8 Applications of Plane Stress (Pressure Vessels, Beams, and Combined Loadings)

Transcription

1 8 pplications of Plane Stress (Pressure Vessels, Beams, and omined oadings) Sperical Pressure Vessels Wen solving te prolems for Section 8., assume tat te given radius or diameter is an inside dimension and tat all internal pressures are gage pressures. Prolem 8.-1 large sperical tank (see figure) contains gas at a pressure of 400 psi. Te tank is 45 ft in diameter and is constructed of ig-strengt steel aving a ield stress in tension of 80 ksi. Determine te required tickness (to te nearest 1/4 inc) of te wall of te tank if a factor of safet of 3.5 wit respect to ielding is required. Solution 8.-1 ROSS SETION Sperical tank Yield stress: 80 ksi (steel) actor of safet: n 3.5 p Radius: r (45 ft) 70 in. Internal pressure: p 400 psi MINIMUM W THIKNESS t min rom Eq. (8-1): s ma pr t t prn.36 in. s Use te net iger 14 inc: t min.50 in. or s pr n t Prolem 8.- Solve te preceding prolem if te internal pressure is 3.5 MPa, te diameter is 18 m, te ield stress is 550 MPa, and te factor of safet is 3.0. Determine te required tickness to te nearest millimeter. Solution 8.- ROSS SETION Sperical tank Yield stress: 550 MPa actor of safet: n 3.0 Radius: r (18 m) 9 m Internal pressure: p 3.5 MPa p MINIMUM W THIKNESS t min rom Eq. (8-1): s ma pr t t prn 85.9 mm s or Use te net iger millimeter: t min 86 mm s pr n t

2 498 HPTER 8 pplications of Plane Stress Prolem 8.-3 emisperical window (or viewport) in a decompression camer (see figure) is sujected to an internal air pressure of 80 psi. Te port is attaced to te wall of te camer 18 olts. ind te tensile force in eac olt and te tensile stress in te viewport if te radius of te emispere is 7.0 in. and its tickness is 1.0 in. Solution 8.-3 Hemisperical viewport REE-BODY DIGRM p T total tensile force in 18 olts a T p T p HORIZ p(r ) force in one olt T (pr ) 684 l Radius: r 7.0 in. Internal pressure: p 80 psi Wall tickness: t.0 in. 18 olts TENSIE STRESS IN VIEWPORT (EQ. 8-1) s pr 80 psi t Prolem 8.-4 ruer all (see figure) is inflated to a pressure of 60 kpa. t tat pressure te diameter of te all is 30 mm and te wall tickness is 1. mm. Te ruer as modulus of elasticit E 3.5 MPa and Poisson s ratio.45. Determine te maimum stress and strain in te all. Solution 8.-4 ROSS-SETION Ruer all p Radius: r (30 mm) 15 mm Internal pressure: p 60 kpa Wall tickness: t. mm Modulus of elasticit: E 3.5 MPa (ruer) Poisson s ratio: 0.45 (ruer) MXIMUM STRESS (EQ. 8-1) s ma pr (60 kpa)(115 mm) t (1. mm).88 MPa MXIMUM STRIN (EQ. 8-4) e ma pr (60 kpa)(115 mm) (1 n) te (1. mm)(3.5 MPa) (0.55).45

3 SETION 8. Sperical Pressure Vessels 499 Prolem 8.-5 Solve te preceding prolem if te pressure is 9.0 psi, te diameter is 9.0 in., te wall tickness is 0.05 in., te modulus of elasticit is 500 psi, and Poisson s ratio is Solution 8.-5 ROSS-SETION Ruer all p MXIMUM STRESS (EQ. 8-1) s ma pr (9.0 psi)(4.5 in.) t (0.05 in.) 405 psi Radius: r (9.0 in.) 4.5 in. Internal pressure: p 9.0 psi Wall tickness: t.05 in. Modulus of elasticit: E 500 psi (ruer) Poisson s ratio: 0.45 (ruer) MXIMUM STRIN (EQ. 8-4) e ma pr (9.0 psi)(4.5 in.) (1 n) te (0.05 in.)(500 psi) (0.55).446 Prolem 8.-6 sperical steel pressure vessel (diameter 480 mm, tickness 8.0 mm) is coated wit rittle lacquer tat cracks wen te strain reaces (see figure). Wat internal pressure p will cause te lacquer to develop cracks? (ssume E 5 GPa and.30.) racks in coating Solution 8.-6 ROSS-SETION Sperical vessel wit rittle coating p racks occur wen e ma rom Eq. (8-4): e ma pr (1 ) te p te e ma r(1 ) p (8.0 mm)(05 GPa)( ) (40 mm)(0.70).93 MPa r 40 mm E 5 GPa (steel) t 8.0 mm 0.30

