Stress and Strain Stress Stress is defined as frce per unit area. It has the same units as pressure, and in fact pressure is ne special variety f stress. When a bdy is subjected t a set f frces and is in equilibrium, then the internal frce develped per unit area is called the stress. Types f Stresses: Stresses are three types such as nrmal (tensile r cmpressive), shearing (direct r trsinal), and bearing stress. Nrmal Stress: Stress that acts perpendicular t a surface. t equilibrium, the intensity f the frce perpendicular t r nrmal t the sectin pint f the bdy is called the nrmal stress at a pint. It can be either tensin r cmpressin. It is dented by. Tensin Stress that acts t lengthen an bject is termed as tensinal stress. Cmpressin Stress that acts t shrten an bject is termed as cmpressinal stress. Shearing Stress: Stress that acts parallel t a surface is termed as shearing stress. It can cause ne bject t slide ver anther. It als tends t defrm riginally rectangular bjects int parallelgrams. The mst general definitin is that shear acts t change the angles in an bject. Shearing stress is dented by τ (ta). It may be direct r trsinal. Bearing stress: Bearing stress is the stress due t the frce that acts perpendicular t the cmmn plane. It is the cntact pressure between separate bdies. s fr eample, sil pressure, pressure between rivet and the cntact surface. Unit f Stress: (a) SI Unit: kg/cm, tn/cm, N/mm (Ma), kn/m (b) British Unit: Ib/in (psi), lb/ft, kip/in (ksi).
Strain: Defrmatin f anybdy per unit length is termed as strain. Strain is defined as the amunt f defrmatin an bject eperiences cmpared t its riginal size and shape. Fr eample, if a blck L cm n a side is defrmed s that it becmes l cm lng, then the elngatin is (Ll) and the strain is /L. Nte that strain is dimensinless. Strain Types f Strain: Strain is tw types: i) Linear Strain ii) Shearing Strain Linear r Lngitudinal Strain: Strain that changes the length f a line withut changing its directin is termed as linear strain. It can be either cmpressinal r tensinal. Linear strain is dented by ε. Cmpressin Lngitudinal r linear strain that shrtens an bject. Tensin Lngitudinal r linear strain that lengthens an bject. Shearing strain Strain that changes the angles f an bject. Shear causes lines t rtate. Shearing strain is assciated with the change in angles. It is dented by γ. Thermal strain Strain due t change in temperature. typical stress-strain diagram: The lad per unit area r stress was pltted against the elngatin per unit length r strain is termed as stress-strain diagram. Figure shwn in belw is a typical stress-strain diagram. rprtinal Limit In the stress-strain diagram, prprtinal limit up t that pint where linear variatin can be btained i.e. stress is prprtinal t strain.
Figure: typical stress-strain diagram Elastic Limit The elastic limit is the stress beynd which the material will nt return t its riginal shape when unladed. ermanent Set It is the pint after elastic limit at which the material des nt return in its riginal psitin. There is a permanent defrmatin called permanent set. Yield int The yield pint is the pint at which there is an appreciable elngatin r yielding f the material withut any crrespnding increase f lad; indeed, the lad may actually decrease while the yielding ccurs. Hwever, the phenmenn f yielding is peculiar t structural steel; ther grades f steel and steel allys r ther materials d nt pssess it, as is indicated by the typical stress-strain curves f these materials shwn in the fllwing figures. Figure: Stress-strain diagram f different materials 3
