160 CHAPTER 2 Axially Loaded Numbers. Stress Concentrations

Transcription

1 160 HTR xially oae Numers Stress oncentrations The prolems for Section.10 are to e solve y consiering the stress-concentration factors an assuming linearly elastic ehavior. rolem.10-1 The flat ars shown in parts (a) an () of the figure are sujecte to tensile forces.0 k. ach ar has thickness t 0.5 in. (a) For the ar with a circular hole, etermine the maximum stresses for hole iameters 1 in. an in. if the with 6.0 in. () For the steppe ar with shouler fillets, etermine the maximum stresses for fillet raii R 0.5 in. an R 0.5 in. if the ar withs are 4.0 in. an c.5 in. (a) R c ros an.10- Solution.10-1 Flat ars in tension R = raius c () (a).0 k t 0.5 in. (a) R WITH IRUR HO ( 6 in.) Otain K from Fig. -6 FOR 1 in.: c 5 in. s nom ct.0 k.40 ksi (5 in.)(0.5 in.) 1 / K.60 6 max k nom 6. ksi () STD R WITH SHOUDR FITS 4.0 in. c.5 in.; Otain k from Fig. -64 s nom ct.0 k 4.80 ksi (.5 in.)(0.5 in.) FOR R 0.5 in.: R/c 0.1 /c 1.60 k.0 max K nom 11.0 ksi FOR R 0.5 in.: R/c 0. /c 1.60 K 1.87 max K nom 9.0 ksi For in.: c 4 in. s nom ct.0 k.00 ksi (4 in.)(0.5 in.) 1 / K.1 max K nom 6.9 ksi

2 STION.10 Stress oncentrations 161 rolem.10- The flat ars shown in parts (a) an () of the figure are sujecte to tensile forces.5 kn. ach ar has thickness t 5.0 mm. (a) For the ar with a circular hole, etermine the maximum stresses for hole iameters 1 mm an 0 mm if the with 60 mm. () For the steppe ar with shouler fillets, etermine the maximum stresses for fillet raii R 6 mm an R 10 mm if the ar withs are 60 mm an c 40 mm. Solution.10- Flat ars in tension R = raius c () (a).5 kn t 5.0 mm (a) R WITH IRUR HO ( 60 mm) Otain k from Fig. -6 FOR 1 mm: c 48 mm s nom ct.5 kn 10.4 Ma (48 mm)(5 mm) 1 / K.51 5 max K nom 6 Ma () STD R WITH SHOUDR FITS 60 mm c 40 mm; Otain K from Fig. -64 s nom ct.5 kn 1.50 Ma (40 mm)(5 mm) FOR R 6 mm: R/c 0.15 /c 1.5 K.00 max K nom 5 Ma FOR R 10 mm: R/c 0.5 /c 1.5 K 1.75 max K nom Ma FOR 0 mm: c 40 mm s nom ct.5 kn 1.50 Ma (40 mm)(5 mm) 1 / K.1 max K nom 9 Ma

3 16 HTR xially oae Numers rolem.10- flat ar of with an thickness t has a hole of iameter rille through it (see figure). The hole may have any iameter that will fit within the ar. What is the maximum permissile tensile loa max if the allowale tensile stress in the material is t? Solution.10- Flat ar in tension t thickness t allowale tensile stress Fin max Fin K from Fig. -64 K * max s nom ct s max K s t K t 1 ecause t,, an t are constants, we write: * max s t t 1 K 1 ct s t K ( )t We oserve that max ecreases as increases. Therefore, the maximum loa occurs when the hole ecomes very small. ( S 0&K S ) max s t t rolem.10-4 roun rass ar of iameter 1 0 mm has upset ens of iameter 6 mm (see figure). The lengths of the segments of the ar are 1 0. m an 0.1 m. Quarter-circular fillets are use at the shoulers of the ar, an the moulus of elasticity of the rass is 100 Ga. If the ar lengthens y 0.1 mm uner a tensile loa, what is the maximum stress max in the ar? 1 1 ros an.10-5

