Mechanical Properties of Solids: Class 11 Physics Chapter 8 Hinglish mein

Physics Part 2 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 22 Oct 2026 · Updated 22 Oct 2026

80 min
23 practice questions
26 MCQs

Is chapter mein aap kya seekhenge

Elastic aur plastic behaviour mein fark karo aur samjhao ki atoms ke beech ke forces se stress kaise paida hota hai
Tensile, compressive, shearing aur hydraulic stress, aur unse jude longitudinal, shearing aur volume strains ko define karo
Hooke's law batao aur stress–strain curve ka har region padho, proportional limit se lekar fracture tak
Young's modulus, shear modulus, bulk modulus, compressibility aur Poisson's ratio ko define karo, aur inse deformations calculate karo
Stretched wire ki elastic potential energy aur uski energy density ½ × stress × strain derive karo
Elastic properties ko design mein use karo: rope ki motaai, beam ka sag aur I-section, pillars aur mountains ki height

NCERT Physics Class 11, Chapter 8, pp. 167–179 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

8.1 Introduction

Chapter 6 mein humne ek wheel ya rod ko rigid body maana tha, jiska shape kitna bhi zor lagao, kabhi nahi badalta. Yeh ek kaam ka model tha, sach nahi. Steel ki bench par khade ho jao to woh millimetre ke ek hisse jitna jhuk jaati hai; sitar ka taar chhedo to woh lamba ho jaata hai; eraser mein angootha dabao to usmein gaddha pad jaata hai. Har asli solid par force lagne se uska size ya shape badalta hai, chahe yeh badlaav dekhne ke liye bahut chhota ho.

Jab forces hata liye jaate hain to kya hota hai? Yahan solids do bade groups mein bant jaate hain. Steel spring, rubber band ya diving board apne original size aur shape mein waapas aa jaate hain: inmein elasticity (प्रत्यास्थता) hoti hai, aur inka deformation elastic tha. Geeli mitti ka gola jis shape mein dabao, wahi shape rakh leta hai. Aise materials mein plasticity (सुघट्यता) hoti hai; goondha hua atta aur modelling clay lagbhag perfectly plastic hote hain (Fig. 8.1).

3/3Spring apni pehli length par laut aata hai (elastic); atta khincha hi reh jaata hai (plastic).

Fig. 8.1Elastic aur plastic behaviour: spring waapas aata hai, atta nahi.

Do cheezein saath-saath rakhi hain: deewar se juda ek steel spring aur board par atte ki ek roll, dono ki shuruaati length same hai, jise ek dashed line se dikhaya gaya hai. Pehle step mein dono rest par hain. Doosre step mein dono ko right ki taraf kheencha jaata hai aur dono saaf taur par lambe ho jaate hain. Teesre step mein pull hata diya jaata hai: spring waapas dashed line tak aa jaata hai, jis par springs back, elastic likha hai, jabki atta lamba hi rehta hai, jis par stays long, plastic likha hai.

Definition

Elasticity

Kisi deformed body ki woh tendency jisse woh deform karne wale forces hatne par apne original size aur shape mein waapas aa jaati hai.

ExampleKheencha hua rubber band chhodne par waapas chhota ho jaata hai.

Engineers is fark par nirbhar karte hain. Jammu aur Kashmir ke Chenab rail bridge ka deck riverbed se 359 m upar hai. Iska steel order karne se pehle designers ko jaanna tha ki har grade ka steel load ke neeche kitna khinchta, mudta aur aenthta hai, aur kis load par woh waapas aana band kar dega. Yahi samajh tay karti hai ki lift cable kitni moti honi chahiye, railway rail ki kamar patli aur flanges chaude kyun hote hain, aur glass ka gilaas girne par kyun toot jaata hai jabki steel ka sirf pichak jaata hai. Yeh chapter aise sawaalon ke jawaab dene ke liye zaroori ideas banata hai.

Quick check

2 mein se 1

Ek kumhaar geeli mitti ke ek lump mein angootha dabata hai aur hata leta hai. Gaddha bana rehta hai. Mitti ke baare mein kya sahi hai?

elastic, kyunki woh deform huirigid, kyunki woh tooti nahiplastic, kyunki deformation bana rahaelastic, kyunki force lagana pada

Answer dekho

Answer (C) plastic, kyunki deformation bana raha

Plastic material force hatne ke baad bhi deformation rakhta hai. Har solid force se deform hota hai, isliye sirf deform hone se koi elastic nahi ho jaata: elastic body waapas aa jaati.

8.2 Stress aur Strain

Ek steel wire se paani ki balti latkao. Wire thoda lamba hota hai aur phir ruk jaata hai: balti equilibrium mein hai. Ab wire ke andar kuch na kuch balti ko uske weight ke barabar force se upar kheench raha hoga. Woh hai padosi atoms ke beech ke forces. Kheenchne se atoms apni natural spacing se thoda door ho jaate hain, aur unke beech ke bonds chhote khinche hue springs ki tarah waapas kheenchte hain (Fig. 8.2). Agar solid ko dabao, to atoms jitna pasand karte hain usse zyada paas aa jaate hain, isliye bonds waapas dhakka dete hain.

2/3Alag kheenchne par bonds waapas kheenchte hain: restoring force stretch ka virodh karta hai.

Fig. 8.2Solid ka spring model: bonds atoms ki spacing mein kisi bhi badlaav ka virodh karte hain.

Paanch-paanch atoms ki teen rows horizontally zigzag springs se aur vertically seedhe bonds se judi hain. Pehle step mein atoms apni natural spacing par hain aur koi net force nahi hai. Doosre step mein bahar ki taraf arrows siron ko alag kheenchte hain, spacing badh jaati hai aur restoring likhe green arrows andar ki taraf point karte hain: bonds waapas kheenchte hain. Teesre step mein andar ki taraf arrows block ko dabate hain, spacing kam ho jaati hai aur restoring arrows bahar ki taraf point karte hain: bonds waapas dhakka dete hain.

Yeh andar ke forces restoring forces hain. Equilibrium mein kisi bhi cross-section par restoring force us par lagne wale external force ko balance karta hai. Material ke liye sirf force maayne nahi rakhta, balki yeh ki woh kitne area par faila hai: cotton ke dhaage se latakta 100 N, steel bar par lage 100 N se bahut alag hai. Isliye hum stress (प्रतिबल) ko us surface ke per unit area restoring force ke roop mein define karte hain jis par woh lagta hai:

stress = \frac{F}{A}

(8.1)

jahan

$F$: restoring force, jo equilibrium mein lagaye gaye force ke barabar hota hai
$A$: us surface ka area jis par yeh lagta hai

Stress ka unit pressure wala hi hai, pascal: $1 P a = 1 N m^{- 2} .$ Iska dimensional formula hai $[M L^{- 1} T^{- 2}] .$ Engineering stresses millions ya sau millions pascals tak jaate hain, isliye megapascal ( $1 M P a = 10^{6} P a$ ) aur gigapascal ( $10^{9} P a$ ) kaam ke units hain.

Forces kaise lagaye gaye hain, iske hisaab se solid khinchta ya dabta hai, shear hota hai, ya har taraf se dabaya jaata hai. Har case ka apna stress aur deformation ka apna maap hota hai, jise strain (विकृति) kehte hain: kisi dimension mein badlaav ko uski original value se divide karna. Fig. 8.3 in sabko ek hi block par dikhata hai.

1/4End faces ke right angle par barabar aur opposite pull: length badhti hai.

Fig. 8.3Teen tarah ke stress: longitudinal (tensile ya compressive), shearing aur hydraulic.NCERT Fig. 8.1 · p. 168

Ek square block jiske original shape ki dashed outline bani hai. Pehle step mein forces uske top aur bottom ko bahar kheenchte hain aur woh lamba aur thoda patla ho jaata hai: tensile stress, length grows. Doosre step mein forces andar dhakka dete hain aur woh chhota aur chauda ho jaata hai: compressive stress, length shrinks. Teesre step mein uska base floor par hai aur top face ke along ek force use side mein khiska kar tircha shape de deta hai: shear, shape changes. Chauthe step mein block paani mein hai, chaaron sides par arrows dabate hain aur woh barabar se sikudta hai: hydraulic stress, volume shrinks.

Length ke along kheenchna ya dhakelna. Ek rod ko uske siron se pakdo aur kheencho. Forces uske axis ke along, end faces ke right angle par lagte hain, aur andar ka stress tensile hota hai. Forces ko ulta kar do taaki woh siron ko paas dhakelein, to stress compressive ho jaata hai. Dono hi case mein yeh length ke along lagta hai, isliye dono ko longitudinal stress kehte hain. Length $L$ se $L \pm Δ L$ ho jaati hai, aur deformation ko is se maapte hain:

longitudinal strain = \frac{Δ L}{L}

(8.2)

Ek face ko doosre ke paas se khiskana. Ab do forces ko do opposite faces ke aar-paar nahi, balki unke along lagne do, jaise tumhari hatheli ek eraser ke top ko dhakel rahi ho jiska base desk se chipka hai (Fig. 8.4). Us face ke per unit area restoring force ko shearing stress kehte hain, ise tangential stress bhi kehte hain. Top face bottom face ke relative $Δ x$ side mein khisakta hai, jo usse $L$ distance neeche hai, isliye har vertical edge ek angle se jhuk jaata hai, jise kehte hain $θ .$ Shearing strain hai

3/3Shearing strain $= Δ x / L = \tan θ$ , chhote angles ke liye $θ$ ke kareeb.

Fig. 8.4Shearing strain: top face,

L

distance door wale face ke relative

Δ x

khisakta hai.