Yield Strength It is clsely assciated with yield pint. In which lad yield pint created that is called yield strength f the materials. Fr materials like timber, aluminum, which des nt have a welldefined yield pint, yield strength is determined by the 0.% ffset methd. This cnsists f drawing a line parallel t the initial tangent f the stress-strain curve, this line being started at an arbitrary ffset- strain, usually f 0.% (0.00 mm/mm r 0.00 in/in.). s shwn in the fllwing figure, the intersectin f that line with the stress-strain curve is called the yield strength. Figure: Offset methd t determine yield strength Ultimate Strength The ultimate stress, r ultimate strength as it is mre cmmnly called, is the highest rdinate n the stress-strain curve. Rupture Strength The rupture strength is the stress at failure. Fr structural steel it is smewhat lwer than ultimate strength because the rupture strength is cmputed by dividing the rupture lad by the riginal crss-sectinal area, which, althugh cnvenient, is incrrect. The errr is caused by a phenmenn knwn as necking. s failure ccurs, the material stretches very rapidly and simultaneusly narrws dwn as shwn in the fllwing figure. 4
Hke's Law: Fr hmgeneus istrpic materials i.e. materials having the same prperties in all directins, stress is directly prprtinal t strain up t prprtinal limit r elastic limit. i.e. α ε r ε.e ------------------- (i) Where, E Mdulus f elasticity Nw, if ε r unit strain then stress, E S mdulus f elasticity may be defined as the stress f a bdy fr its unit strain. It can als be defined as the slpe f straight-line prtin f the stress-strain diagram. The unit f E is the unit f. cnvenient variatin f Hke's law is btained by replacing by its equivalent and replacing ε by, s that equatin (i) becmes- Or, --------------- (ii) Equatin (ii) epresses the relatin amng the ttal defrmatin, the applied lad, the ttal length L, the crss-sectinal area, and the mdulus f elasticity E. The unit defrmatin has the same unit as length L, since the units f and E, being equivalent. Nte that, equatin (ii) is subject t the restrictins mentined belw:. The lad must be aial.. The bar must have a cnstant crss-sectin. 3. Material must be hmgeneus and istrpic. 4. The stress must nt eceed the prprtinal limit. Mdulus f Rigidity Fr hmgeneus, istrpic materials shearing stress is directly prprtinal t shearing strain up t prprtinal limit. i.e. Or,, Where, G Mdulus f rigidity Nw, if γ r unit shearing strain then τ G, S mdulus f rigidity may be defined as the shearing stress f a bdy fr its unit shearing strain. The unit f G is the unit f τ. 5
issn's Rati Eperiment shw that if a bar is lengthened by aial tensin, there is a reductin in the transverse dimensins. Simen D. issn shwed that the rati f the unit defrmatins r strains in these directins is cnstant fr stresses within the prprtinal limit. ccrdingly, this ratin is named after him is issn's rati. issn's rati is the rati f the lateral strain t aial strain. It is dented by ν (neu). Mathematically, ν Where, is aial strain in X-directin and r is lateral strain in Y-directin and Z- directin. Cmmn values f issn's rati are 0.0 fr cncrete, 0.5 t 0.30 fr steel and 0.33 fr mst ther materials. Bi-aial and tri-aial strain issn's rati permits t etend Hke's law f uniaial stress t the case f bi-aial and triaial stresses. Fllwing figure shws that an element is subjected simultaneusly t tensile stress in the and y directin. The strain in the -directin due t tensile stress is, simultaneusly the tensile stress will prduce lateral cntractin in the -directin f the amunt -ν r -νε. 0 + 0 0 0 S, the resultant unit defrmatin r strain in the -directin will be, + But and -ν ν Similarly the resultant unit defrmatin r strain in the y-directin will be, + But ν and 6 ν fr tw dimensinal bdy, ν ν and ν ν further etensin f this discussin results in the fllwing epressins fr strain caused by simultaneus actin f triaial tensile stresses: ν( + ), ν( + ), and ν( + )