4 500 HPTER 8 pplications of Plane Stress Prolem 8.-7 sperical tank of diameter 50 in. and wall tickness.0 in. contains compressed air at a pressure of 400 psi. Te tank is constructed of two emisperes joined a welded seam (see figure on te net page). (a) Wat is te tensile load f (l per in. of lengt of weld) carried te weld? (See te figure on te net page.) () Wat is te maimum sear stress ma in te wall of te tank? (c) Wat is te maimum normal strain in te wall? (or steel, assume E psi and.9.) Weld Solution 8.-7 Welded tank (sperical) Welded seam f T c p(r ) r 30.0 kin. pr (400 psi)(5 in.) r 5 in. E psi t.0 in. 0.9 (Steel) p 400 psi (a) TENSIE OD RRIED BY WED T total load f load per inc T p p(r ) c ircumference of tank r () MXIMUM SHER STRESS IN W (EQ. 8-3) t ma pr (400 psi)(5 in.) 4t 4( in.) 7500 psi (c) MXIMUM NORM STRIN IN W (EQ. 8-4) pr(1 n) (400 psi)(5 in.)(0.71) e ma te (.0 in.)( psi) Prolem 8.-8 Solve te preceding prolem for te following data: diameter 1.0 m, tickness 50 mm, pressure 4 MPa, modulus 10 GPa, and Poisson s ratio 0.9. Solution 8.-8 Welded tank (sperical) r 500 mm E 0 GPa t 50 mm 0.9 (Steel) p 4 MPa Welded seam (a) TENSIE OD RRIED BY WED T total load f load per inc T p p(r ) c ircumference of tank r f T c p(r ) pr (4 MPa)(500 mm) r (c) MXIMUM NORM STRIN IN W (EQ. 8-4) 6.0 MNm e ma pr (1 n) te () MXIMUM SHER STRESS IN W (EQ. 8-3) (4 MPa)(500 mm) (1.9) t ma pr (4 MPa)(500 mm) (50 mm)( MPa) t 4(50 mm) MPa

5 SETION 8. Sperical Pressure Vessels 501 Prolem 8.-9 sperical stainless-steel tank aving a diameter of 0 in. is used to store propane gas at a pressure of 400 psi. Te properties of te steel are as follows: ield stress in tension, 140,000 psi; ield stress in sear, 65,000 psi; modulus of elasticit, psi; and Poisson s ratio, 0.8. Te desired factor of safet wit respect to ielding is.75. lso, te normal strain must not eceed Determine te minimum permissile tickness t min of te tank. Solution 8.-9 ROSS-SETION Propane tank (steel) r 0 in. E psi p 400 psi ,000 psi n.75 65,000 psi ma MINIMUM W THIKNESS t (1) TENSION (EQ. 8-1) t 1 pr pr s ma (s n).36 in. p s ma pr t 1 (400 psi)(10 in.) (140,000 psi).75 () SHER (EQ. 8-3) t pr 4t n.54 in. (3) STRIN (EQ. 8-4) t 3 pr (1 n) e ma E t ma pr 4t (400 psi)(10 in.) 4(65,000 psi).75 e ma pr (1 n) t 3 E (400 psi)(10 in.) (1.8) ( )(300 6 psi).88 in. Strain governs since t 3 t 1 and t 3 t t min.88 in. Prolem Solve te preceding prolem if te diameter is 500 mm, te pressure is 16 MPa, te ield stress in tension is 950 MPa, te ield stress in sear is 450 MPa, te factor of safet is.4, te modulus of elasticit is 00 GPa, Poisson s ratio is 0.8, and te normal strain must not eceed Solution ROSS-SETION p MINIMUM W THIKNESS t (1) TENSION (EQ. 8-1) t 1 pr pr s ma (s n) mm Propane tank (steel) r 50 mm E 0 GPa p 6 MPa MPa n MPa ma s ma pr t 1 (16 MPa)(50 mm) (950 MPa).4 () SHER (EQ. 8-3) t pr 4t n mm (3) STRIN (EQ. 8-4) t 3 pr (1 n) e ma E (16 MPa)(50 mm) ( )(00 GPa) (10.8) mm Strain governs since t 3 t 1 and t 3 t t min 6.0 mm t ma pr 4t (16 MPa)(50 mm) 4(450 MPa).4 e ma pr (1 n) t 3 E

6 50 HPTER 8 pplications of Plane Stress Prolem ollow pressuried spere aving radius r 4.8 in. and wall tickness t.4 in. is lowered into a lake (see figure). Te compressed air in te tank is at a pressure of 4 psi (gage pressure wen te tank is out of te water). t wat dept D 0 will te wall of te tank e sujected to a compressive stress of 90 psi? d Solution Pressuried spere under water ROSS-SETION r 4.8 in. p 1 4 psi t.4 in. densit of water 6.4 lft 3 (1) IN IR: p 1 4 psi p 1 (1) IN IR ompressive stress in tank wall equals 90 psi. (Note: is positive in tension.) s pr t (p 1 p )r t 90 psi 90 psi (4 psi D 0)(4.8 in.) (0.4 in.) D 0 Solve for D 0 : D in. 90 ft () UNDER WTER: p 1 4 psi p p 1 () UNDER WTER D 0 dept of water (in.) 6.4 lft3 p gd in. 3 ft 3 D D 0 (psi)