Ek rectangular eraser desk par khada hai aur uska base chipka hua hai. Pehle step mein woh rest par hai. Doosre step mein force F uske top face ko right ki taraf dhakelta hai aur desk base ko left ki taraf barabar force se pakde rakhti hai, isliye block parallelogram ban kar jhuk jaata hai. Teesre step mein original vertical edge dashed dikhaya gaya hai, top face ka side shift delta x se, faces ke beech ki height L se, aur dashed edge aur tirche edge ke beech chhota angle theta bottom par mark kiya gaya hai.

shearing strain = \frac{Δ x}{L} = \tan θ

(8.3)

Asal mein $θ$ bahut chhota hota hai, isliye hum small-angle approximation use kar sakte hain, $θ$ radians mein lekar:

shearing strain \approx θ

(8.4)

Yeh kitna sahi hai? $θ = 5^{\circ} = 0.0873 r a d$ par $\tan θ = 0.0875$ milta hai: dono mein 0.3 % se bhi kam fark hai. Load wale metal parts mein shear strains aksar isse bhi hazaaron guna chhote hote hain.

Har taraf se dabana. Ek steel ball ko gehri jheel ke tal tak gira do. Paani uski surface ke har hisse par dabata hai, hamesha right angle par aur har jagah barabar zor se, isliye ball bahut thodi chhoti ho jaati hai lekin bilkul gol rehti hai. Per unit area restoring force, yaani hydraulic stress, seedhe paani ke extra pressure $p$ ke barabar hota hai, aur upar laane par ball apna size waapas pa leti hai. Yahan strain hai

volume strain = \frac{Δ V}{V}

(8.5)

Har strain do lengths ya do volumes ka ratio hai (ya shear ke liye radians mein ek angle), isliye strain ek pure number hai jiska koi unit aur koi dimensions nahi hote.

Quick check

3 mein se 1

0.50 m m^{2}

cross-sectional area wali nylon fishing line par ek machhli latki hai, jiska weight hai

40 N .

Line mein tensile stress MPa mein kitna hai?

MPa

Answer dekho

Answer 80 MPa

σ = F / A = 40 / (0.50 \times 10^{- 6}) P a = 8.0 \times 10^{7} P a = 80 M P a .

Aam galti yeh bhool jaana hai ki

1 m m^{2} = 10^{- 6} m^{2} .

8.3 Hooke's Law

Ek wire, spring ya rod par dheere se load lagao aur maapo ki woh kitna deform hota hai. Load double karo to deformation double ho jaata hai; teen guna karo to teen guna (Fig. 8.5). Jab tak deformation chhota rehta hai, stress aur strain ek fixed ratio mein saath-saath badhte aur ghatte hain. Is rule ko Hooke's law kehte hain.

LawHooke's law

Jab tak deformation chhota hai, body mein stress aur usse bana strain ek constant ratio rakhte hain.

stress = k \times strain

(8.6)

jahan

$k$: material ka modulus of elasticity

3/3Points origin se guzarti seedhi line par hain: Hooke's law.

Fig. 8.5Load mein barabar badhotari se extension mein barabar badhotari hoti hai.

Left mein ek spring bracket se latka hai, uske paas ek vertical scale hai, aur spring ke neeche ke sire par ek pointer hai. Pehle step mein us par ek mass hai aur uska point right wale load vs extension graph par plot hota hai. Doosre step mein do masses hain, spring double khinchta hai aur doosra point double height aur double distance par aata hai. Teesre step mein teen masses se teesra point milta hai, aur origin se guzarti ek dashed seedhi line teenon points se hokar jaati hai, jis par straight line likha hai.

Constant $k$ ko modulus of elasticity (प्रत्यास्थता गुणांक) kehte hain. Strain ka koi unit nahi hota, isliye $k$ ka unit stress wala hi hai, pascal. Bada modulus matlab stiff material, jise chhote se strain ke liye bhi bada stress chahiye. Har tarah ke stress ka apna modulus hota hai; inse hum Section 8.5 mein milenge.

Hooke's law ke sach hone ki koi gehri wajah nahi hai: yeh empirical hai, yaani maapne se mila ek pattern. Metals ise achhe se follow karte hain, lekin sirf ek had tak ke stress tak, jabki rubber aur body ke kai tissues lagbhag shuru se hi seedhi line se hat kar mud jaate hain. Agla section dikhata hai ki law kahan kaam karna band karta hai.

Try karke dekho

Ghar par Hooke's law test karo

Tumhe chahiye

ek halka spring (hardware shop se, ya pen ka spring)
ek paper cup aur thoda dhaaga
das ek jaise ₹1 ke sikke
millimetre marks wala ruler
ek rubber band aur tape

Pehle guess karo

Do sikke spring ko 6 mm kheenchte hain. Agar spring Hooke's law follow karta hai, to chhe sikke use kitna kheenchenge?

12 mm18 mm36 mm6 mm

Ek halke spring ke upar wale sire ko shelf ke kinaare par tape se chipkao taaki woh khula latke, aur uske neeche ke sire se dhaage ki madad se ek paper cup latkao.
Uske paas ek ruler seedha khada chipkao aur cup ke bottom ka level note karo.
Ek baar mein do sikke daalo. Har baar cup ko rukne do aur naya level record karo.
Har load ke liye extension nikaalo aur extension vs sikkon ki ginti ka graph plot karo.
Spring ki jagah rubber band lekar dobara karo (khinche hue band se apna chehra door rakho).

Table 8.1

Sikke	Spring ka extension (mm)	Rubber band ka extension (mm)
2
4
6
8
10

Tumhari entries apne aap save hoti hain.

Spring ke liye points origin se guzarti seedhi line ke kareeb aate hain: extension load ke proportional hai, jaisa Hooke's law kehta hai. Rubber band ke points mudte hain, kyunki rubber pehle aasaani se khinchta hai aur phir stiff ho jaata hai. Spring par zyada load daalo to woh apni pehli length par waapas nahi aayega: tum uski elastic limit paar kar chuke ho. Section 8.4 wala tensile test lab tumhe wire ko isse kahin aage tak kheenchne deta hai.

Quick check

3 mein se 1

Apne proportional region mein ek wire

0.40 m m

khinchta hai jab load hota hai

50 N .

Kitna load use

1.0 m m

kheenchega?

100 N250 N200 N125 N

Answer dekho

Answer (D) 125 N

Load extension ke proportional hai:

50 \times 1.0 / 0.40 = 125 N .

100 N chunna aisa maanna hai ki extension double hua; asal mein yeh 2.5 guna badha hai.

8.4 Stress–Strain Curve

Halki si kheench se lekar tootne tak koi material kaise behave karta hai? Ek sample, maan lo ek wire, ko testing machine mein clamp karo aur use dheere-dheere kheencho, jabki machine do numbers record karti hai: kheenchne wala force aur extension. Ise tensile test kehte hain. Har pair ko stress (force ÷ original area) aur strain (extension ÷ original length) mein badal kar plot karne se material ka stress–strain curve milta hai, jo uske deform hone ka fingerprint hai. Sample ko kheenchne ki jagah dabane ya shear karne se isi tarah ke doosre curves milte hain.

Fig. 8.6 mild steel jaise ductile metal ka curve dikhata hai. Iske pehle hisse mein strains bahut chhote hain, 1 % se bhi kaafi kam, isliye drawing mein us hisse ko faila kar dikhaya gaya hai. Origin O se curve ko follow karo:

5/5Sample F par toot jaata hai.

Fig. 8.6Ductile metal ke liye stress vs strain, pehle load se fracture tak (elastic hissa scale se kahin zyada chauda banaya gaya hai).NCERT Fig. 8.2 · p. 169

Ek stress vs strain graph. Pehle step mein origin O se point P tak ek seedhi line tezi se upar jaati hai, jis par Hooke's law holds likha hai. Doosre step mein line yield point Y tak badhte hue thodi mudti hai, jis par still elastic likha hai; ek dashed level yield strength sigma y ko mark karta hai. Teesre step mein curve dheere se ek point K tak uthta hai, aur K se OP ke parallel kheenchi dashed line O ke right mein strain axis tak pahunchti hai; O se is gap par permanent set likha hai. Chauthe step mein curve apne sabse oonche point U par pahunchta hai, ultimate strength sigma u par. Paanchve step mein curve neeche girkar fracture point F tak aata hai, jise cross se mark kiya gaya hai.

O se P: proportional region. Ek seedhi line: Hooke's law sahi hai. Iske along kahin bhi load hata do to sample bilkul pehle jitna lamba ho jaata hai. P proportional limit ko mark karta hai.
P se Y: elastic lekin proportional nahi. Graph seedhi line se hat kar mudta hai, lekin abhi koi nuksaan nahi hua: unload karo to sample apna raasta waapas O tak tay karta hai. Is hisse ka ant, Y, yield point ya elastic limit hai, aur wahan ka stress material ki yield strength $σ_{y}$ hai.
Y ke aage: plastic flow. Curve chapta ho jaata hai: atoms ki layers ek doosre ke paas se khisakti hain, aur thode se extra stress par sample bahut khinchta hai. K par load hata do to sample curve ke along nahi, balki OP ke parallel line ke along waapas aata hai, aur shuruaat se lamba reh jaata hai. Yeh bacha hua strain permanent set hai, aur deformation plastic hai.
U: ultimate tensile strength. Curve ultimate tensile strength $σ_{u}$ par peak karta hai. Yahan se sample ek jagah par patla hone lagta hai, ek neck banti hai, aur woh jagah ghatte force mein bhi khinchti rehti hai.
F: fracture. Sample neck ke sabse patle hisse par toot jaata hai.

U aur F ke beech ka gap batata hai ki material kaise fail hota hai. Agar F, U se kaafi aage hai, to material tootne se pehle wires mein kheencha ya mod kar shape diya ja sakta hai: yeh ductile (तन्य) hai, jaise copper, gold ya mild steel. Agar U aur F lagbhag ek hi jagah hain, to material elastic na rehne ke thodi der baad hi, bina zyada warning ke, toot jaata hai: yeh brittle (भंगुर) hai, jaise glass, cast iron ya china ka cup (Fig. 8.7).