rblem: piece f 5 cm by 5 cm by mm steel plate is subjected t unifrmly distributed stresses alng its edges shwn in the fllwing Figure. (a) If 0t and y 0t, what change in thickness will ccur due t the applicatin f these frces? (b) T cause the same change in thickness as in (a) by alne, what must be its magnitude? Let, E 0 5 kg/cm and ν 0.5. Slutin: We have, ν( + ) We get, 666.67 /,. 666.67. /, and 0 () ν( + ) 7.77 0 / 0 0.5 (666.67+666.67) Change in thickness 7.77 0. 33.33 0 (cntractin) (ns) (b) We have, 7.77 0, 0,? We get, ν( + ) 7.77 0 33.68 / 0 0.5 ( +0) gain we get, 33.68 33.68 5. 4000 4 (ns) Bulk Mdulus: The rati f the hydrstatic cmpressive stress t the decrease in vlume is called mdulus f cmpressin r bulk mdulus. 7
Relatin between bulk mdulus, mdulus f elasticity and issn's rati: Cnsidering an infinitesimal elements whse sides are d, dy and dz. S, the initial vlume f the element will be ddydz. fter straining the sides will becme +, +, and +. They may be written as d(+ ), (+ ), dz(+ ) respectively. fter subtracting the initial vlume f the strained element, the change in vlume is determined. The change in vlume is d(+ )+ dz(+ )- ddydz (+ + + + + + + ) (+ + + ) [Nte: the prduct f strain + + + are t small are neglected] (+ + + ) +( + + ) ( + + ) S, the change in vlume per unit vlume will be ( ) ( + + ) Therefre, in the infinitesimal strain thery, the change in vlume per unit vlume, ften referred as dilatatin, e and is defined as e + +. Based n the generalized Hke's law, the dilatatin can be fund in terms f stresses and material cnstants. Fr this purpse, the fllwing three equatins must be added tgether. ν( + ), ν( + ), and ν( + ) They yields, + + (ν) ( + + ). Which means that dilatatin is prprtinal t the algebraic sum f all nrmal stresses. If an elastic bdy is subjected t hydrstatic pressure f unifrm intensity p, s that yz, then we get ν( ) 3 ν() (ν) The quantity k represents the rati f the hydrstatic cmpressive stress t the decrease in vlume and is called bulk mdulus. S, the relatin yields, (ν). ssignment: Shw that fr any material, relatin between mdulus f elasticity, mdulus f rigidity, and issn's rati is, (ν ). 8
Deflectin (defrmatin) f an ially Laded Member: When the deflectin f an aially laded member is a design parameter, it is necessary t determine the defrmatins. ial defrmatins are als required in the analysis f statically indeterminate bars. The deflectin characteristics f bars als prvide necessary infrmatin fr determining the stiffness f systems in mechanical vibratin analysis. Cnsider the aially laded bar shwn in the fllwing figure fr deriving a relatin fr aial bar defrmatin. The applied frces,, and 3 are held in equilibrium by the frce 4. The crss-sectinal area f the bar is permitted t change gradually. The change in length that takes place in the bar between pints B and D due t the applied frce is t be determined. n arbitrary element cut frm the bar is als shwn in figure. Frm free-bdy cnsideratins, this element is subjected t a pull (), which, in general, is a variable quantity. The infinitesimal defrmatin dδ that takes place within this element upn applicatin f the frces is equal t the strainε multiplied by the length d. The ttal defrmatin between any tw given pints n a bar is simply the sum f the element defrmatins. Nw, dδ ε dδ ε d d δ dδ δ ε d The magnitude f the strain ε depends n the magnitude f the stress. The is fund in general by dividing the variable frce () by the crrespnding area () i.e. δ E linearly elastic materials, accrding t Hke's law fr uniaial stress, ε E δ ( ) ( ) d E ( ) ( ). Fr 9