7 SETION 8.3 lindrical Pressure Vessels 503 lindrical Pressure Vessels Wen solving te prolems for Section 8.3, assume tat te given radius or diameter is an inside dimension and tat all internal pressures are gage pressures. Prolem scua tank (see figure) is eing designed for an internal pressure of 1600 psi wit a factor of safet of.0 wit respect to ielding. Te ield stress of te steel is 35,000 psi in tension and 16,000 psi in sear. If te diameter of te tank is 7.0 in., wat is te minimum required wall tickness? Solution Scua tank s allow s 7,500 psi n ind required wall tickness t. t allow t 8,000 psi n lindrical pressure vessel p 600 psi n.0 d 7.0 in. r 3.5 in. 35,000 psi 6,000 psi (1) BSED ON TENSION (EQ. 8-5) s ma pr t t 1 pr (1600 psi)(3.5 in.).30 in. s allow 17,500 psi () BSED ON SHER (EQ. 8-10) t ma pr t t pr (1600 psi)(3.5 in.).350 in. t allow (8,000 psi) Sear governs since t t 1 t min.350 in. Prolem 8.3- tall standpipe wit an open top (see figure) as diameter d.0 m and wall tickness t 8 mm. (a) Wat eigt of water will produce a circumferential stress of 10.9 MPa in te wall of te standpipe? () Wat is te aial stress in te wall of te tank due to te water pressure? d

8 504 HPTER 8 pplications of Plane Stress Solution 8.3- Vertical standpipe d (a) HEIGHT O WTER: s 1 pr (EQ. 8-5) t ( )(1.0 m) 0.9 MPa m Solve for (meters): (10.9)(0.018) ( )(1.0).0 m d.0 m r.0 m t 8 mm weigt densit of water 9.81 knm 3 eigt of water (meters) p water pressure (9810 Nm 3 )()(10 6 ) MPa (MPa) () XI STRESS IN THE W DUE TO WTER PRESSURE ONE Because te top of te tank is open, te internal pressure of te water produces no aial (longitudinal) stresses in te wall of te tank. ial stress equals ero. Prolem n inflatale structure used a traveling circus as te sape of a alf-circular clinder wit closed ends (see figure). Te faric and plastic structure is inflated a small lower and as a radius of 40 ft wen full inflated. longitudinal seam runs te entire lengt of te ridge of te structure. If te longitudinal seam along te ridge tears open wen it is sujected to a tensile load of 540 pounds per inc of seam, wat is te factor of safet n against tearing wen te internal pressure is 0.5 psi and te structure is full inflated? ongitudinal seam Solution Inflatale structure Half-circular clinder r 40 ft 480 in. Internal pressure p.5 psi T tensile force per unit lengt of longitudinal seam Seam tears wen T T ma 540 lin. ind factor of safet against tearing. IRUMERENTI STRESS (EQ. 8-5) s 1 pr were t tickness of faric t ctual value of T due to internal pressure t T t pr (0.5 psi)(480 in.) 40 lin. TOR O SETY n T ma T 540 lin lin.

9 SETION 8.3 lindrical Pressure Vessels 505 Prolem tin-walled clindrical pressure vessel of radius r is sujected simultaneousl to internal gas pressure p and a compressive force acting at te ends (see figure). Wat sould e te magnitude of te force in order to produce pure sear in te wall of te clinder? Solution lindrical pressure vessel OR PURE SHER, te stresses and must e equal in magnitude and opposite in sign (see, e.g., ig in Section 7.3). r Radius p Internal pressure OR pr t pr t Solve for : 3pr rt STRESSES (SEE EQS. 8-5 ND 8-6): s 1 pr t s pr t pr t rt Prolem strain gage is installed in te longitudinal direction on te surface of an aluminum everage can (see figure). Te radius-to-tickness ratio of te can is 00. Wen te lid of te can is popped open, te strain canges Wat was te internal pressure p in te can? (ssume E psi and.33.) 1 OZ (355 m) Solution luminum can STRIN IN ONGITUDIN DIRETION (EQ. 8-11a) e pr (1 n) te or p tee r(1 n) 1 OZ (355 m) strain gage r 0 E psi.33 t e 0 cange in strain wen pressure is released ind internal pressure p. e e 0 p tee 0 (r)(1 n) Ee 0 (rt)(1 n) Sustitute numerical values: p (10 06 psi)( ) (00)(1.66) 50 psi

10 506 HPTER 8 pplications of Plane Stress Prolem circular clindrical steel tank (see figure) contains a volatile fuel under pressure. strain gage at point records te longitudinal strain in te tank and transmits tis information to a control room. Te ultimate sear stress in te wall of te tank is 8 MPa and a factor of safet of is required. t wat value of te strain sould te operators take action to reduce te pressure in te tank? (Data for te steel are as follows: modulus of elasticit E 5 GPa and Poisson s ratio.30.) lindrical tank Pressure relief valve Solution lindrical tank Strain gage UT 8 MPa E 5 GPa.30 n (factor of safet) t ma t UT 41 MPa n ind maimum allowale strain reading at te gage. s 1 pr t s pr t rom Eq. (8-10): t ma s 1 pr t p ma tt ma r rom Eq. (8-11a): e pr (1 n) te (e ) ma p mar te (1 n) t ma (1 n) E 41 MPa e ma (1.60) GPa Prolem clinder filled wit oil is under pressure from a piston, as sown in te figure. Te diameter d of te piston is 1.80 in. and te compressive force is 3500 l. Te maimum allowale sear stress allow in te wall of te clinder is 5500 psi. Wat is te minimum permissile tickness t min of te clinder wall? (See te figure on te net page.) p linder Piston Solution linder wit internal pressure p d.80 in. r.90 in l allow 5500 psi ind minimum tickness t min. Pressure in clinder: p r Maimum sear stress (Eq. 8-10): t ma pr t rt Minimum tickness: t min r t allow Sustitute numerical values: 3500 l t min.113 in. (0.90 in.)(5500 psi)