1/3Copper: U aur F kaafi door hain, isliye tootne se pehle yeh bahut khinchta hai.

Fig. 8.7Ductile, brittle aur elastomeric behaviour (curves ke strain scales alag hain).NCERT Fig. 8.3 · p. 170

Har step mein stress aur strain axes par ek curve hai. Pehle step mein copper: tez seedhi shuruaat, phir sabse oonche point U tak dheere uthta lamba curve, phir right mein kaafi door fracture point F tak girna; U aur F ke beech dashed bracket par far apart likha hai. Doosre step mein cast iron: lagbhag seedhi line jo achaanak ek cross par khatam hoti hai, jis par U and F together likha hai. Teesre step mein rubber: ek curve jo bahut bade strain tak dheere uthta hai aur phir tootne se pehle tezi se upar mudta hai, jis par huge strain, no permanent set likha hai. Teenon steps mein strain axis ka scale same nahi hai.

Rubber ke rules alag hain. Rubber band ko uski length ke teen-chaar guna tak kheencho, phir bhi woh apni pehli length par laut aata hai, lekin uska graph lagbhag pehle pal se hi mudta hai, aur tootne se pehle woh kabhi plastic stage se nahi guzarta. Jis material ka elastic stretch bahut bada ho lekin proportional bilkul na ho, use elastomer kehte hain. Tumhari body mein bhi aise hain: aorta ki deewar, yaani dil se nikalne wali badi artery, har dhadkan par gubbare ki tarah phoolti hai. Iska curve pehle dheere aur phir tezi se uthta hai, isliye artery pehle aasaani se failti hai aur phir virodh karti hai.

Table 8.2Kuch materials ki typical strengths (asli samples composition aur treatment ke hisaab se badalte hain)NCERT Table 8.1 · p. 170

Material	Yield strength $σ_{y}$ (MPa)	Ultimate strength $σ_{u}$ (MPa)	Behaviour
Structural (mild) steel	250	400–550	ductile
Aluminium alloy 6061	275	310	ductile
Copper (annealed)	70	220	bahut ductile
Grey cast iron	—	150–300	brittle
Soda-lime glass	—	lagbhag 50	brittle
Insaan ki haddi (compact)	—	lagbhag 130	kaafi brittle

InteractiveTensile test lab

Ek testing machine sample ko do grips ke beech pakadti hai, aur paas mein uska stress–strain graph hai. Tum mild steel, copper, cast iron ya rubber chunte ho aur wire ka diameter aur length set karte ho. Slider sample ko step-by-step kheenchta hai (Play test yeh apne aap karta hai) aur point material ka curve banata hai, saath mein force, extension, stress aur shuruaati slope ki readings dikhti hain. Release load sample ko unload karta hai: elastic region mein yeh zero par laut aata hai, yield point ke aage seedhi line ke along waapas aata hai aur permanent set rakhta hai. Kaafi kheencho to sample mein neck banti hai aur woh toot jaata hai. Ek toggle graph ko small-strain region par zoom karta hai.

Fig. 8.8Ek material chuno, phir slider drag karo ya Play test dabao taaki sample khinche. Release load dikhata hai ki woh waapas aata hai ya nahi.

Quick check

3 mein se 1

Ek steel wire par uske yield point se zyada load lagaya jaata hai aur phir load hata diya jaata hai. Kya hota hai?

woh kuch extra length rakh leta hai, yaani permanent setwoh usi curve ke along apni original length par laut aata haiload hatate hi woh toot jaata haiuska Young's modulus zero ho jaata hai

Answer dekho

Answer (A) woh kuch extra length rakh leta hai, yaani permanent set

Yield point ke aage deformation kuch had tak plastic hota hai. Unload karne par wire elastic line ke parallel line ke along chalta hai, isliye kuch strain bacha rehta hai. Usi curve ke along waapas aana sirf elastic limit se neeche hota hai.

8.5 Elastic Moduli

Designers poori koshish karte hain ki structures curve ke seedhi line wale hisse mein, proportional limit se neeche rahein, jahan deformations chhote aur poori tarah reversible hote hain. Wahan stress aur strain ka ratio constant hota hai. Yeh modulus of elasticity material ki property hai (temperature ke saath thoda badalta hai), sample ke size ya shape ki nahi. Stress aur strain kai tarah ke hote hain, isliye kai moduli hain; Table 8.5 mein yeh sab ek jagah hain.

8.5.1 Young's Modulus

Ek chhoti steel rod ko kheencho, ya utne hi stress se dabao, dono tarah se uski length utni hi badalti hai. Isliye dono ke liye ek hi constant kaam karta hai, Young's modulus $Y$ (यंग गुणांक): longitudinal stress $σ$ divided by usse bana longitudinal strain $ε$ .

Y = \frac{σ}{ε}

(8.7)

$L$ length aur $A$ cross-sectional area wali rod ko agar force $F$ se $Δ L$ kheencha jaaye, to (8.1) aur (8.2) se milta hai

\begin{aligned} Y & = \frac{F / A}{Δ L / L} \\ = \frac{F L}{A Δ L} \end{aligned}

(8.8)

Iska unit stress ki tarah pascal hai. Isi equation ko ulta karke kisi bhi wire ka extension nikal sakte hain, $Δ L = F L / (A Y)$ : chhote modulus wale material ka lamba, patla wire sabse zyada khinchta hai.

Table 8.3Room temperature par kuch materials ke elastic constants (typical values)NCERT Table 8.1 · p. 170

Material	$Y$ (GPa)	$G$ (GPa)	$B$ (GPa)	Poisson's ratio
Steel	200	79	160	0.29
Aluminium	70	26	76	0.33
Copper	120	45	140	0.34
Glass	70	29	40	0.22
Lead	16	5.6	46	0.44
Tungsten	411	161	310	0.28
Rubber	0.01–0.1	—	lagbhag 2	lagbhag 0.5

Crane ke hooks, rails aur building frames steel ke kyun bante hain? 1 m lambe aur $1 m m^{2}$ cross-section wale wires se same 100 N load latkao: steel wire 0.50 mm khinchta hai, copper wala 0.83 mm aur aluminium wala 1.4 mm. Steel ka modulus aam materials mein sabse bade moduli mein se hai, isliye yeh sabse kam deform hota hai. Glass lagbhag aluminium jitna stiff hai; concrete, haddi aur lakdi naram hain, lagbhag 10–30 GPa; rubber ka modulus hazaaron guna chhota hai.

Example 8.1NCERT Example 8.1 · p. 171

Footbridge ke hanger ka khinchna

Ek vyast sadak ke upar bana footbridge vertical steel rods se latka hai. Ek rod $2.5 m$ lambi aur $16 m m$ diameter ki hai, aur us par tension hai $30 k N .$ Steel ke liye $Y = 2.0 \times 10^{11} P a$ lekar nikaalo (a) rod mein stress, (b) strain aur (c) extension. (d) Steel ki yield strength hai $250 M P a .$ Stress isse kitne guna kam hai? Rod ka apna weight ignore karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 8.2NCERT Example 8.2 · p. 171

Ek ke baad ek do wires

$6.0 k g$ mass ka ek garden lamp $2.0 m$ lambe steel wire se latka hai, jo sire se sira jod kar $1.5 m$ lambe aluminium wire se juda hai; dono $1.0 m m$ diameter ke hain. Maan lo steel ke liye $Y_{s} = 2.0 \times 10^{11} P a$ , aluminium ke liye $Y_{a} = 7.0 \times 10^{10} P a$ aur $g = 9.8 m / s^{2} .$ (a) Har wire ka extension aur lamp ka total neeche aana nikaalo. (b) Kitne mass ka lamp total extension $3.0 m m$ karega?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 8.3NCERT Example 8.3 · p. 171

Ek coolie ki shin bone kitni chhoti hoti hai?

Ek railway coolie dono pairon par khade hokar sir par $40 k g$ ka suitcase uthata hai. Har shin bone (tibia) lagbhag $38 c m$ lambi hai, aur uska effective cross-sectional area hai $3.2 c m^{2} .$ Compression mein haddi ke liye $Y = 1.6 \times 10^{10} P a$ aur $g = 9.8 m / s^{2}$ lekar nikaalo (a) suitcase ki wajah se har shin bone ka extra compression, (b) strain aur (c) compression jab woh aage kadam badhane ke liye ek pair uthata hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Ek hi material ke do wires same load uthate hain. Wire Q, wire P se double lamba hai aur uska diameter bhi double hai. Q ka extension P ke extension ke mukable kitna hai?

chaar gunadoublesameaadha

Answer dekho

Answer (D) aadha

Δ L = F L / (A Y)

aur

A \propto d^{2}

L

double karne se extension double hota hai,

d

double karne se 4 se divide hota hai, isliye Q,

2 / 4 = 1 / 2

guna khinchta hai. 'Double' wala answer bhool jaata hai ki area diameter ke square par depend karta hai.

8.5.2 Shear Modulus

Shearing mein stress us face ke per unit area tangential force hai jiske along woh lagta hai, aur strain jhukaav ka angle hai. Inka ratio shear modulus $G$ (अपरूपण गुणांक) hai, jise modulus of rigidity bhi kehte hain:

\begin{aligned} G & = \frac{σ_{s}}{Δ x / L} \\ = \frac{F L}{A Δ x} \end{aligned}

(8.10)

(8.4) se strain seedhe $θ$ hai, isliye

G = \frac{F}{A θ}

(8.11)

σ_{s} = G θ

$G$ ka unit bhi pascal hai. Table 8.3 ka $G$ column dikhata hai ki kisi material ke liye shear modulus Young's modulus se chhota hota hai, aksar uske ek-tihaai se do-paanchve ke beech: atoms ki layers ko ek doosre par khiskana unhe alag kheenchne se aasaan hai. Shear modulus sirf solids ka hota hai. Liquid ka apna koi shape nahi hota, isliye halke se halka side push bhi use ek fixed angle par tikne ki jagah beh jaane deta hai.