rblem: Find the deflectin f the free end caused by the applicatin f a cncentrated frce. The elastic mdulus f the material is E (neglect the weight f the bar). The crsssectin area f the bar is. B d Slutin : We get, ( ) and ε ( ) E E ( ) ( ) d We knw, δ + C E t 0, δ, then C 0 L t L, δ ns. E ( ) δ δ d δ + C E E E rblem: cpper aerial wire 60 ft. lng, weighting 0.0396 lb per ft. and having a crss sectinal area f 0.008 sq. in. is suspended vertically frm its upper end; E 5 0 6 psi (i) Calculate the ttal amunt f elngatin f this wise because f its wn weight (ii) Find the ttal elngatin if a weight f 50 lb is attached t its lwer end (iii) Determine what maimum weight W, this wire can safely supprt at its lwer en if the unit tress in it must nt eceed its elastic limit f 0,000 psi. dy 60 0.008 sq.in. 0
Slutin: Here, 6 ( 0.0396 y) lb 0.0396y lb, E 5 0 psi, L dy, and 0.008sq. in. L L 0.096ydy (i) We get, δ E δ 0 0.058 5 0 60 3.85 3.85 y 60 3.85 δ ydy 6 6 6 5 0 5 0 0 5 0 0 60 (.) 3 δ 3.87 0 ft 0.0394inch ns 6 ( 60) (ii) Here, 50 lb, L 60, 0.008 sq. in., E 5 6 6 psi L 50 60 δ 0.058 ft 0.65inch. 6 E 0.08 5 0 Ttal elngatin δ +δ 0.394+0.650.66 inch (ns.) W 0.0396 60 (iii) Stress due t self wt 0.08 Maimum stress, 0,000 psi. - 0,-66 9384 psi W W 9384 W 965lb ns 0.008 (.) 66psi W? ssignment: steel bar 500 ft lng, suspended vertically frm its ne end, supprts a lad at its lwer end. The maimum stress in the bar is 8000 psi caused by the lad and by its wn weight. If E 30 0 6 psi and steel weigh 490 lb/ft, find the ttal elngatin f the bar. (ns δ.43 inch). ssignment: steel rd having a crss-sectinal area / in and a length f 600 ft is suspended vertically. It supprts a lad f 5000 at its lwer end. If steel weigh 490 lb/ft 3 and E 30 0 6 psi and, find the ttal elngatin in the rd. (ns δ.64 inch). rblem : Tw bars are t be cut frm a cm thick metal plate s that bth bars have a cnstant thickness f cm. Bar is t have a cnstant width f 4 cm thrughut its entire length. Bar B is t be 6 cm in wide at the tp and cm wide at the bttm. Each bar is t be subjected t the same lad. Determine the rati amunt. Neglect the weight f the bar. L L B s that bth bars will stretch the same
6cm B B 4cm L LB y LB cm cm Slutin: Fr Bar, Frm Bar B, gain, 3LB LB LB / L L E L L E 8E X X + ( 4) X 6 3 L B LB y 4 y L B L B... ( ) ; rea ( ) ( ) E L d 8E B lg e 3... L LB Nw, L LB lge 3 8E 8E ( ) 4 8 y L B L B L lg e 3. L B ( ns.) X rblem: The dimensins f a frustum f a right circular supprted at the large end n a rigid base are shwn in the fllwing figure. Determine the deflectin f the tp due t the weight f the bdy. The unit weight f material is γ; the elastic mdulus is E. cm cm X X 4 cm d 4 cm 6 cm 6 cm
Slutin : + 4 cm and 3 r r π π ( r ) π, and 3 πγ π π γ 44 3 3 44 L 0 L E 36 d E 36 3 πγ 3 44 γ 44 d + π 3E E 44 36 60γ E ( Shrening ) ( ns. ) rblem: The cmpsite rd shwn in the fllwing figure is firmly attached t unyielding supprts. Cmpute the stress in each material caused by the applicatin f aial lad 00 kn. (rea f aluminum bar is 9 cm and E 70 Ga. rea f steel bar is cm and E 00 Ga.) Slutin: R a Here, + R 00 B ala aea ( i) R 0 RB 30 and s 9 7000 0,000 R 0 30 RB a s R B.539 R - (ii) 9 7000 0,000 Slving (i) and (ii) R 56.5 56.5kN a 6.8 kn / cm 9 43.5 RB 00 56.5 43.5kN s.96 kn / cm 3