11 SETION 8.3 lindrical Pressure Vessels 507 Prolem Solve te preceding prolem if d 90 mm, 4 kn, and allow 40 MPa. Solution linder wit internal pressure d 90 mm r 45 mm 4.0 kn allow 40 MPa ind minimum tickness t min. Pressure in clinder: p r p Maimum sear stress (Eq. 8-10): t ma pr t rt Minimum tickness: t min r t allow Sustitute numerical values: 4.0 kn t min 3.71 mm (45 mm)(40 MPa) Prolem standpipe in a water-suppl sstem (see figure) is 1 ft in diameter and 6 inces tick. Two oriontal pipes carr water out of te standpipe; eac is ft in diameter and 1 inc tick. Wen te sstem is sut down and water fills te pipes ut is not moving, te oop stress at te ottom of te standpipe is 130 psi. (a) Wat is te eigt of te water in te standpipe? () If te ottoms of te pipes are at te same elevation as te ottom of te standpipe, wat is te oop stress in te pipes? Solution Vertical standpipe Sustitute numerical values: (130 psi)(6 in.) 300 in. 5 ft 6.4 l 178 in. 3 (7 in.) HORIZONT PIPES d 1 ft 4 in. r 1 in. t 1.0 in. d ft 44 in. r 7 in. t 6 in. g 6.4 lft lin.3 oop stress at ottom of standpipe 30 psi (a) IND HEIGHT O WTER IN THE STNDPIPE p pressure at ottom of standpipe rom Eq. (8-5): or s 1t s 1 pr gr t t gr () IND HOOP STRESS IN THE PIPES Since te pipes are ft in diameter, te dept of water to te center of te pipes is aout 4 ft. 1 4 ft 88 in. p l s 1 p 1r 1 g 1r in. 3 (88 in.)(1 in.) t 1 t in. 5 psi Based on te average pressure in te pipes: 5 psi

12 508 HPTER 8 pplications of Plane Stress Prolem clindrical tank wit emisperical eads is constructed of steel sections tat are welded circumferentiall (see figure). Te tank diameter is 1. m, te wall tickness is 0 mm, and te internal pressure is 1600 kpa. (a) Determine te maimum tensile stress in te eads of te tank. () Determine te maimum tensile stress c in te clindrical part of te tank. (c) Determine te tensile stress w acting perpendicular to te welded joints. (d) Determine te maimum sear stress in te eads of te tank. (e) Determine te maimum sear stress c in te clindrical part of te tank. Welded seams Solution lindrical tank Welded seams Hemispere (c) TENSIE STRESS IN WEDS (EQ. 8-6) s w pr 4.0 MPa t d. m t mm r.6 m p 600 kpa (a) MXIMUM TENSIE STRESS IN HEMISPHERES (EQ. 8-1) s pr (1600 kpa)(0.6 m) 4.0 MPa t (0 mm) (d) MXIMUM SHER STRESS IN HEMISPHERES (EQ. 8-3) t pr 4t s.0 MPa (e) MXIMUM SHER STRESS IN YINDER (EQ. 8-10) t c pr t s c 4.0 MPa () MXIMUM STRESS IN YINDER (EQ. 8-5) s c pr t s 48.0 MPa Prolem clindrical tank wit diameter d 6 in. is sujected to internal gas pressure p 400 psi. Te tank is constructed of steel sections tat are welded circumferentiall (see figure). Te eads of te tank are emisperical. Te allowale tensile and sear stresses are 8000 psi and 300 psi, respectivel. lso, te allowale tensile stress perpendicular to a weld is 6400 psi. Determine te minimum required tickness t min of (a) te clindrical part of te tank, and () te emisperical eads.

13 SETION 8.3 lindrical Pressure Vessels 509 Solution Welded seams lindrical tank Hemispere WED: s pr (EQ. 8-6) t d 6.0 in. r 8.0 in. p 400 psi allow 8000 psi (tension) allow 300 psi (sear) Weld: allow 6400 psi (tension) (a) IND MINIMUM THIKNESS O YINDER TENSION: s ma pr (EQ. 8-5) t t min pr (400 psi)(8.0 in.).40 in. s allow 8000 psi SHER: t ma pr (EQ. 8-10) t t min pr (400 psi)(8.0 in.).50 in. t allow ()(300 psi) t min pr (400 psi)(8.0 in.).5 in. s allow (6400 psi) Sear governs. t min.50 in. () IND MINIMUM THIKNESS O HEMISPHERES TENSION: s ma pr (EQ. 8-1) t t min pr (400 psi)(8.0 in.).0 in. s allow (8000 psi) SHER: t ma pr (EQ. 8-3) 4t t min pr (400 psi)(8.0 in.).5 in. 4t allow 4(300 psi) Sear governs. t min.5 in. Prolem pressuried steel tank is constructed wit a elical weld tat makes an angle 60 wit te longitudinal ais (see figure). Te tank as radius r.5 m, wall tickness t 5 mm, and internal pressure p.4 MPa. lso, te steel as modulus of elasticit E 0 GPa and Poisson s ratio.30. Determine te following quantities for te clindrical part of te tank: (a) te circumferential and longitudinal stresses, () te maimum in-plane and out-of-plane sear stresses, (c) te circumferential and longitudinal strains, and (d) te normal and sear stresses acting on planes parallel and perpendicular to te weld (sow tese stresses on a properl oriented stress element). Helical weld Solution lindrical pressure vessel Helical weld 60º r.5 m t 5 mm p.4 MPa E 0 GPa.30 (a) IRUMERENTI STRESS s 1 pr (4 MPa)(0.5 m) 80 MPa t (15 mm) ONGITUDIN STRESS s pr t s 1 40 MPa () IN-PNE SHER STRESS t 1 s 1 s OUT-O-PNE SHER STRESS t s 1 pr MPa 4t pr 40 MPa t