Shear engineering mein har jagah milta hai: us bolt mein jo do plates ko ek doosre par khisakne se rokta hai, aenthe hue drive shaft mein, aur coil spring ke wire mein, jo spring ke khinchne par khinchta nahi balki aenthta hai.

Example 8.4NCERT Example 8.4 · p. 172

Rubber pads par bridge

Kai bridges moti rubber (elastomeric) bearing pads par tike hote hain jo deck ko temperature ke saath failne aur sikudne dete hain. Ek pad plan mein $50 c m \times 40 c m$ aur height mein $6.0 c m$ hai, aur uska base pier se fixed hai. Garam dopahar mein failta deck pad ke top ko side mein kheenchta hai, aur yeh force hai $18 k N .$ Rubber ka shear modulus hai $0.90 M P a .$ Nikaalo (a) shearing stress, (b) shearing strain aur (c) pad ke top ka side shift.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

2 mein se 1

Liquid ka shear modulus kyun nahi hota?

kisi bhi side force se woh fixed strain tak pahunchne ki jagah beh jaata haiuske molecules ek doosre par koi force nahi lagateuski density bahut kam haiuska volume badal nahi sakta

Answer dekho

Answer (A) kisi bhi side force se woh fixed strain tak pahunchne ki jagah beh jaata hai

Shear modulus ke liye diye gaye shearing stress par ek fixed strain chahiye. Liquid tab tak behta rehta hai jab tak side force lagta hai. Uske molecules ek doosre ko attract zaroor karte hain, isiliye liquids judi rehti hain.

8.5.3 Bulk Modulus

Kisi fluid mein rakhi body, jiska pressure $p$ badh jaaye, har taraf se barabar dabti hai, jaise Section 8.2 mein: hydraulic stress $p$ hai aur volume strain hai $Δ V / V .$ Inka ratio bulk modulus $B$ (आयतन गुणांक) define karta hai:

B = - \frac{p}{Δ V / V}

(8.12)

Dabane se stable body hamesha sikudti hai, isliye $Δ V$ aur $p$ ke signs ulte hote hain, aur minus sign $B$ ko positive number banata hai. Iska reciprocal, compressibility $k$ (संपीड्यता), ulta sawaal answer karta hai: har pascal extra pressure par volume kitne fraction se sikudta hai?

k = \frac{1}{B} = - \frac{1}{Δ p} \frac{Δ V}{V}

(8.13)

$Y$ aur $G$ ke ulat, bulk modulus solids ke saath liquids aur gases par bhi laagu hota hai. Fluid ka apna shape nahi hota, lekin volume hota hai, aur dabane se woh volume kam hota hai. Table 8.4 aur Fig. 8.9 matter ki teenon states ki tulna karte hain.

Table 8.4Matter ki teen states mein bulk modulus aur compressibilityNCERT Table 8.3 · p. 173

Substance	$B$ (Pa)	$k = 1 / B$ $(P a^{- 1})$
Steel	$1.6 \times 10^{11}$	$6.3 \times 10^{- 12}$
Mercury	$2.85 \times 10^{10}$	$3.5 \times 10^{- 11}$
Water	$2.2 \times 10^{9}$	$4.5 \times 10^{- 10}$
Constant temperature par hawa, 1 atm	$1.0 \times 10^{5}$	$1.0 \times 10^{- 5}$

2/2Ek extra atmosphere steel ko 0.00006 %, paani ko 0.005 % aur hawa ko (constant temperature par) 50 % sikod deta hai.

Fig. 8.9Ek atmosphere pressure badhane par: steel aur paani mushkil se badalte hain, hawa ka volume aadha ho jaata hai.

Pistons wale teen ek jaise cylinders saath-saath khade hain, jinmein ek steel block, paani aur bikhre molecules ke roop mein dikhayi hawa bhari hai. Pehle step mein koi weight nahi hai. Doosre step mein har piston par barabar weight rakha jaata hai, jisse same extra pressure milta hai. Steel aur paani ke pistons dikhne laayak nahi hilte; hawa ka piston apni aadhi height tak gir jaata hai aur molecules neeche wale aadhe hisse mein simat jaate hain. Volume ke badlaav par steel ke liye 0.00006 percent, paani ke liye 0.005 percent aur hawa ke liye 50 percent likha hai.

Same extra pressure par paani steel se lagbhag 70 guna zyada sikudta hai, aur constant temperature par hawa paani se 20 000 guna se bhi zyada. Itna bada fark kyun? Fig. 8.2 ke spring model par waapas jao. Solid mein har atom pehle se hi har taraf stiff bonds se daba hua hai, isliye atoms ko thoda sa bhi paas laane ke liye bahut bada pressure chahiye. Liquid ke molecules bhi ek doosre ko chhoote hain, lekin dheele bandhe hote hain. Gas mostly khaali jagah hai, aur use dabane se molecules bas us jagah ke chhote hisse mein simat jaate hain; isiliye gas ki compressibility uske pressure aur temperature par bahut depend karti hai.

Table 8.5Kis deformation ke liye kaun sa modulusNCERT Table 8.4 · p. 173

Deformation	Kya badalta hai	Strain	Modulus	Kis par laagu
Length ke along kheenchna ya dabana	length	$Δ L / L$	Young's, $Y = F L / (A Δ L)$	solids
Ek face ko doosre ke paas se khiskana	sirf shape	$θ \approx Δ x / L$	shear, $G = F / (A θ)$	solids
Har taraf se barabar dabana	sirf volume	$Δ V / V$	bulk, $B = - p / (Δ V / V)$	solids, liquids aur gases

Example 8.5NCERT Example 8.5 · p. 174

Samudra ke tal par dabaav

Ek research submersible ek solid steel test ball lekar $6000 m$ ki gehraai tak jaati hai. Seawater ki density $1030 k g / m^{3}$ , $g = 9.8 m / s^{2}$ , steel ke liye $B = 1.6 \times 10^{11} P a$ aur seawater ke liye $B = 2.3 \times 10^{9} P a$ lo. Surface ke mukable volume mein fractional kami nikaalo, (a) steel ball ki aur (b) us gehraai par seawater ki.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Aage ki baat: Chapter 14 mein tum dekhoge ki yahi moduli sound ki speed bhi tay karte hain. Stiff, halka medium sound ko tez le jaata hai, isiliye sound paani mein hawa se lagbhag chaar guna tez chalti hai, aur steel mein usse bhi tez.

Quick check

3 mein se 1

B = - p / (Δ V / V)

mein minus sign kyun hai?

kyunki pressure ek negative quantity haitaaki B positive ho, kyunki pressure badhne se volume ghatta haikyunki volume strain ka koi unit nahi haikyunki B sirf gases ke liye define hai

Answer dekho

Answer (B) taaki B positive ho, kyunki pressure badhne se volume ghatta hai

Pressure badhne (

p > 0

) se volume ghatta hai (

Δ V < 0

), isliye ratio

p / (Δ V / V)

negative hai; minus sign

B

ko positive banata hai.

8.5.4 Poisson's Ratio

Ek chauda rubber band kheencho aur uski chaudaai dekho: lamba hote hue woh patla hota jaata hai (Fig. 8.10). Steel wire bhi yahi karta hai, bas bahut kam dikhta hai. Force ke along strain, yaani longitudinal strain, ke saath uske right angle par bhi ek strain aata hai, lateral strain. Agar $d$ diameter aur $L$ length wala wire $Δ L$ lamba ho aur uska diameter $Δ d$ ghat jaaye, to lateral strain hai $Δ d / d .$ Elastic limit se neeche side wala strain length wale strain ke saath-saath chalta hai, isliye inka ratio material ki fixed property hai, jiska naam French mathematician Siméon Denis Poisson (1781–1840) ke naam par rakha gaya: Poisson's ratio (पॉयसन अनुपात):

2/2Length mein badhotari $Δ L$ hai, aur diameter mein kami $Δ d .$

Fig. 8.10Kheenchi gayi rod lambi aur patli hoti hai; dono strains ka ratio Poisson's ratio hai.

Pehle step mein ek rubber rod horizontally padi hai, uski length L neeche aur diameter d beech mein mark hai. Doosre step mein forces uske siron ko bahar kheenchte hain: rod lambi ho jaati hai, jise L plus delta L se mark kiya gaya hai, aur patli, jise d minus delta d se mark kiya gaya hai, aur ek dashed outline uska original size dikhati hai.

ν = \frac{Δ d / d}{Δ L / L} = \frac{Δ d}{Δ L} \cdot \frac{L}{d}

jahan

$ν$: Poisson's ratio

Divide karne par koi unit nahi bachta, isliye Poisson's ratio bas ek number hai. Zyadatar metals ke liye yeh lagbhag 0.25 aur 0.35 ke beech hota hai: steel ke liye 0.29 ke kareeb aur aluminium ke liye 0.33 (Table 8.3). Rubber ke liye yeh lagbhag 0.5 hai, woh value jis par material khinchne par bhi volume same rehta hai. Seekh general hai: ek direction mein force doosri directions mein bhi strains paida karta hai, isliye ek akela modulus solid ka sab kuch describe nahi kar sakta. Jo material har direction mein ek jaisa behave karta hai, uske elastic behaviour ko poori tarah describe karne ke liye do independent constants chahiye, jaise $Y$ aur Poisson's ratio.

