVERNS t min 0.15 in. IXIL SRESS 1 roblem 8.5-8 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
SECIN 8.5 Combined Loadings 531 Solution 8.5-8 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 1.5664 mm 4 W Mg (60 kg)(9.81 ms ) 588.6 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 (80 10 6 a)f L I L ma 80 ma Units: Ma ma radius ENSILE SRESS W W I p s W 46.839 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 5316 6400 f ma allow 100 Ma 100 Ma 3.40 (3.40) 6400 f ma (100 3.40) (3.40) 6400 f ma ma 0.9114 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 46.839 Ma 80 ma (Ma)
53 CHER 8 pplications of lane Stress roblem 8.5-9 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 1 5.0 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 8.5-9 edal crank 37,730 psi d b 1 b 9431psi 160 lb b 1 5.0 in. d 0.6 in. b.5 in. SRESS RESULNS on cross section at point : orque: b 400 lb-in. Moment: M b 1 800 lb-in. Shear force: V 160 lb SRESSES IN 16 t 3 9431 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.
SECIN 8.5 Combined Loadings 533 roblem 8.5-10 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 8.5-10 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 14.147 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 117.893 106 Units: Ma N m Substitute into Eq. (1): ma allow 30 Ma (1.5 Ma) (117.893 10 6 ) Square both sides, rearrange, and solve for : (30) (1.5) (117.893 10 6 ) 743.750 13,899 10 53,51 1 106 (Nˇ ˇm) 31.3 10 3 N m ma 31 kn m s pr 50 Ma t (1) s 1, s s s s t 37.5 18.878 Ma 1 56.4 Ma 18.6 Ma ma 56.4 Ma Maimum IN-LNE SHER SRESS s s t ma t 18.9 Ma
534 CHER 8 pplications of lane Stress roblem 8.5-11 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 8.5-11 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.) 6.381 in. For bending: I 1 1 (bh3 ) 1 (b t)(h t)3 1 1 1 (.0 in.)(3.5 in.)3 1 (1.75 in.)(3.5 in.)3 1.1396 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.) 0 454 psi 844 psi RINCIL SRESSES ND MXIMUM SHER SRESS s 1, s s 17 psi 3097 psi 1 434 psi 1870 psi s s t ma t 3097 psi MXIMUM ENSILE SRESS: t 430 psi s s t
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 8.5-1 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 8.5-1 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 4 64 64qR 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 3 18.97 qr d 3
536 CHER 8 pplications of lane Stress roblem 8.5-13 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 8.5-13 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 4 3 64 d d shear stress due to torque (d ) t I p (d ) d 4 3 16 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)
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 3 0.8356 lbin. 3 L 0 in. 1 93 psi d 1.50 in. roblem 8.5-14 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 8.5-14 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) 94.48 mm I r 3 t (50 mm) 3 (3.0 mm).356 10 6 mm 4 SRESSES IN HE WLL F HE NK MXIMUM ENSILE SRESS s ma s allow 7 Ma s s s s t 7 43.750 (530.5 10 6 ) [14.583ˇ ˇ(530.5 10 6 ) ] ˇ ˇ(9.5493) 8.50 0.0005305 (14.583 0.0005305 ) 91.189 Square both sides and simplif: 494.1 0.014501 SLVE FR 34,080 N R ma 34.1 kn s pr t (3.5 Ma)(50 mm) (3.0 mm) 9.167 Ma 1.0610 10 3 Units: Ma, newtons s pr 58.333 Ma t t r (450 ˇm)(50 mm) Nˇ I.356 10 6 mm 4 9.5493 Ma 94.48 mm
538 CHER 8 pplications of lane Stress roblem 8.5-15 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 8.5-15 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. 3 7500 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
SECIN 8.5 Combined Loadings 539 roblem 8.5-16 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 8.5-16 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) 3.375 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) 7593.8 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 (5.796 10 6 mm 4 ) 0.380 Ma s s t ma t (65.507 Ma) (0.380 Ma) 68.60 Ma 68.6 Ma RERIES F HE UULR CRSS SECIN I 64 (d 4 d 1 4 ).8981 10 6 mm 4 SRESSES IN I I 5.796 10 6 mm 4 Q 3 (r 3 r 1 3 ) 1 1 (d 3 d 1 3 ) 40.667 10 3 mm 3 (From Eq. 5-43b, Chapter 5) SRESSES IN 0 s Mc Md I I (7593.8 ˇm)(0.1 m) Nˇ (.8981 10 6 mm 4 ) 131.01 Ma
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 0.380 Ma VQ Ib (50 N)(40.667 103 mm 3 ) (.8981 10 6 mm 4 )(0 mm) 1.5786 Ma 0.380 Ma 1.5786 Ma 18.80 Ma ure shear. 18.8 Ma SRESSES IN C 0 0 t r VQ I Ib 0.380 Ma 1.5786 Ma 1.96 Ma ure shear. C.0 Ma C roblem 8.5-17 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 8.5-17 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 0.08333 psi horizontal wind force on the sign p (0.08333 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.)
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) 89.81 in. 4 I p I 579.6 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 1 4165 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. 89.81 in. 4 148.5 psi 4,304. psi 4156 psi (,140 lb-in.)(10 in.) t 191 psi (579.6 in. 4 ) roblem 8.5-18 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 0 0.4 m C p 0 1 60 mm Cross section at
54 CHER 8 pplications of lane Stress Solution 8.5-18 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 17.3 10 3 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 4 1.0858 Ma 57.909 Ma 58.995 Ma t (808 ˇm)(30 mm) 4(00 N) Nˇ 17.3 10 3 mm 4 3(87.4 mm ) 19.051 Ma 0.953 Ma 0.004 Ma MXIMUM SRESSES IN s 1, s s s s t 9.498 Ma 35.641 Ma 1 6.1 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
SECIN 8.5 Combined Loadings 543 roblem 8.5-19 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 8.5-19 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) 4787.0 in. 4 I p I 9574.0 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 6509.1 psi 3969.6 psi 1 10,479 psi 540 psi MXIMUM SHER SRESSES In-plane: 3970 psi ut-of-plane: t s 1 ors t s 1 540 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 5175.0 psi 506.8 psi I 7681.8 psi s pr 10,350 psi t t r 100.7 psi I p Stress element on the top of the clinder as seen from above. s pr t Mr 3 5175.0 psi 506.8 psi I 668. psi s pr 10,350 psi t z 0 0 t r 100.7 psi I p
544 CHER 8 pplications of lane Stress RINCIL SRESSES s 1, s s 9015.9 psi 1668.9 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 1 5340 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 8.5-0 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 8.5-0 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
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 314.16 mm 4 0 (a) SRESSES IN z 0 0 0 0 0 t d 4V 76.394 Ma 4.44 Ma I p 3 7.15 Ma MXIMUM SRESSES IN s Md 15.79 Ma I 0 t d 76.394 Ma I p MXIMUM SRESSES IN s 1, s s s s t 76.40 Ma 108.04 Ma 1 31.64 Ma 184.44 Ma Element is in URE SHER. 1 7. 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 108.04 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.