14 510 HPTER 8 pplications of Plane Stress (c) IRUMERENTI STRIN 80 MPa e 1 (.30) (00 GPa) ONGITUDIN STRIN e e s (1 n) E (40 MPa) [1 (0.30) ] (00 GPa) (d) STRESSES ON WED 70 MPa 19 MPa e 1 s 1 ( n) OR 30º E 80 MPa s 1 s s s s cos u t sin u 60 MPa (0 MPa)(cos 60º) 60 MPa 0 MPa 50 MPa t 1 1 s s sin u t cos u (0 MPa)(sin 60º) 7.3 MPa s 1 s s s 1 40 MPa 80 MPa 50 MPa 70 MPa O 30 60º 90º 30º 40 MPa 80 MPa Prolem Solve te preceding prolem for a welded tank wit 65, r 8 in., t.6 in., p 0 psi, E psi, and.30. Solution lindrical tank Helical weld () IN-PNE SHER STRESS t 1 s 1 s pr 500 psi 4t OUT-O-PNE SHER STRESS t s 1 pr 3000 psi t 65º r 8 in. t.6 in. p 0 psi E psi.30 (a) IRUMERENTI STRESS s 1 pr (00 psi)(18 in.) 6000 psi t (0.6 in.) ONGITUDIN STRESS s pr t s psi (c) IRUMERENTI STRIN e 1 s 1 ( n) E 6000 psi e 1 (.30) ( psi) ONGITUDIN STRIN e s (1 n) E 3000 psi e [1 (0.30) ] psi

15 SETION 8.4 Maimum Stresses in Beams 511 (d) STRESS ON WED 5460 psi 1150 psi 3540 psi 5 O 65º 90º 5º 3000 psi 6000 psi OR 5º s 1 s s s s cos u t sin u 4500 psi (1500 psi)(cos 50º) 4500 psi 960 psi 3540 psi t 1 1 s s sin u t cos u (1500 psi)(sin 50º) 150 psi s 1 s s s psi 6000 psi 3540 psi 5460 psi Maimum Stresses in Beams Wen solving te prolems for Section 8.4, consider onl te in-plane stresses and disregard te weigts of te eams. Prolem cantilever eam of rectangular cross section is sujected to a concentrated load P 5 k acting at te free end (see figure). Te eam as widt 4 in. and eigt 0 in. Point is located at distance c ft from te free end and distance d 3 in. from te ottom of te eam. alculate te principal stresses and and te maimum sear stress ma at point. Sow tese stresses on sketces of properl oriented elements. P c d Solution antilever eam P 5 k c ft 4 in. 4 in. d 3 in. 0 in. STRESSES T POINT I in.4 1 M Pc l-in. V P 5,000 l P c d s M ( l-in.)(.0 in.) I in psi t VQ I t (15,000 l)(4.0 in.3 ) (333.3 in. 4 )(4 in.) 160 psi 47.5 psi Q d d 4.0 in psi 47.5 psi 160 psi d.0 in.

16 51 HPTER 8 pplications of Plane Stress PRINIP STRESSES t tan u p s s p 3.63º and p 11.81º p 56.37º and p 78.19º s 1 s s s s cos u t sin u or p 3.63º: s 1 59 psi or p 56.37º: s psi Terefore, 99 psi and u p psi and u p psi 1180 psi O 1080 psi MXIMUM SHER STRESSES 1080 psi psi O 98.8 psi 78. s s t ma t 179 psi B u s1 u p and 180 psi u s u s and 1180 psi s avg s s 1080 psi Prolem 8.4- Solve te preceding prolem for te following data: P 0 kn, 00 mm, 0 mm, c.5 m, and d 50 mm. Solution 8.4- P 0 kn 0 mm d 50 mm STRESSES T POINT antilever eam 00 mm c.5 m I mm 4 M Pc 60,000 N m V P 0 kn P c d s M (60,000 ˇm)(50 mm) Nˇ I mm MPa t VQ I t (10 kn)(375,000 mm3 ) 6.75 MPa ( mm 4 )(100 mm) 45.0 MPa 6.75 MPa Q d d 375,000 mm MPa 45.0 MPa d 50 mm