Example

Copper wire patla hota hai

$3.0 m$ lamba aur $2.0 m m$ diameter wala copper wire elastically kheencha jaata hai, aur uska extension hai $1.5 m m .$ Copper ka Poisson's ratio 0.34 hai. Uska diameter kitna ghatta hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

2 mein se 1

0.30 Poisson's ratio wali steel rod ko itna kheencha jaata hai ki uski length 0.10 % badh jaaye. Uska diameter kitne percent ghatta hai?

Answer dekho

Answer 0.03 %

Lateral strain

= 0.30 \times 0.10 % = 0.030 % .

Aam galti hai 0.30 se multiply karne ki jagah divide karna.

8.5.5 Stretched Wire mein Elastic Potential Energy

Wire ko lamba kheenchne ka matlab hai uske atoms ko unke bonds ke khilaaf alag kheenchna, aur iske liye work lagta hai. Elastic limit se neeche iska kuch bhi barbaad nahi hota: wire ise elastic potential energy ke roop mein rakhta hai aur chhota hote waqt waapas de deta hai. Isi tarah dhanush teer chhodta hai aur gulel (catapult) kankad phenkti hai. Wire kitni energy rakhta hai?

DerivationStretched wire mein store energy

Maan lo natural length $L$ aur cross-sectional area $A$ wale wire ko dheere-dheere kheencha jaata hai, taaki woh hamesha equilibrium mein rahe. Jab uska extension $x$ hai, to (8.8) use pakde rakhne ke liye zaroori force deta hai:

F (x) = \frac{Y A}{L} x

Force extension ke proportional badhta hai, jaise ek spring jiska constant hai $Y A / L .$ Ise thoda aur $d x$ kheenchne mein work $d W = F d x$ lagta hai, jo force–extension line ke neeche ek patli strip hai (Fig. 8.11). $x = 0$ se final extension $l$ tak strips jodne par:

\begin{aligned} W & = \int_{0}^{l} \frac{Y A}{L} x d x \\ = \frac{1}{2} \frac{Y A}{L} l^{2} \end{aligned}

Yeh line ke neeche triangle ka area hai, $\frac{1}{2} F l$ , jahan $F = Y A l / L$ final force hai. Strain $l / L$ aur volume $A L$ ke terms mein, $Y (l / L) = σ$ use karke:

\begin{aligned} W & = \frac{1}{2} Y {(\frac{l}{L})}^{2} (A L) \\ = \frac{1}{2} σ ε (A L) \end{aligned}

Work elastic potential energy ke roop mein store hota hai, $U = W .$ Volume $A L$ se divide karne par per unit volume store energy milti hai, yaani energy density:

u = \frac{1}{2} σ ε

(8.14)

3/3Saari strips milkar triangle banati hain: $W = \frac{1}{2} F l .$

Fig. 8.11Wire kheenchne mein kiya gaya work uski force–extension line ke neeche ka area hai.

Force vs extension ka graph. Pehle step mein ek seedhi line origin se extension l par final force F tak uthti hai, jis par F equals YA over L times l likha hai. Doosre step mein line ke neeche ek patli vertical strip highlight hai, jis par dW equals F dl likha hai. Teesre step mein line ke neeche poora triangle shaded hai aur us par W equals one half F l likha hai.

Kyunki $σ = Y ε$ , energy density ko aise bhi likh sakte hain: $u = \frac{1}{2} Y ε^{2}$ ya $u = σ^{2} / 2 Y .$ Diye gaye stress par chhote Young's modulus wala material per unit volume zyada energy store karta hai. Achhe spring materials mein high yield strength aur moderate modulus dono hote hain, isliye woh permanently mude bina kaafi energy store kar sakte hain.

Example

Stretched steel wire mein energy

$2.0 m$ lamba aur $1.0 m m^{2}$ cross-section wala steel wire $2.0 m m$ kheencha jaata hai ( $Y = 2.0 \times 10^{11} P a$ ). Nikaalo (a) final force, (b) store energy aur (c) per unit volume energy.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

2 mein se 1

Elastic limit ke andar rakhte hue wire ka extension double kiya jaata hai. Store elastic energy ho jaati hai

doubleaadhisamechaar guna

Answer dekho

Answer (D) chaar guna

U = \frac{1}{2} (Y A / L) l^{2}

extension ke square ki tarah badhta hai. 'Double' wala answer bhool jaata hai ki force bhi double hota hai.

8.6 Materials ke Elastic Behaviour ke Applications

Har bridge, building, crane aur aircraft is chapter ke ideas se design hota hai. Designer ko pata hona chahiye ki har part apne upar aane wale loads se kitna deform hoga, aur yeh pakka karna hota hai ki koi bhi part kabhi apni elastic limit se zyada stress na jhele, achanak aane wali cheezon ke liye achha-khaasa margin rakh kar. Chaar examples yeh soch dikhate hain.

Lifting ropes. Crane ka hook ya lift ka cage steel rope se latakta hai. Rope ko kabhi uske plastic region tak nahi kheenchna chahiye, warna woh hamesha ke liye lambi aur kamzor ho jaayegi. Isliye sabse bhaari load $M$ se bana stress $M g / A$ yield strength $σ_{y}$ se neeche rehna chahiye, jo ek minimum cross-sectional area tay karta hai:

A \geq \frac{M g}{σ_{y}}

(8.15)

Example

Crane ki rope kitni moti ho?

Pune ki ek building site par tower crane $4.0$ tonnes ( $1 t = 1000 k g$ ) uthane ke liye rated hai. Uski rope steel ki hai, jiski yield strength hai $250 M P a .$ (a) Ek solid rope ka sabse kam cross-sectional area aur radius nikaalo jo bas elastic rahe. (b) Safety ke liye rope ko rated load ke 5 guna ke liye design kiya jaata hai. Iske liye kitna radius chahiye?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

2/27 strands mein 49 patle wires: same steel area, kahin zyada flexible.

Fig. 8.12Wire rope utne hi steel area wali solid rod jitna load uthati hai, lekin mud sakti hai.

Pehle step mein ek solid gol steel rod ka cross-section dikhaya gaya hai, jis par one solid rod, strong but stiff likha hai. Doosre step mein rod dhundhli ho jaati hai aur uske paas ek dashed outline ke andar saat strands wali rope ka cross-section aata hai, har strand mein saat gol wires, jis par 7 strands of 7 wires, same steel, bends easily likha hai.

Beams. Beam kisi gap ke aar-paar load uthata hai: floor joist, bridge girder, naali par rakha takhta. Iska kaam hai kam se kam mudna. Breadth $b$ aur depth $d$ wale rectangular cross-section ki bar, jo $l$ distance door supports par tiki ho aur beech mein weight $W$ se loaded ho, uska centre par sag hai

δ = \frac{W l^{3}}{4 b d^{3} Y}

(8.16)

jahan

$δ$: centre par sag (depression)
$b$: breadth, yaani cross-section ki horizontal chaudaai
$d$: depth, yaani cross-section ki vertical height
$l$: supports ke beech ka span

Ise derive karne ke liye abhi humare paas jitna calculus hai usse zyada chahiye, lekin hum ise padh sakte hain. Sag span ke saath tezi se badhta hai, $l^{3}$ ki tarah, isliye lambe spans ko extra support chahiye. Stiff material (bada $Y$ ) madad karta hai. Sabse khaas hai shape ka role: sag breadth ke saath $1 / b$ ki tarah ghatta hai, lekin depth ke saath $1 / d^{3} .$ Breadth double karne se sag aadha hota hai; depth double karne se aathvaan hissa reh jaata hai. Isliye beams chaude nahi balki gehre banaye jaate hain (Fig. 8.13).

1/3Chauda aur patla: takhta bahut jhukta hai.

Fig. 8.13Sag

1 / (b d^{3})

par depend karta hai: depth breadth se kahin zyada maayne rakhti hai.NCERT Fig. 8.7 · p. 175

Ek beam do triangular supports par tika hai, uske centre par weight W hai aur right mein uska cross-section bana hai. Pehle step mein beam chapta rakha takhta hai, chauda aur patla, aur saaf dikhne laayak jhukta hai; ek dashed outline uski bina load wali position dikhati hai aur ek arrow sag delta ko mark karta hai. Doosre step mein wahi takhta apne edge par khada hai, ooncha aur patla, aur mushkil se jhukta hai, jis par 25 times less sag likha hai. Teesre step mein I-shape ka steel beam, jiske chaude top aur bottom flanges ek patle web se jude hain, lagbhag bina jhuke load uthata hai, jis par deep but light likha hai.

Steel girders ka cross-section I-shape ka kyun hota hai? Jab beam jhukta hai, to uski upar ki layers dabti hain aur neeche ki layers khinchti hain, jabki beech ke paas ki layers mushkil se strain hoti hain, isliye beech se door ka metal zyadatar kaam karta hai. I-section apna zyadatar steel do chaude flanges mein rakhta hai, upar aur neeche, jo ek patle web se jude hote hain: gehra aur stiff, phir bhi halka aur sasta. Chaude flanges ise palatne aur side mein mudne se bhi rokte hain, is failure ko buckling kehte hain, jo ek oonche, patle takhte ko tab khatra deta hai jab load thoda sa bhi centre se hata ho, jaise bridge par chalti gaadiyan. Railway rails isi idea ko thode badle roop mein use karti hain.