4 STION.10 Stress oncentrations 16 Solution.10-4 Roun rass ar with upset ens = 6 mm 1 = 0 mm 100 Ga 0.1 mm 0.1 m 1 0. m 1 R raius of fillets 1 1 Solve for : mm 0 mm mm Use Fig. -65 for the stress-concentration factor: s nom ( 1 ) 1 SUSTITUT NUMRI VUS: R mm 0.15 D 1 0 mm Use the ashe curve in Fig K 1.6 s max Ks nom (1.6)(8.68 Ma) 46 Ma ( 1 ) 1 (0.1 mm)(100 Ga) s nom 8.68 Ma (0.1 m)( 0 6) 0. m rolem.10-5 Solve the preceing prolem for a ar of monel metal having the following properties: in., 1.4 in., in., 5.0 in., an psi. lso, the ar lengthens y in. when the tensile loa is applie. Solution.10-5 Roun ar with upset ens = 1.4 in. 1 = 1.0 in psi in. 1 0 in. 5 in. 1.4 in. 1.0 in. R raius of filletsr 0. in Solve for : Use Fig. -65 for the stress-concentration factor. s nom ( 1 ) 1 SUSTITUT NUMRI VUS: s nom ( in.)(5 106 psi) (5 in.)( ) 0 in. R 0. in. 0. D in. Use the ashe curve in Fig K 1.5 s max Ks nom (1.5)(984 psi) 6100 psi ( 1 ) 1,984 psi

5 164 HTR xially oae Numers rolem.10-6 prismatic ar of iameter 0 0 mm is eing compare with a steppe ar of the same iameter ( 1 0 mm) that is enlarge in the mile region to a iameter 5 mm (see figure). The raius of the fillets in the steppe ar is.0 mm. 1 (a) Does enlarging the ar in the mile region make it stronger than the prismatic ar? Demonstrate your answer y etermining the maximum permissile loa 1 for the prismatic ar an the maximum permissile loa for the enlarge ar, assuming that the allowale stress for the material is 80 Ma. () What shoul e the iameter 0 of the prismatic ar if it is to have the same maximum permissile loa as oes the steppe ar? Soluton.10-6 rismatic ar an steppe ar mm 1 0 mm 5 mm Fillet raius: R mm llowale stress: t 80 Ma (a) OMRISON OF RS rismatic ar: 1 s t 0 s 0 t 4 1 Steppe ar: See Fig. -65 for the stressconcentration factor. R.0 mm D 1 0 mm D 5 mm RD D 1D 1.5K 1.75 s nom s nom 1 s max K 80 Ma 1.75 (0 mm) kn s nom s max K 1 s t K 1 nlarging the ar makes it weaker, not stronger. The ratio of loas is 1 K 1.75 () DIMTR OF RISMTI R FOR TH SM OW OD 1 s 0 t 4 s t K K mm 15.1 mm K 1.75 (80 Ma) 4 (0 mm) 5.1 kn

6 STION.10 Stress oncentrations 165 rolem.10-7 steppe ar with a hole (see figure) has withs.4 in. an c 1.6 in. The fillets have raii equal to 0. in. What is the iameter max of the largest hole that can e rille through the ar without reucing the loa-carrying capacity? c Solution 10-7 Steppe ar with a hole c.4 in. c 1.6 in. Fillet raius: R 0. in. Fin max SD UON FITS (Use Fig. -64).4 in. c 1.6 in. R 0. in. R/c 0.15 /c 1.5 K.10 max s nom ct s max K 0.17 t s max ct s max K c (t) SD UON HO (Use Fig. -6).4 in. iameter of the hole (in.) c 1 max s nom c,t s max ( )t K 1 K 1 ts max (in.) / K max ts max ase upon hole ase upon fillets max t max 0.1 max 0.51 in. 0.0 (in.)