17 SETION 8.4 Maimum Stresses in Beams 513 PRINIP STRESSES t tan u p.3000 s s p 6.70º and p 8.35º p 96.7º and p 98.35º s 1 s s s s cos u t sin u or p 6.70º: s MPa or p 96.70º: s MPa Terefore, 46.0 MPa and u p MPa and u p MPa O 46.0 MPa MPa MXIMUM SHER STRESSES s s t ma t B 3.5 MPa u s1 u p and 3.5 MPa u s u s and 3.5 MPa O s avg s s.5 MPa.5 MPa MPa Prolem simple eam of rectangular cross section (widt 4 in., eigt 8 in.) carries a uniform load of 1000 l/ft on a span of 10 ft (see figure). ind te principal stresses and and te maimum sear stress ma at a cross section 1 ft from te left-and support at eac of te following locations: (a) te neutral ais, () in. aove te neutral ais, and (c) te top of te eam. (Disregard te direct compressive stresses produced te uniform load earing against te top of te eam.) 1 ft 1000 l/ft 10 ft 4 in. 8 in. Solution Simpl supported eam q 000 l/ft 8 in. 4 in. c I in in. M qc c.0 ft (a) NEUTR XIS qc 54,000 l-in. 0 ft V q qc 4,000 l t 3V psi Pure sear: 88 psi, 188 psi, ma 88 psi

18 514 HPTER 8 pplications of Plane Stress () in. BOVE THE NEUTR XIS s M I t VQ l)(4 in.)( in.)(3 in.) (4000 I ( in. 4 )(4 in.) psi rom Eq. (8-): (54,000 l-in.)( in.) 63.8 psi in. 4 s 1, s ; B s t ; 346. psi 30 psi 663 psi rom Eq. (8-4): t ma s B t 346 psi (c) TOP O THE BEM s Mc (54,000 l-in.)(4 in.) 166 psi I in. 4 Uniaial stress:, 166 psi, ma 633 psi Prolem n overanging eam B of rectangular cross section supports a concentrated load P at te free end (see figure). Te span lengt from to B is, and te lengt of te overang is /. Te cross section as widt and eigt. Point D is located midwa etween te supports at a distance d from te top face of te eam. Knowing tat te maimum tensile stress (principal stress) at point D is 36.1 MPa, determine te magnitude of te load P. Data for te eam are as follows:.5 m, 45 mm, 80 mm, and d 30 mm. D d B P Solution D Overanging eam.5 m 45 mm 80 mm d 30 mm Maimum principal stress at point D: 36.1 MPa Determine te load P units: P newton M D P P ˇm) (Nˇ V D P.5 P (N) STRESSES T POINT D 3 I mm 4 d B P d 60 mm s M (0.375 P)(60 mm) I mm 4,08.81 P (Pa) Q d d 01,50 mm3 t VQ I (0.5 P)(101,50 mm3 ) ( mm 4 )(45 mm) P (Pa) PRINIP STRESS s 1 s s B s s t s B s t P ( P) ( P) P (Pa) But 36.1 MPa P 35,000 N P 35.0 kn

19 SETION 8.4 Maimum Stresses in Beams 515 Prolem Solve te preceding prolem if te stress and dimensions are as follows: 30 psi, 75 in.,.0 in., 9.0 in., and d.5 in. Solution Overanging eam d P D B 75 in..0 in. 9.0 in. d.5 in. Maimum principal stress at point D: 30 psi Determine te load P units: P pounds M D P P (l-in.) 4 V D P.5 P (l) STRESSES T POINT D PRINIP STRESS s 1 s s B s B s t.3148 P ( P) ( P).4641 P (psi) But 30 psi s s t P 4999 l P 5.00 k 3 I 1.5 in.4 1 d 3.0 in. s M (18.75 P)(3.0 in.) I 11.5 in P (psi) Q d d 1.5 in.3 t VQ I (0.5 P)(11.5 in.3 ) (11.5 in. 4 )(.0 in.) P (psi) Prolem eam of wide-flange cross section (see figure) as te following dimensions: 0 mm, t 0 mm, 300 mm, and 1 60 mm. Te eam is simpl supported wit span lengt 3.0 m. concentrated load P 0 kn acts at te midpoint of te span. t a cross section located 1.0 m from te left-and support, determine te principal stresses and and te maimum sear stress ma at eac of te following locations: (a) te top of te eam, () te top of te we, and (c) te neutral ais. t 1

20 516 HPTER 8 pplications of Plane Stress Solution P 0 kn c.0 m 3.0 m M Pc 60 knˇ ˇm V P 60 kn 0 mm t 0 mm 300 mm 1 60 mm I 3 1 ( t) mm 4 1 (a) TOP O THE BEM (POINT ) c Simpl supported eam B t P 1 () TOP O THE WEB (POINT B) s M (60 knˇ ˇm)(130 mm) I mm MPa Q () 1 t VQ (60 kn)( mm 3 ) It ( mm 4 )(10 mm) MPa s 1, s s s s t B MPa 4.5 MPa, 76.1 MPa s s t ma t B 40.3 MPa (c) NEUTR XIS (POINT ) Q 4 ( t) mm 3 t VQ (60 kn)( mm 3 ) It ( mm 4 )(10 mm) 3.17 MPa Pure sear: 3. MPa, 3. MPa ma 3. MPa mm 3 4 s Mc (60 knˇ ˇm)(150 mm) I mm MPa Uniaial stress: 8.7 MPa ma 41.3 MPa