Example

Naali par ek takhta

Monsoon mein $20 c m$ chauda aur $4.0 c m$ mota lakdi ka takhta ek naali ke aar-paar rakha jaata hai, supports $2.0 m$ door hain, taaki log us par se chal sakein. Lakdi ke liye $Y = 1.0 \times 10^{10} P a$ lo. (a) Jab $600 N$ weight wala insaan beech mein khada ho, to takhta kitna jhukta hai? (b) Agar wahi takhta edge par rakha jaaye to sag kitna hoga?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Pillars. Columns aur pillars compression mein load uthate hain, aur unke sire maayne rakhte hain. Gol siron wala pillar jise support karta hai use chhote area par chhoota hai, isliye wahan stress bada hota hai, aur pillar aasaani se jhuk kar buckle ho sakta hai. Jis pillar ke sire chaude base aur chaude top mein faile hon, woh load ko bade area par baantta hai aur line mein mazbooti se tika rehta hai, isliye kahin zyada load utha sakta hai. Kai purani buildings ke stone pillars ke chaude base aur bracket wale top bilkul yahi kaam karte hain.

Mountain kitna ooncha ho sakta hai? Elasticity mountains ki height ko bhi seemit karti hai. $h$ height wale mountain ki choti se uske base tak rock ka ek vertical column socho (Fig. 8.14). Iska weight, iske cross-section par faila hua, base ke rock par lagbhag $h ρ g$ stress se dabata hai, jahan $ρ$ rock ki density hai. Base har taraf se barabar nahi dabta, kyunki slopes par bahar se kuch nahi dabata, isliye is stress mein bada shearing hissa hota hai. Agar yeh us stress se zyada ho jaaye jis par garam, gehra rock behne lagta hai, to base dheere-dheere side mein nikalne lagta hai aur mountain tab tak dhansta hai jab tak stress us limit se neeche na aa jaaye.

2/3Agar $h ρ g$ rock ke flow stress se zyada ho, to base bahar nikalta hai aur mountain dhans jaata hai.

Fig. 8.14Mountain ka base us stress se neeche rehna chahiye jis par rock behne lagta hai.

Pehle step mein ek mountain bana hai jismein choti se base tak rock ka ek patla vertical column hai aur uski height h mark hai; base par stress lagbhag h rho g hai. Doosre step mein mountain ke upar ek oonchi dashed outline uthti hai, jis par too tall, base rock flows likha hai, aur base par arrows bahar ki taraf point karte hain. Teesre step mein do mountains saath-saath hain: Earth par chhota wala, jis par about 9 km likha hai, aur Mars par kahin ooncha wala, jis par about 25 km likha hai.

Example

Sabse ooncha mumkin mountain

Crustal rock ki density $2.7 \times 10^{3} k g / m^{3}$ aur jis stress par woh behne lagta hai, use maan lo lagbhag $2.5 \times 10^{8} P a .$ Ek jaisa rock maan kar mountain ki sabse zyada height ka andaaza lagao (a) Earth par ( $g = 9.8 m / s^{2}$ ) aur (b) Mars par ( $g = 3.7 m / s^{2}$ ).

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Ek rectangular beam ki depth double ki jaati hai jabki uski breadth, span, material aur load same rehte hain. Uska sag ho jaata hai

aadhaek-chauthaiaathvaan hissadouble

Answer dekho

Answer (C) aathvaan hissa

δ \propto 1 / d^{3}

, isliye

d

double karne se sag utne guna ghatta hai:

2^{3} = 8.

'Aadha' breadth double karne ke liye sahi hota.

Recap

Summary aur key points

Asli solids forces se deform hote hain. Elastic body forces hatne par apna size aur shape waapas pa leti hai; plastic body deformed hi rehti hai (Section 8.1).
Stress, yaani har unit area par restoring force ( $F / A$ ), pascals mein maapa jaata hai; strain, yaani dimension mein badlaav divided by original dimension, ka koi unit nahi hota (Section 8.2).
Tensile ya compressive stress se longitudinal strain $Δ L / L$ milta hai; shearing stress se shearing strain $Δ x / L = \tan θ \approx θ$ ; hydraulic stress se volume strain $Δ V / V$ (Section 8.2).
Hooke's law: jab tak deformations chhote hain, stress strain ke proportional hai, aur constant modulus of elasticity hai (Law).
Metal ke stress–strain curve mein proportional limit, yield point (elastic limit), permanent set ke saath plastic flow, ultimate tensile strength aur fracture dikhte hain. Ductile materials ultimate strength se kaafi aage tootte hain, brittle materials uske kareeb (Section 8.4).
Rubber jaise elastomers bahut bade strains tak elastically khinchte hain lekin Hooke's law follow nahi karte (Section 8.4).
Young's modulus $Y = F L / (A Δ L)$ ; metals ka, sabse zyada steel ka, $Y$ bada hota hai, aur physics mein bada modulus matlab 'zyada elastic' (Section 8.5.1).
Shear modulus $G = F / (A θ)$ , aksar $Y$ ka ek-tihaai se do-paanchva hissa; $Y$ aur $G$ sirf solids ke hote hain (Section 8.5.2).
Bulk modulus $B = - p / (Δ V / V)$ matter ki saari states par laagu hota hai; compressibility $k = 1 / B$ ; gases solids se das lakh guna se bhi zyada compressible hain (Section 8.5.3).
Poisson's ratio lateral strain divided by longitudinal strain hai: metals ke liye lagbhag 0.3 aur rubber ke liye 0.5 ke kareeb (Section 8.5.4).
Stretched wire elastic energy $\frac{1}{2} F l$ store karta hai; per unit volume $u = \frac{1}{2} σ ε$ (Section 8.5.5).
Design in ideas ko use karta hai: safety factor ke saath rope area $A \geq M g / σ_{y}$ , beam sag $δ = W l^{3} / (4 b d^{3} Y)$ aur I-section, faile siron wale pillars, aur Earth par mountain ki sabse zyada height lagbhag 10 km (Section 8.6).

Practice · 23 questions

Practice questions answers ke saath (NCERT “Exercises”)