7 166 HTR xially oae Numers Nonlinear ehavior (hanges in engths of ars) rolem.11-1 ar of length an weight ensity hangs vertically uner its own weight (see figure). The stress-strain relation for the material is given y the Ramerg-Osgoo equation (q. -71): Derive the following formula for the elongation of the ar. 0 m 0 (m 1) m 0 0 Solution.11-1 ar hanging uner its own weight STRIN T DISTN x x et cross-sectional area et N axial force at istance x e s s 0 s m gx s 0 s m 0 gx s 0 x N x s N gx ONGTION OF R gx e x x s g s m 0 (m 1) g Q..D. s 0 0 gx m x s 0 rolem.11- prismatic ar of length 1.8 m an cross-sectional area 480 mm is loae y forces 1 0 kn an 60 kn (see figure). The ar is constructe of magnesium alloy having a stress-strain curve escrie y the following Ramerg-Osgoo equation: , ( Ma) in which has units of megapascals. (a) alculate the isplacement of the en of the ar when the loa 1 acts alone. () alculate the isplacement when the loa acts alone. (c) alculate the isplacement when oth loas act simultaneously. Solution.11- xially loae ar m 480 mm 1 0 kn 60 kn RamergOsgoo quation: e s Ma) 45,000 1 s (s Fin isplacement at en of ar.

8 STION.11 Nonlinear ehavior 167 (a) 1 TS ON s 1 0 kn 6.5 Ma 480 mm e c e 1.67 mm () TS ON s 60 kn 15 Ma 480 mm e c e 5.1 mm (c) OTH 1 ND R TING s 1 e e mm s 60 kn 15 Ma 480 mm e e 1.71 mm 90 kn Ma 480 mm mm (Note that the isplacement when oth loas act simultaneously is not equal to the sum of the isplacements when the loas act separately.) rolem.11- circular ar of length in. an iameter 0.75 in. is sujecte to tension y forces (see figure). The wire is mae of a copper alloy having the following hyperolic stress-strain relationship: 18, ( ksi) 1 00 (a) Draw a stress-strain iagram for the material. () If the elongation of the wire is limite to 0.5 in. an the maximum stress is limite to 40 ksi, what is the allowale loa? Solution.11- in. 4 opper ar in tension 0.75 in in. () OW OD Max. elongation max 0.5 in. Max. stress max 40 ksi ase upon elongation: e max max 0.5 in in. (a) STRSS-STRIN DIGRM s 18,000e 0 e 0.0(s ksi) 1 00e (ksi) Slope = 18,000 ksi symptote equals 60 ksi s max 18,000 e max 1 00 e max 4.06 ksi SD UON STRSS: s max 40 ksi Stress governs. max (40 ksi)( in. ) 17.7 K

9 168 HTR xially oae Numers rolem.11-4 prismatic ar in tension has length.0 m an cross-sectional area 49 mm. The material of the ar has the stress-strain curve shown in the figure. Determine the elongation of the ar for each of the following axial loas: 10 kn, 0 kn, 0 kn, 40 kn, an 45 kn. From these results, plot a iagram of loa versus elongation (loa-isplacement iagram). 00 (Ma) Solution.11-4 ar in tension m 49 mm (kn) 0 0 STRSS-STRIN DIGRM (See the prolem statement for the iagram) OD-DISMNT DIGRM (mm) / e e (kn) (Ma) (from iagram) (mm) NOT: The loa-isplacement curve has the same shape as the stress-strain curve. rolem.11-5 n aluminum ar sujecte to tensile forces has length 150 in. an cross-sectional area.0 in. The stress-strain ehavior of the aluminum may e represente approximately y the ilinear stress-strain iagram shown in the figure. alculate the elongation of the ar for each of the following axial loas: 8 k, 16 k, 4 k, k, an 40 k. From these results, plot a iagram of loa versus elongation (loa-isplacement iagram). 1,000 psi 1 = psi = psi 0

10 STION.11 Nonlinear ehavior 169 Solution.11-5 luminum ar in tension OD-DISMNT DIGRM 150 in..0 in. / e e (k) (psi) (from q. 1 or q. ) (in.) 8 4, , STRSS-STRIN DIGRM 4 1, , , k k 0 1 (k) in psi in psi 1 1,000 psi e 1 s 1 1,000 psi psi For 0 s s 1 : (in.) e s s (s psi) psi q. (1) For s s 1 : e e 1 s s 1 s 1, s 0.008(s psi) q. ()