21 SETION 8.4 Maimum Stresses in Beam 517 Prolem eam of wide-flange cross section (see figure) as te following dimensions: 5 in., t.5 in., in., and in. Te eam is simpl supported wit span lengt 0 ft and supports a uniform load q 6 k/ft. alculate te principal stresses and and te maimum sear stress ma at a cross section located 3 ft from te left-and support at eac of te following locations: (a) te ottom of te eam, () te ottom of te we, and (c) te neutral ais. Solution Simpl supported eam q 6.0 kft c 3 ft 36 in. V q c 0 ft 0 in. qc,000 l M qc qc 5.0 in. t.5 in. in in. I 3 1 ( t) in. 4 1 (a) BOTTOM O THE BEM (POINT ) s Mc (756,000 l-in.)(6.0 in.) I in. 4 5,866 psi Uniaial stress: 5,870 psi, ma 7930 psi q 756,000 l-in. () BOTTOM O THE WEB (POINT B) s M (756,000 l-in.)(5.5 in.) I in. 4 3,883 psi t VQ (1,000 l)(1.094 in.3 ) It (85.89 in. 4 )(0.5 in.) 1771 psi s 1, s s s s t B psi 4,100 psi, 0 psi s s t ma t B 7160 psi (c) NEUTR XIS (POINT ) Q 1 Q 4 ( t) in. 3 t VQ (1,000 l)(7.984 in.3 ) It (85.89 in. 4 )(0.5 in.) 349 psi Pure sear: 350 psi, 350 psi, ma 350 psi in. 3 4 t B 1

22 518 HPTER 8 pplications of Plane Stress Prolem W 1 4 wide-flange eam (see Tale E-l, ppendi E) is simpl supported wit a span lengt of 8 ft (see figure). Te eam supports a concentrated load of 0 kips at midspan. t a cross section located ft from te left-and support, determine te principal stresses and and te maimum sear stress ma at eac of te following locations: (a) te top of te eam, () te top of te we, and (c) te neutral ais. W 1 4 D ft ft 8 ft 0 k 4 ft Solution t Section D-D: M P ( ft) k-ft V P 0 k ft t f D D Simpl supported eam 40 k-in. W1 4 I 88.6 in in. t w.00 in. t f.5 in in. 1 t f in. D ft 8 ft 96 in. B t w P k W ft 1 () TOP O THE WEB (POINT B) s M M( 1) 15,50 psi I I t VQ (10 k)(5.19 in.3 ) It w (88.6 in. 4 )(0.00 in.) 950 psi s 1, s s Q () 1 s s t B 7,760 psi 8,300 psi 540 psi, 16,060 psi s s t ma t B 8,300 psi (c) NEUTR XIS (POINT ) Q 4 ( t w) in. 3 t VQ (10 k)(8.50 in.3 ) It w (88.6 in. 4 )(0.00 in.) 4,800 psi in. 3 4 Pure sear: ma 4,800 psi, 4,800 psi, ma 4,800 psi (a) TOP O THE BEM (POINT ) s Mc M() I I 16,130 psi Uniaial stress:, s t ma 8,070 psi (40 k-in.)(5.955 in.) 88.6 in. 4 16,130 psi,

23 SETION 8.4 Maimum Stresses in Beam 519 Prolem W 8 8 wide-flange eam (see Tale E-l, ppendi E) is simpl supported wit a span lengt of 10 in. (see figure). Te eam supports two smmetricall placed concentrated loads of 6.0 k eac. t a cross section located 0 in. from te rigt-and support, determine te principal stresses and and te maimum sear stress ma at eac of te following locations: (a) te top of te eam, () te top of te we, and (c) te neutral ais. W in. 6.0 k 6.0 k D 40 in. 0 in. 0 in. 10 in. Solution Simpl supported eam 10 in. P 6.0 k t Section D-D: M P(0 in.) 0 k-in. V P 6.0 k W 8 8 I 98.0 in in. t w.85 in. t f.465 in in. 1 t f 7.13 in. (a) TOP O THE BEM (POINT ) s Mc M() I I 4435 psi Uniaial stress:, t ma W 8 8 t f 40 in. B P D t w 4930 psi, s 470 psi 40 in. 1 P D 0 in. 0 in. (10 k-in.)(4.03 in.) 98.0 in. 4 () TOP O THE WEB (POINT B) s M M( 1) I I 4365 psi t VQ (6.0 k)( in.3 ) It w (98.0 in. 4 )(0.85 in.) 479 psi s 1, s s Q () 1 s s t B 18 psi 3303 psi 10 psi, 5480 psi s s t ma t B 3300 psi (c) NEUTR XIS (POINT ) Q 4 ( t w) in. 3 (10 k-in.)(3.565 in.) 98.0 in. 4 t VQ (6.0 k)( in.3 ) It w (98.0 in. 4 )(0.85 in.) 870 psi in. 3 4 Pure sear: ma 870 psi, 870 psi, ma 870 psi