NumericalEasyNCERT Exercise 8.1
Wire P $2.4 m$ lamba hai aur uska cross-section $2.0 \times 10^{- 6} m^{2}$ hai; wire Q $3.6 m$ lamba hai aur uska cross-section hai $1.5 \times 10^{- 6} m^{2} .$ Same load par dono bilkul utna hi khinchte hain. Unke Young's moduli ka ratio $Y_{P} / Y_{Q}$ nikaalo.
Solution dikhao
$\frac{F L_{P}}{A_{P} Y_{P}} = \frac{F L_{Q}}{A_{Q} Y_{Q}}$
$\frac{Y_{P}}{Y_{Q}} = \frac{L_{P} / A_{P}}{L_{Q} / A_{Q}} = \frac{1.2 \times 10^{6}}{2.4 \times 10^{6}}$
$= 0.50$ : wire P ka material wire Q ke material se aadha stiff hai.
2
NumericalMediumNCERT Exercise 8.2
Ek metal wire par tensile test table wali readings deta hai. Chaaho to inhe plot karo.
Table 8.6
Strain ( $\times 10^{- 3}$ ) Stress (MPa)
0 0
0.5 35
1.0 70
1.5 105
2.0 140
2.5 158
3.0 166
3.5 170
1. Material ka Young's modulus pascals mein nikaalo.
  Pa
  Solution dikhao
  $2.0 \times 10^{- 3}$ strain tak har $0.5 \times 10^{- 3}$ strain par stress 35 MPa badhta hai, isliye
  $Y = \frac{7.0 \times 10^{7} P a}{1.0 \times 10^{- 3}} = 7.0 \times 10^{10} P a$
2. Material ki yield strength ka andaaza lagao.
  Solution dikhao
  Answerlagbhag $1.5 \times 10^{8} P a$
  Seedhi line $2.5 \times 10^{- 3}$ strain par 175 MPa deti, lekin reading sirf 158 MPa hai: material 140 aur 158 MPa ke beech apne seedhi line wale region se bahar nikal gaya. Uski yield strength hai lagbhag $1.5 \times 10^{8} P a .$
3
Short answerEasyNCERT Exercise 8.3
Do materials, P aur Q, ke stress–strain curves same axes aur scales par hain.
Fig. 8.15Materials P aur Q ke stress–strain curves.NCERT Fig. 8.10 · p. 178
Do curves origin se shuru hote hain. Curve P, dashed, ek seedhi line mein tezi se uthta hai aur sirf thode se strain ke baad moderate stress par ek cross ke saath khatam hota hai. Curve Q kam tezi se uthta hai, mudta hai, dheere-dheere ek zyada bade highest stress tak uthta rehta hai aur phir thoda neeche aakar bade strain par ek cross ke saath khatam hota hai.
1. Kis material ka Young's modulus zyada hai?
  PQdono barabar hain
  Answer dekho
  Answer (A) P
  P ka seedha hissa zyada steep hai, isliye diye gaye strain ke liye zyada stress chahiye.
2. Kaun sa material zyada strong hai, jo tootne se pehle zyada stress jhel sake?
  PQ
  Answer dekho
  Answer (B) Q
  Q ka sabse bada stress (ultimate tensile strength) P ke kisi bhi stress se zyada hai. Stiffness aur strength alag properties hain.
3. Kaun sa material brittle hai?
  PQ
  Answer dekho
  Answer (A) P
  P apne seedhe hisse ke turant baad toot jaata hai, lagbhag bina plastic stretching ke; Q tootne se pehle kaafi khinchta hai.
4
Short answerMediumNCERT Exercise 8.4
Batao ki har statement true hai ya false, aur reason do.
1. Physicist ke matlab mein, rubber band same size ke steel wire se zyada elastic hai.
  SahiGalat
  Answer dekho
  Answer Galat
  False. Physics mein 'zyada elastic' matlab bada modulus. Same stress par steel rubber se kahin kam deform hota hai, isliye steel zyada elastic hai.
2. Jab coil spring kheencha jaata hai, to uski stiffness mainly uske wire ke Young's modulus se tay hoti hai, kyunki wire lamba hota hai.
  SahiGalat
  Answer dekho
  Answer Galat
  False. Wire ki total length mushkil se badalti hai. Coil ka har turn aenthta hai, isliye wire shear mein hai, aur spring ki stiffness mainly shear modulus par depend karti hai.
5
NumericalMediumNCERT Exercise 8.5
Ek latakne wale planter mein do wires hain, har ek $0.80 m m$ diameter ka. $1.8 m$ lamba steel wire ek hook se latka hai aur apne neeche ke sire par $2.0 k g$ ka gamla pakde hai; us gamle ke bottom se $0.80 m$ lamba copper wire doosra gamla pakde hai, jiska mass hai $3.0 k g .$ Maan lo steel ke liye $Y = 2.0 \times 10^{11} P a$ , copper ke liye $Y = 1.2 \times 10^{11} P a$ aur $g = 9.8 m / s^{2} .$
1. Steel wire ka extension mm mein nikaalo.
  mm
  Solution dikhao
  Area: $A = π (4.0 \times 10^{- 4})^{2} = 5.03 \times 10^{- 7} m^{2} .$ Steel wire dono gamle pakadta hai: $T = (5.0) (9.8) = 49 N .$
  $Δ L_{s} = \frac{T L_{s}}{A Y_{s}} = 8.8 \times 10^{- 4} m$
2. Copper wire ka extension mm mein nikaalo.
  mm
  Solution dikhao
  Copper wire sirf neeche wala gamla pakadta hai: $T = (3.0) (9.8) = 29.4 N .$
  $Δ L_{c} = \frac{T L_{c}}{A Y_{c}} = 3.9 \times 10^{- 4} m$
NumericalMediumNCERT Exercise 8.6
Ek anti-vibration mount rubber ka cube hai, jiska edge hai $5.0 c m .$ Ek face vertical deewar se chipka hai, aur saamne wale face par $4.0 k g$ ka instrument fixed hai, isliye uska weight us face ke along lagta hai. Rubber ka shear modulus hai $0.40 M P a .$ Bahar wala face kitna neeche jhukta hai, mm mein? ( $g = 9.8 m / s^{2}$ )
mm
Solution dikhao
Weight $F = 39.2 N$ , $A = 2.5 \times 10^{- 3} m^{2}$ area wale face ke along lagta hai, aur faces $L = 0.050 m$ door hain.
$Δ x = \frac{F L}{A G}$
$= 1.96 \times 10^{- 3} m \approx 2.0 m m$ , yaani lagbhag 0.039 ka shearing strain.
NumericalMediumNCERT Exercise 8.7
Chhe ek jaise khokhle steel columns ek overhead paani ki tanki ko uthaye hue hain, jiska total mass hai $3.0 \times 10^{5} k g .$ Har column ka outer radius $40 c m$ hai aur inner radius $32 c m .$ Agar load barabar bantta hai, to har column mein compressive strain nikaalo. ( $Y = 2.0 \times 10^{11} P a$ , $g = 9.8 m / s^{2}$ )
Solution dikhao
Har column par load: $F = (3.0 \times 10^{5}) (9.8) / 6 = 4.9 \times 10^{5} N .$
$A = π (0.16 - 0.1024) m^{2} = 0.181 m^{2}$
$ε = \frac{F}{A Y} = 1.35 \times 10^{- 5}$
NumericalEasyNCERT Exercise 8.8
$12.0 m m \times 25.0 m m$ rectangular cross-section wali aluminium busbar ko $21.0 k N$ force se kheencha jaata hai, jo uski elastic limit ke kaafi andar hai. Strain nikaalo. ( $Y = 7.0 \times 10^{10} P a$ )
Solution dikhao
$A = (0.0120) (0.0250) = 3.00 \times 10^{- 4} m^{2}$
$σ = \frac{2.10 \times 10^{4}}{3.00 \times 10^{- 4}} = 7.0 \times 10^{7} P a$
$ε = \frac{σ}{Y} = 1.0 \times 10^{- 3}$
NumericalEasyNCERT Exercise 8.9
Ek mountain ropeway ki haul cable ka radius hai $2.0 c m .$ Safety ke liye usmein stress kabhi is value se zyada nahi hona chahiye: $1.5 \times 10^{8} P a .$ Woh sabse zyada kitna tension utha sakti hai, newtons mein?
N
Solution dikhao
$F_{\max} = σ_{\max} π r^{2}$
$= (1.5 \times 10^{8}) π (0.020)^{2} N = 1.9 \times 10^{5} N$ , lagbhag 19 tonnes ka weight.
10
NumericalHardNCERT Exercise 8.10
$24 k g$ mass ka ek uniform signboard teen vertical wires se seedha latka hai, har wire $1.5 m$ lamba hai aur board ke dono siron aur beech mein juda hai. Dono siron wale wires brass ke hain ( $Y = 1.0 \times 10^{11} P a$ ) aur beech wala wire steel ka hai ( $Y = 2.0 \times 10^{11} P a$ ). Har wire ko same tension uthana hai.
1. Har wire mein tension kitna hai, newtons mein?
  N
  Solution dikhao
  Teen barabar tensions weight ko support karte hain: $T = (24) (9.8) / 3 = 78.4 N .$
2. Brass wire ke diameter aur steel wire ke diameter ka ratio nikaalo.
  Solution dikhao
  Barabar tension aur barabar length ke saath $Δ L = T L / (A Y)$ teenon ke liye tabhi same hoga jab $A Y$ same ho:
  $A_{b} Y_{b} = A_{s} Y_{s}$
  $\frac{d_{b}}{d_{s}} = \sqrt{\frac{Y_{s}}{Y_{b}}} = \sqrt{2} \approx 1.41$
11
NumericalHardNCERT Exercise 8.11
$2.0 k g$ ki metal ball ek steel wire se bandhi hai jiski natural length $0.80 m$ aur cross-section $2.0 m m^{2}$ hai, aur use vertical circle mein ghumaya jaata hai. Sabse neeche wale point par woh 2.0 revolutions per second se ghoomti hai. Wire ka mass ignore karo aur maan lo $r = 0.80 m$ , $Y = 2.0 \times 10^{11} P a$ aur $g = 9.8 m / s^{2} .$
1. Sabse neeche wale point par wire mein tension nikaalo, newtons mein.
  N
  Solution dikhao
  $T = m g + m ω^{2} r$
  $ω = 2 π (2.0) = 12.6 r a d / s$ ke saath:
  $T = (2.0) (9.8 + 126.3) N = 272 N$
2. Wahan wire ka extension mm mein nikaalo.
  mm
  Solution dikhao
  $Δ L = \frac{T L}{A Y}$
  $= 5.4 \times 10^{- 4} m \approx 0.54 m m .$ Stress, $1.4 \times 10^{8} P a$ , steel ki yield strength se safely neeche hai.
12
NumericalMediumNCERT Exercise 8.12
Ek laboratory test mein mazboot cylinder mein rakhe $2.000 L$ hydraulic oil par pressure $200 a t m$ badhaya jaata hai, aur volume mein kami hoti hai $26.0 m L .$ Maan lo $1 a t m = 1.013 \times 10^{5} P a .$
1. Oil ka bulk modulus pascals mein nikaalo.
  Pa
  Solution dikhao
  $p = (200) (1.013 \times 10^{5}) = 2.03 \times 10^{7} P a$
  $\frac{∣ Δ V ∣}{V} = \frac{26.0}{2000} = 0.0130$
  $B = \frac{p}{∣ Δ V ∣ / V} = 1.56 \times 10^{9} P a$
2. Ise constant temperature par hawa ke bulk modulus, lagbhag $1.0 \times 10^{5} P a$ , se compare karo, aur roz ki bhasha mein batao ki itna bada fark kyun hai.
  Solution dikhao
  Answerlagbhag $1.6 \times 10^{4}$ guna bada
  $B_{oil} / B_{air} \approx 1.6 \times 10^{4}$ : oil ko compress karna lagbhag 16 000 guna mushkil hai. Uske molecules pehle se ek doosre ko chhoo rahe hain, isliye unhe aur paas dabane ka matlab hai strong repulsive forces ke khilaaf dhakelna. Hawa ke molecules average mein lagbhag das molecular diameters door hote hain aur mushkil se interact karte hain, isliye gas aasaani se dab jaati hai.
NumericalMediumNCERT Exercise 8.13
Ocean mein gehraai par pressure atmospheric se $400 a t m$ zyada hai. Surface par seawater ki density $1.025 \times 10^{3} k g / m^{3}$ hai aur bulk modulus $2.3 \times 10^{9} P a .$ Us gehraai par uski density kg/m³ mein nikaalo. ( $1 a t m = 1.013 \times 10^{5} P a$ )
kg/m³
Solution dikhao
Fixed mass ka mass same rehta hai jabki uska volume ghat kar $V (1 - p / B)$ ho jaata hai, isliye $ρ^{'} = ρ / (1 - p / B) .$
$\frac{p}{B} = \frac{4.05 \times 10^{7}}{2.3 \times 10^{9}} = 0.0176$
$ρ^{'} = \frac{1025}{1 - 0.0176} = 1.043 \times 10^{3} k g / m^{3}$
Wahan paani surface se lagbhag 1.8 % zyada dense hai.
NumericalEasyNCERT Exercise 8.14
Ek glass ka kancha pressure vessel mein rakha hai jismein hydraulic pressure mein badhotari hai $60 a t m .$ Glass ke liye $B = 4.0 \times 10^{10} P a$ lekar uske volume mein fractional kami nikaalo. ( $1 a t m = 1.013 \times 10^{5} P a$ )
Solution dikhao
$\frac{∣ Δ V ∣}{V} = \frac{p}{B}$
$= (6.08 \times 10^{6}) / (4.0 \times 10^{10}) = 1.5 \times 10^{- 4}$ , yaani lagbhag 0.015 % ki kami.
NumericalMediumNCERT Exercise 8.15
$5.0 c m$ radius wala solid aluminium sphere hydraulic pressure mein rakha jaata hai, jo hai $1.2 \times 10^{7} P a .$ Uska volume kitna ghatta hai, cm³ mein? ( $B = 7.6 \times 10^{10} P a$ )
cm³
Solution dikhao
$V = \frac{4}{3} π (0.050)^{3} = 5.24 \times 10^{- 4} m^{3}$
$∣ Δ V ∣ = \frac{p V}{B} = 8.3 \times 10^{- 8} m^{3}$
Yeh hai lagbhag $0.083 c m^{3} .$
NumericalEasyNCERT Exercise 8.16
$5.0 L$ mercury par pressure kitna badhana hoga taaki woh 0.050 % compress ho? Answer atmospheres mein do. ( $B = 2.85 \times 10^{10} P a$ , $1 a t m = 1.013 \times 10^{5} P a$ )
atm
Solution dikhao
$p = B \frac{∣ Δ V ∣}{V} = (2.85 \times 10^{10}) (5.0 \times 10^{- 4})$
$= 1.43 \times 10^{7} P a \approx 141 a t m .$ 5.0 L volume se koi fark nahi padta.
Quick answerMedium
Ek metal wire ko die se kheencha jaata hai taaki uski length double ho jaaye jabki volume same rahe. Same load par uska extension ho jaata hai
2 guna8 guna4 gunasame
Answer dekho
Answer (C) 4 guna
Constant volume par length double karne se area aadha ho jaata hai, aur $Δ L = F L / (A Y)$ itne guna badhta hai: $2 \times 2 = 4.$
Quick answerMedium
Assertion (A): Steel rubber se zyada elastic hai. Reason (R): Same stress par steel mein strain rubber se kahin kam hota hai.
A aur R true hain, aur R, A ko explain karta haiA aur R true hain, lekin R, A ko explain nahi kartaA true hai lekin R false haiA false hai lekin R true hai
Answer dekho
Answer (A) A aur R true hain, aur R, A ko explain karta hai
Dono true hain. Bada modulus (same stress par kam strain) hi physics mein 'zyada elastic' ka matlab hai, isliye R, A ko explain karta hai.
Quick answerMedium
Ek steel wire ( $Y = 2.0 \times 10^{11} P a$ ) par tensile stress hai $1.0 \times 10^{8} P a .$ Per unit volume store elastic energy J/m³ mein nikaalo.
J/m³
Answer dekho
Answer 25000 J/m³
$u = σ^{2} / 2 Y = (1.0 \times 10^{16}) / (4.0 \times 10^{11}) = 2.5 \times 10^{4} J / m^{3} .$
Quick answerMedium
Ek material thoda kheenchne par bilkul same volume rakhta hai. Uska Poisson's ratio hai
00.250.51
Answer dekho
Answer (C) 0.5
$Δ V / V = Δ L / L + 2 Δ d / d = ε (1 - 2 ν)$ , jo zero hota hai jab $ν = 0.5.$ Rubber iske kareeb aata hai.
Quick answerHard
Assertion (A): Constant temperature par rakhi gas ka bulk modulus uska pressure badhne par badhta hai. Reason (R): Constant temperature par gas ke liye $p V$ constant hai, isliye uska bulk modulus uske pressure ke barabar hai.
A aur R true hain, aur R, A ko explain karta haiA aur R true hain, lekin R, A ko explain nahi kartaA true hai lekin R false haiA false hai lekin R true hai
Answer dekho
Answer (A) A aur R true hain, aur R, A ko explain karta hai
$p V =$ constant se $V d p + p d V = 0$ , isliye $B = - V d p / d V = p .$ Zyada pressure wali gas ko compress karna mushkil hai, aur R, A ko explain karta hai.
22
Quick answerEasy
Har situation ko us elastic constant se match karo jo mainly use tay karta hai.
Left mein koi item tap karo, phir right mein uska match.
Answer dekho
Answer
lift cable ka khinchna : Young's modulus
drive shaft ka aenthna : shear modulus
gehre samudra mein doobti ball : bulk modulus
kheencha hua wire patla hona : Poisson's ratio
Kheenchna longitudinal strain (Y) hai, aenthna shear (G) hai, har taraf ka pressure volume badalta hai (B), aur side mein patla hona lateral strain (Poisson's ratio) hai.
Quick answerMedium
Wires A aur B ek hi material aur length ke hain, aur A ka diameter B ka double hai. Same force dono ko kheenchta hai. Store elastic energies ka ratio $U_{A} / U_{B}$ kya hai?
421/21/4
Answer dekho
Answer (D) 1/4
Same force ke liye $U = \frac{1}{2} F Δ L = F^{2} L / (2 A Y) \propto 1 / A$ ; A ka area chaar guna hai, isliye woh ek-chauthai energy store karta hai.