11 170 HTR xially oae Numers rolem.11-6 rigi ar, pinne at en, is supporte y a wire D an loae y a force at en (see figure). The wire is mae of high-strength steel having moulus of elasticity 10 Ga an yiel stress Y 80 Ma. The length of the wire is 1.0 m an its iameter is mm. The stress-strain iagram for the steel is efine y the moifie power law, as follows: Y 0 Y n Y Y D (a) ssuming n 0., calculate the isplacement at the en of the ar ue to the loa. Take values of from.4 kn to 5.6 kn in increments of 0.8 kn. () lot a loa-isplacement iagram showing versus. Solution.11-6 Rigi ar supporte y a wire From q. ():e s Y s 1n s Y (5) xial force in wire: F Wire: 10 Ga D Stress in wire: s F RODUR: ssume a value of alculate from q. (6) alculate e from q. (4) or (5) alculate from q. () (6) Y 80 Ma 1.0 m mm mm 4 STRSS-STRIN DIGRM s e(0 s s Y ) s s e n Y (s s s Y )(n 0.) Y (1) () (Ma) e (mm) (kn) q. (6) q. (4) or (5) q. () For Y 80 Ma: e kn 5.86 mm (a) DISMNT T ND OF R () OD-DISMNT DIGRM elongation of wire e () 8 Otain e from stress-strain equations: From q. (1):e s (0 s s Y) (4) (kn) 6 4 =.86 kn Y = 80 Ma = 5.86 mm (mm)

12 STION.1 lastoplastic nalysis 171 lastoplastic nalysis The prolems for Section.1 are to e solve assuming that the material is elastoplastic with yiel stress Y, yiel strain Y, an moulus of elasticity in the linearly elastic region (see Fig. -70). rolem.1-1 Two ientical ars an support a vertical loa (see figure). The ars are mae of steel having a stress-strain curve that may e iealize as elastoplastic with yiel stress Y. ach ar has cross-sectional area. Determine the yiel loa Y an the plastic loa. Solution.1-1 Two ars supporting a loa Structure is statically eterminate. The yiel loa Y an the plastic lea occur at the same time, namely, when oth ars reach the yiel stress. JOINT F vert 0 ( Y ) sin Y s Y sin u rolem.1- steppe ar with circular cross sections is hel etween rigi supports an loae y an axial force at milength (see figure). The iameters for the two parts of the ar are 1 0 mm an 5 mm, an the material is elastoplastic with yiel stress Y 50 Ma. Determine the plastic loa. 1

13 17 HTR xially oae Numers Solution.1- ar etween rigi supports mm 5 mm Y 50 Ma DTRMIN TH STI OD : at the plastic loa, all parts of the ar are stresse to the yiel stress. oint : F F SUSTITUT NUMRI VUS: (50 Ma) 4 ( 1 ) (50 Ma) 4 [(0 mm) (5 mm) ] 01 kn F Y 1 F Y F F s Y 1 s Y s Y ( 1 ) rolem.1- horizontal rigi ar supporting a loa is hung from five symmetrically place wires, each of cross-sectional area (see figure). The wires are fastene to a curve surface of raius R. (a) Determine the plastic loa if the material of the wires is elastoplastic with yiel stress Y. () How is change if ar is flexile instea of rigi? (c) How is change if the raius R is increase? R Solution.1- Rigi ar supporte y five wires (a) STI OD F F F F F t the plastic loa, each wire is stresse to the yiel stress. 5s Y F Y () R IS FXI t the plastic loa, each wire is stresse to the yiel stress, so the plastic loa is not change. (c) RDIUS R IS INRSD t the plastic loa, each wire is stresse to the yiel stress, so the plastic loa is not change.

14 STION.1 lastoplastic nalysis 17 rolem.1-4 loa acts on a horizontal eam that is supporte y four ros arrange in the symmetrical pattern shown in the figure. ach ro has cross-sectional area an the material is elastoplastic with yiel stress Y. Determine the plastic loa. Solution.1-4 eam supporte y four ros F F F F t the plastic loa, all four ros are stresse to the yiel stress. F Y Sum forces in the vertical irection an solve for the loa: F F sin s Y (1 sin ) rolem.1-5 The symmetric truss D shown in the figure is constructe of four ars an supports a loa at joint. ach of the two outer ars has a cross-sectional area of 0.07 in., an each of the two inner ars has an area of in. The material is elastoplastic with yiel stress Y 6 ksi. Determine the plastic loa. 1 in. 54 in. 1 in. D 6 in.