24 50 HPTER 8 pplications of Plane Stress Prolem cantilever eam of T-section is loaded an inclined force of magnitude 7.5 kn (see figure). Te line of action of te force is inclined at an angle of 60 to te oriontal and intersects te top of te eam at te end cross section. Te eam is.0 m long and te cross section as te dimensions sown. Determine te principal stresses and and te maimum sear stress ma at points and B in te we of te eam. B.0 m kn 5 mm 75 mm mm 75 5 mm 150 mm Solution B antilever eam of T-section.0 m P 7.5 kn.0 m (150 mm)(5 mm) 7500 mm 50 mm t 5 mm kn R c 1 c 150 mm B 5 mm 150 mm R 5 mm OTION O ENTROID rom Eq. (1-7) in apter 1: c a i i Use te ase of te cross section as te reference ais (line R-R) or te we: (75 mm) (5 mm)(150 mm) 81,50 mm 3 or te flange: (16.5 mm) (150 mm)(5 mm) 609,375 mm 3 c 81,50 mm3 609,375 mm mm mm c 1 75 mm c 56.5 mm MOMENT O INERTI I 3 tc 3 3 c ( t)(c 1 t) mm 4 EQUIVENT ODS T REE END O BEM P H P cos 60º 3.75 kn P V P sin 60º kn M P H c N m c 1 P V M c P H STRESS RESUTNTS T IXED END O BEM B R V N P H 3.75 kn V P V kn M M c P V 13,01 N m (Note tat M is negative) M N

25 SETION 8.4 Maimum Stresses in Beam 51 STRESS T POINT (BOTTOM O WEB) s N Mc I.50 MPa 7.64 MPa 7.14 MPa Uniaial stress:, 7.1 MPa, s t ma 36.1 MPa STRESS T POINT B (TOP O WEB) s N M(c 1 t) I.50 MPa 9.11 MPa 9.61 MPa t VQ I t (6.495 kn)( mm 3 ) ( mm 4 )(5 mm) 1.97 MPa Q t c 1 t mm 3 s 1, s s s s t B 9.81 MPa 0.00 MPa 9.8 MPa, 0. MPa s s t ma t B 0.0 MPa Prolem simple eam of rectangular cross section as span lengt 60 in. and supports a concentrated load P 8 k at te midpoint (see figure). Te eigt of te eam is 6 in. and te widt is in. Plot graps of te principal stresses and and te maimum sear stress ma, sowing ow te var over te eigt of te eam at cross section mn, wic is located 0 in. from te left-and support. m n P Solution Simple eam c m P n P 8 k 60 in. c in. in. 6 in.

26 5 HPTER 8 pplications of Plane Stress ROSS SETION mn Pc M 80 k-in. V P 9 k s M 1 M 5000 (1) I 3 Units: in., psi () Q 1 4 t VQ I 6V (9 ) Units: in., psi s 1 s B s t s s B s t t ma s B t G 3 in. B in. 3 O B D E G (3) (4) (5) (6) POINT ( in.) Eq. (1): 5000 psi Eq. (3): 1000 psi Eqs. (4), (5), and (6): 90 psi, 5190 psi, ma 690 psi POINT D ( ),, Pure sear: 15 psi, 115 psi, ma 15 psi POINT E ( 1 in.) t 3V Eq. (1): 5000 psi Eq. (3): 1000 psi Eqs. (4), (5), and (6): 5190 psi, 190 psi, ma 690 psi POINT ( in.) Eq. (1): 0,000 psi Eq. (3): 65 psi Eqs. (4), (5), and (6): 0,040 psi, 40 psi, ma 5040 psi 3V 115 psi POINT G ( 3 in.) Eq. (1): 5,000 psi Uniaial stress: 5,000 psi,, ma 7500 psi E in. 6 D POINT ( 3 in.) Eq. (1): 15,000 psi Uniaial stress:, 15,000 psi, ma 7500 psi GRPHS O STRESSES 0 15,000 B 40 10, D E , G (psi) 1 15,000 0 (psi) (psi) ma POINT B ( in.) Eq. (1): 10,000 psi Eq. (3): 65 psi Eqs. (4), (5), and (6): 40 psi, 10,040 psi, ma 5040 psi

27 SETION 8.4 Maimum Stresses in Beam 53 Prolem Solve te preceding prolem for a cross section mn located 0.15 m from te support if.7 m, P 44 kn, 0 mm, and mm. Solution c m Simple eam P 44 kn.7 m c.15 m mm 0 mm ROSS SETION mn Pc M V P 7 kn 0.8 ˇm knˇ s M 1 M 3.75 (1) I 3 Units: mm, MPa () Q 1 4 t VQ I 6V Units: mm, MPa s 1 s B s t s s B s t n t ma s B t O B D E G P G 60 mm B 40 mm 3 E mm 6 D POINT ( 60 mm) Eq. (1): 5 MPa Uniaial stress:, 5 MPa, ma 1 MPa (3) (4) (5) (6) POINT B ( 40 mm) Eq. (1): 150 MPa Eq. (3): 5 MPa Eqs. (4), (5), and (6): 4 MPa, 154 MPa, ma 79 MPa POINT ( mm) Eq. (1): 75 MPa Eq. (3): 40 MPa Eqs. (4), (5), and (6): 7 MPa, 9 MPa, ma 55 MPa POINT D ( ),, Pure sear: 45 MPa, 45 MPa, ma 45 MPa POINT E ( 0 mm) Eq. (1): 75 MPa Eq. (3): 40 MPa Eqs. (4), (5), and (6): 9 MPa, 17 MPa, ma 55 MPa POINT ( 40 mm) Eq. (1): 50 MPa Eq. (3): 5 MPa Eqs. (4), (5), and (6): 54 MPa, 4 MPa, ma 79 MPa POINT G ( 60 mm) Eq. (1): 5 MPa Uniaial stress: 5 MPa,, ma 1 MPa GRPHS O STRESSES t 3V 3V 45 MPa B D E G (MPa) 1 (MPa) (MPa) ma