Strain ( $\times 10^{- 3}$ )	Stress (MPa)
0	0
0.5	35
1.0	70
1.5	105
2.0	140
2.5	158
3.0	166
3.5	170

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 8.1	Practice Q1
NCERT Exercise 8.2	Practice Q2
NCERT Exercise 8.3	Practice Q3
NCERT Exercise 8.4	Practice Q4
NCERT Exercise 8.5	Practice Q5
NCERT Exercise 8.6	Practice Q6
NCERT Exercise 8.7	Practice Q7
NCERT Exercise 8.8	Practice Q8
NCERT Exercise 8.9	Practice Q9
NCERT Exercise 8.10	Practice Q10
NCERT Exercise 8.11	Practice Q11
NCERT Exercise 8.12	Practice Q12
NCERT Exercise 8.13	Practice Q13
NCERT Exercise 8.14	Practice Q14
NCERT Exercise 8.15	Practice Q15
NCERT Exercise 8.16	Practice Q16

Quick check: MCQs

8.1 Introduction2 questions
8.2 Stress aur Strain3 questions
8.3 Hooke's Law3 questions
8.4 Stress–Strain Curve3 questions
8.5.1 Young's Modulus3 questions
8.5.2 Shear Modulus2 questions
8.5.3 Bulk Modulus3 questions
8.5.4 Poisson's Ratio2 questions
8.5.5 Stretched Wire mein Elastic Potential Energy2 questions
8.6 Materials ke Elastic Behaviour ke Applications3 questions

Glossary · 15

Key terms (English + Hindi)

bulk modulusआयतन गुणांकaayatan gunank: Hydraulic stress aur volume strain ka ratio, $B = - p / (Δ V / V)$ ; solids, liquids aur gases sabke liye define hai.Chapter mein pehli baar yahan aaya →

compressibilityसंपीड्यताsampeedyata: Bulk modulus ka reciprocal: pressure mein unit badhotari par volume mein fractional kami.Chapter mein pehli baar yahan aaya →

ductileतन्यtanya: Tootne se pehle plastically kaafi khinch sakne wala, jaise copper; brittle material, jaise glass, apni elastic limit ke thodi der baad hi toot jaata hai.Chapter mein pehli baar yahan aaya →

elastic limitप्रत्यास्थता सीमाpratyasthata seema: Sabse bada stress jo material jhel kar bhi poori tarah waapas aa sake; is yield point par stress ko yield strength kehte hain.Chapter mein pehli baar yahan aaya →
elasticityप्रत्यास्थताpratyasthata: Body ki woh property jisse deform karne wale forces hatne par woh apne original size aur shape mein waapas aa jaati hai.Chapter mein pehli baar yahan aaya →
elastomer: Rubber jaisa material jo bahut bade strains tak elastically khinchta hai lekin Hooke's law follow nahi karta.Chapter mein pehli baar yahan aaya →

modulus of elasticityप्रत्यास्थता गुणांकpratyasthata gunank: Jahan Hooke's law sahi hai, wahan stress aur strain ka ratio; yeh material ki property hai.Chapter mein pehli baar yahan aaya →

permanent set: Woh strain jo elastic limit se zyada stress jhel chuki body ko unload karne par bacha rehta hai.Chapter mein pehli baar yahan aaya →
plasticityसुघट्यताsughatyata: Material ki woh property jisse deform karne wala force hatne ke baad bhi woh diya gaya naya shape rakhta hai, jaise mitti.Chapter mein pehli baar yahan aaya →
Poisson's ratioपॉयसन अनुपातPoisson anupat: Kheenchi gayi body mein lateral strain divided by longitudinal strain; ek pure number, metals ke liye lagbhag 0.3.Chapter mein pehli baar yahan aaya →

shear modulusअपरूपण गुणांकaparupan gunank: Shearing stress aur shearing strain ka ratio, $G = F / (A θ)$ .Chapter mein pehli baar yahan aaya →
strainविकृतिvikriti: Body ke kisi dimension (length, shear ka angle ya volume) mein fractional badlaav; iska koi unit nahi hota.Chapter mein pehli baar yahan aaya →
stressप्रतिबलpratibal: Deformed body ke andar per unit area restoring force, $F / A$ , jo pascals mein maapa jaata hai.Chapter mein pehli baar yahan aaya →

ultimate tensile strength: Tensile test mein material ka sabse bada stress, jiske baad usmein neck banti hai aur woh toot jaata hai.Chapter mein pehli baar yahan aaya →

Young's modulusयंग गुणांकYoung gunank: Longitudinal stress aur longitudinal strain ka ratio, $Y = F L / (A Δ L)$ .Chapter mein pehli baar yahan aaya →

8.1 Introduction

8.2 Stress aur Strain

8.3 Hooke's Law

8.4 Stress–Strain Curve

8.5 Elastic Moduli

8.5.1 Young's Modulus

8.5.2 Shear Modulus

8.5.3 Bulk Modulus

8.5.4 Poisson's Ratio

8.5.5 Stretched Wire mein Elastic Potential Energy

8.6 Materials ke Elastic Behaviour ke Applications

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