rblem: cmpsite bar cnsists f an aluminum sectin rigidly fastened between a brnze sectin and a steel sectin as shwn in fllwing figure. ial lads are applied at the psitin indicated. Determine the stress in each sectin. 4000 lb Brnze (. in. ) luminium.8 in. Steel.6 in. 7000 lb 9000 lb 000 lb.3 ft.6 ft.7 ft Slutin: T calculate the stresses, we must first determine the aial lad n each sectin. The apprpriate free-bdy diagrams are shwn in the fllwing figure. 4000 lb br 4000 lb al 9000 lb 4000 lb 9000 lb st 000 lb Frm the figure we can determine br 4000 lb (tensin, al 5000 lb (cmpressin), and st 7000 lb (cmpressin). The stresses in each sectin are- 4000. br br br 5000.8 al al al 7000.6 st st st 3,330 psi,780 psi 4,380 psi ( ns. ) ( ns. ) ( ns. ) rblem: rd f variable crss-sectin built in at ne end is subjected t three aial frces as shwn in the fllwing figure. Find the maimum nrmal stress..5 cm 5 cm 3 t 9 t 4 8 t
Slutin: a b c d 4 t 3 t 9 t 8 t 4 t 9 t 8 t + + + Reactin at is R a 3-9+8 4t (left ward) Frm the figure, ab 4t, bc 9t and cd 8t ab bc cd ma ab ab bc bc cd cd 4 5 9.5 8.5.64.64 tn / cm 0.75.44 tn / cm tn / cm ( + ) tn / cm ( ns. ) ( + ) ( + ) ssignment: rd is cmpsed f an aluminum sectin rigidly attached between steel and brnze sectins, as shwn in the fllwing figure. ial lads are applied at the psitins indicated. If 3000 lb and the crss sectinal area f the rd is 0.5 in., determine the stress in each sectin. 4 Steel luminum Brnze ft 3 ft.5 ft ssignment: rd is cmpsed f an aluminum sectin rigidly attached between steel and brnze sectins, as shwn in the fllwing figure. ial lads are applied at the psitins indicated. Find the maimum value f that will nt eceed a stress in steel f 40 Ma, in aluminum f90 Ma, r in brnze f 00 Ma. (ns: 0.0 kn) 5
Steel 500 mm luminum 500 mm Brnze 500 mm 4.5 m.0 m.5 m Elastic Strain energy: In mechanics, energy is defined as the capacity t d wrk and wrk is the prduct f a frce and the distance in the directin that the frce mves. ccrding t the principal f cnservatin f energy, the internal wrk dne (i.e. internal frce defrmatin) is equal t the eternal wrk dne by the applied frces. In slid, defrmable bdies, stresses multiplied by their respective areas are frces and defrmatins are distances. The prduct f these tw quantities is the internal wrk dne in a bdy by eternally applied frces. This internal wrk is stred in a bdy as the elastic strain energy r the elastic strain energy. Let us cnsider an element f elastic material. nrmal stress alng directin, and ε linear strain in - directin t any instant, frce acting n any frce BCD ( )(dy.dz) and elngatin ε.d. The element is made f a linearly elastic material, stress is prprtinal t strain. Therefre, if the element is initially free f stress, the frce, which acts n the element, increases linearly frm zer until it attains full value. S, the average frce acting n the element, while defrmatin 0 + dy. dz is taking place dy. dz Nw, infinitesimal wrk dne du dydz ε. d. ε. ε. d. dy. dz dv dv dv E E Where, dv vlume f the element 6
du. ε dv E Strain energy per unit vlume Strain energy density v v. ε Ttal wrk dne, u. dv. dv E Similarly, fr shearing strain it can be shwn that v τ U. dv E rblem: Tw bars made by linearly elastic materials whse prprtins are shwn in the fllwing figures t absrb the same amunt f energy delivered by aial frces. Cmpare the stresses in the tw bars caused by the same input f energy. I 3L/4 L/4 Slutin: Let, and are stresses f member and. We knw, u v dv E v E E E u dv V ( L) gain, u dv E E dv lwer part + E dv upper part u v L 3L L 3L 5L + + E 4 8E 4 E 4 8 E 8 bsrb same amunt f energy u u E E L ( L) ( 5 )) (.).65 ns 5 8 0. 79 7