15 174 HTR xially oae Numers Solution.1-5 Truss with four ars 1 in. 7 in. 7 in. 1 in. D STI OD t the plastic loa, all ars are stresse to the yiel stress in. F Y F Y 6 5 s Y 8 5 s Y SUSTITUT NUMRI VUS: 60 in. JOINT F F 45 in. quilirium: F 5 F 4 5 or 0.07 in in. Y 6 ksi 8 5 (6 ksi)( in. ) 5 (6 ksi)(0.07 in. ) 1.6 k 4.6 k 47.9 k 6 5 F 8 5 F rolem.1-6 Five ars, each having a iameter of 10 mm, support a loa as shown in the figure. Determine the plastic loa if the material is elastoplastic with yiel stress Y 50 Ma. Solution.1-6 Truss consisting of five ars F F F F F t the plastic loa, all five ars are stresse to the yiel stress F Y Sum forces in the vertical irection an solve for the loa: F 1 F 5 F 10 mm 4 Y 50 Ma mm s Y (5 45 5) s Y Sustitute numerical values: (4.01)(50 Ma)(78.54 mm ) 8.5 kn

16 STION.1 lastoplastic nalysis 175 rolem.1-7 circular steel ro of iameter 0.60 in. is stretche tightly etween two supports so that initially the tensile stress in the ro is 10 ksi (see figure). n axial force is then applie to the ro at an intermeiate location. (a) Determine the plastic loa if the material is elastoplastic with yiel stress Y 6 ksi. () How is change if the initial tensile stress is oule to 0 ksi? Solution.1-7 ar hel etween rigi supports OINT : 0.6 in. Y 6 ksi Initial tensile stress 10 ksi s Y ()(6 ksi) (0.60 in.) k (a) STI OD The presence of the initial tensile stress oes not affect the plastic loa. oth parts of the ar must yiel in orer to reach the plastic loa. () INITI TNSI STRSS IS DOUD is not change. rolem.1-8 rigi ar is supporte on a fulcrum at an loae y a force at en (see figure). Three ientical wires mae of an elastoplastic material (yiel stress Y an moulus of elasticity ) resist the loa. ach wire has cross-sectional area an length. (a) Determine the yiel loa Y an the corresponing yiel isplacement Y at point. () Determine the plastic loa an the corresponing isplacement at point when the loa just reaches the value. (c) Draw a loa-isplacement iagram with the loa as orinate an the isplacement of point as ascissa. a a a a

17 176 HTR xially oae Numers Solution.1-8 Rigi ar supporte y wires () STI OD Y Y a a a a (a) YID OD Y Yieling occurs when the most highly stresse wire reaches the yiel stress Y. Y M 0 Y s Y t point : s Y s Y t point : Y s Y Y Y Y t the plastic loa, all wires reach the yiel stress. M 0 4s Y t point : (s Y ) s Y t point : s Y (c) OD-DISMNT DIGRM Y Y 0 Y 4 Y Y rolem.1-9 The structure shown in the figure consists of a horizontal rigi ar D supporte y two steel wires, one of length an the other of length /4. oth wires have cross-sectional area an are mae of elastoplastic material with yiel stress Y an moulus of elasticity. vertical loa acts at en D of the ar. 4 D (a) Determine the yiel loa Y an the corresponing yiel isplacement Y at point D. () Determine the plastic loa an the corresponing isplacement at point D when the loa just reaches the value. (c) Draw a loa-isplacement iagram with the loa as orinate an the isplacement D of point D as ascissa.

18 STION.1 lastoplastic nalysis 177 Solution.1-9 Rigi ar supporte y two wires STRSSS 4 D s F (7) s F s s Wire has the larger stress. Therefore, it will yiel first. (a) YID OD s s Y s s s Y (From q. 7) cross-sectional area Y yiel stress moulus of elasticity DISMNT DIGRM D D OMTIIITY: D FR-ODY DIGRM F F D (1) () F s Y F 1 s Y From q. (): 1 s Y (s Y ) 4 Y s Y From q. (4): F s Y From q. (): D Y s Y () STI OD t the plastic loa, oth wires yiel. Y F F Y From q. (): ( Y ) ( Y ) s Y QUIIRIUM: M 0 F () F () (4) F F 4 () From q. (4): F s Y From q. (): D s Y FOR-DISMNT RTIONS F F 4 Sustitute into q. (1): F 4 F (4, 5) F F (6) (c) OD-DISMNT DIGRM Y 0 Y D 5 4 Y Y

19 178 HTR xially oae Numers rolem.1-10 Two cales, each having a length of approximately 40 m, support a loae container of weight W (see figure). The cales, which have effective cross-sectional area 48.0 mm an effective moulus of elasticity 160 Ga, are ientical except that one cale is longer than the other when they are hanging separately an unloae. The ifference in lengths is 100 mm. The cales are mae of steel having an elastoplastic stress-strain iagram with Y 500 Ma. ssume that the weight W is initially zero an is slowly increase y the aition of material to the container. (a) Determine the weight W Y that first prouces yieling of the shorter cale. lso, etermine the corresponing elongation Y of the shorter cale. () Determine the weight W that prouces yieling of oth cales. lso, etermine the elongation of the shorter cale when the weight W just reaches the value W. (c) onstruct a loa-isplacement iagram showing the weight W as orinate an the elongation of the shorter cale as ascissa. (Hint: The loa isplacement iagram is not a single straight line in the region 0 W W Y.) W Solution W s 1 W 1 (a) YID OD W Y Two cales supporting a loa 40 m 48.0 mm 160 Ga ifference in length 100 mm Y 500 Ma INITI STRTHING OF 1 Initially, cale 1 supports all of the loa. et W 1 loa require to stretch cale 1 to the same length as cale mm (elongation of cale 1 ) 400 Ma (s 1 6 s Y ale 1 yiels first. F 1 Y 4 kn 1Y total elongation of cale 1 1Y F 1 s Y 0.15 m 15 mm Y 1Y 15 mm Y elongation of cale 1Y 5 mm F Y 4.8 kn W Y F 1 F 4 kn 4.8 kn 8.8 kn W kn OK) () STI OD W F 1 s Y F s Y W s Y 48 kn elongation of cale F s Y 0.15 mm 15mm 1 5 mm 1 5 mm (c) OD-DISMNT DIGRM W (kn) W 1 W Y W 1 Y W Y W Y W W Y Y (mm) 0 W W 1 : slope 19,000 N/m W 1 W W Y : slope 84,000 N/m W Y W W : slope 19,000 N/m

20 STION.1 lastoplastic nalysis 179 rolem.1-11 hollow circular tue T of length 15 in. is uniformly compresse y a force acting through a rigi plate (see figure). The outsie an insie iameters of the tue are.0 an.75 in., repectively. concentric soli circular ar of 1.5 in. iameter is mounte insie the tue. When no loa is present, there c in. etween the ar an the rigi plate. oth ar an tue are mae of steel having an elastoplastic stress-strain iagram with 9 10 ksi an Y 6 ksi. T (a) Determine the yiel loa Y an the corresponing shortening Y of the tue. () Determine the plastic loa an the corresponing shortening of the tue. (c) onstruct a loa-isplacement iagram showing the loa as orinate an the shortening of the tue as ascissa. (Hint: The loa-isplacement iagram is not a single straight line in the region 0 Y.) T T Solution.1-11 Tue an ar supporting a loa learance = T T T R: 1.5 in. 15 in in ksi Y 6 ksi TU:.0 in in. T 4 ( 1 ) in in. 4 INITI SHORTNING OF TU T Initially, the tue supports all of the loa. et 1 loa require to close the clearance 1 T 1,87 l et 1 shortening of tue in. s 1 1 T 19,0 psi (s 1 6 s Y OK)

21 180 HTR xially oae Numers (a) YID OD Y ecause the tue an ar are mae of the same material, an ecause the strain in the tue is larger than the strain in the ar, the tue will yiel first. F T Y T 40,644 l TY shortening of tue at the yiel stress TY F T s Y in. T Y TY in. Y shortening of ar TY in. F Y 9,45 l Y F T F 40,644 l 9,45 l 70,097 l Y 70,100 l () STI OD (c) OD-DISMNT DIGRM 100 (kips) y Y 1.1 Y Y 1.49 Y Y 0 1 : slope 180 k/in. 1 Y : slope 5600 k/in. Y : slope 40 k/in (in.) F T Y T F Y F T F s Y ( T ) 104,00 l shortening of ar F s Y in. T in. T in.