Chapter 1. Tension and Compression in Bars

Transcription

1 Chpter 1 Tension nd Compression in Brs 1

2 1 Tension nd Compression in Brs 1.1 Stress Strin Constitutive Lw Singe Br under Tension or Compression Stticy Determinte Systems of Brs Stticy Indeterminte Systems of Brs Suppementry Exmpes Summry Objectives: In this textbook bout the Mechnics of Mteris we investigte the stressing nd the deformtions of estic structures subject to ppied ods. In the first chpter we wi restrict ourseves to the simpest structur members, nmey, brs under tension or compression. In order to tret such probems, we need kinemtic retions nd constitutive w to compement the equiibrium conditions which re known from Voume 1. The kinemtic retions represent the geometry of the deformtion, wheres the behviour of the estic mteri is described by the constitutive w. The students wi ern how to ppy these equtions nd how to sove stticy determinte s we s stticy indeterminte probems. D. Gross et., Engineering Mechnics 2, DOI / _1, Springer-Verg Berin Heideberg 2011

3 1.1 Stress Stress Let us consider stright br with constnt cross-section re A. The ine connecting the centroids of the cross sections is ced the xis of the br. The ends of the br re subjected to the forces whose common ine of ction is the xis (ig. 1.1). The extern od cuses intern forces. The intern forces cn be visuized by n imginry cut of the br (compre Voume 1, Section 1.4). They re distributed over the cross section (see ig. 1.1b) nd re ced stresses. Being re forces, they hve the dimension force per re nd re mesured, for exmpe, s mutipes of the unit MP (1 MP = 1 N/mm 2 ). The unit Psc (1 P = 1 N/m 2 ) is nmed fter the mthemticin nd physicist Bise Psc ( ); the notion of stress ws introduced by Augustin Louis Cuchy ( ). In Voume 1 (Sttics) we ony det with the resutnt of the intern forces (= norm force) wheres now we hve to study the intern forces (= stresses). 1.1 c A ϕ c c d τ σ c A = cos A ϕ b σ e ϕ σ τ ig. 1.1 c N In order to determine the stresses we first choose n imginry cut c c perpendicur to the xis of the br. The stresses re shown in the free-body digrm (ig. 1.1b); they re denoted by σ. We ssume tht they ct perpendicury to the exposed surfce A of the cross section nd tht they re uniformy distributed. Since they re norm to the cross section they re ced norm stresses. Their resutnt is the norm force N shown in ig. 1.1c (compre Voume 1, Section 7.1). Therefore we hve N = σa nd the stresses σ cn be ccuted from the norm force N:

4 8 1 Tension nd Compression in Brs σ = N A. (1.1) In the present exmpe the norm force N is equ to the ppied force. Thus, we obtin from (1.1) σ = A. (1.2) In the cse of positive norm force N (tension) the stress σ is then positive (tensie stress). Reversey, if the norm force is negtive (compression) the stress is so negtive (compressive stress). Let us now imgine the br being sectioned by cut which is not orthogon to the xis of the br so tht its direction is given by the nge ϕ (ig. 1.1d). The intern forces now ct on the exposed surfce A = A/ cos ϕ. Agin we ssume tht they re uniformy distributed. We resove the stresses into component σ perpendicur to the surfce (the norm stress) nd component τ tngenti to the surfce (ig. 1.1e). The component τ which cts in the direction of the surfce is ced sher stress. Equiibrium of the forces cting on the eft portion of the br yieds (see ig. 1.1e) : σa cos ϕ + τa sin ϕ =0, : σa sin ϕ τa cos ϕ =0. Note tht we hve to write down the equiibrium conditions for the forces, not for the stresses. WithA = A/ cos ϕ we obtin σ + τ tn ϕ =, σtn ϕ τ =0. A Soving these two equtions for σ nd τ yieds 1 σ = 1+tn 2 ϕ A, τ = tn ϕ 1+tn 2 ϕ A.

5 1.1 Stress 9 It is prctic to write these equtions in different form. Using the stndrd trigonometric retions 1 1+tn 2 ϕ =cos2 ϕ, cos 2 ϕ = 1 (1 + cos 2 ϕ), 2 sin ϕ cos ϕ = 1 2 sin 2 ϕ nd the bbrevition σ 0 = /A (= norm stress in section perpendicur to the xis) we finy get σ = σ 0 2 (1 + cos 2 ϕ), τ = σ 0 sin 2 ϕ. (1.3) 2 Thus, the stresses depend on the direction of the cut. If σ 0 is known, the stresses σ nd τ cn be ccuted from (1.3) for rbitrry vues of the nge ϕ. The mximum vue of σ is obtined for ϕ =0,inwhichcseσ mx = σ 0 ; the mximum vue of τ is found for ϕ = π/4 forwhichτ mx = σ 0 /2. If we section br ner n end which is subjected to concentrted force (ig. 1.2, section c c) we find tht the norm stress is not distributed uniformy over the cross-section re. The concentrted force produces high stresses ner its point of ppiction (ig. 1.2b). This phenomenon is known s stress concentrtion. It cn be shown, however, tht the stress concentrtion is restricted to sections in the proximity of the point of ppiction of the concentrted force: the high stresses decy rpidy towrds the verge vue σ 0 s we increse the distnce from the end of the br. This fct is referred to s Sint-Vennt s principe (Adhémr Jen Cude Brré de Sint-Vennt, ). c c c c b σ ig. 1.2 c σ

6 10 1 Tension nd Compression in Brs The uniform distribution of the stress is so disturbed by hoes, notches or ny brupt chnges (discontinuities) of the geometry. If, for exmpe, br hs notches the remining cross-section re (section c c ) is so subjected to stress concentrtion (ig. 1.2c). The determintion of these stresses is not possibe with the eementry nysis presented in this textbook. Let us now consider br with ony sight tper (compre Exmpe 1.1). In this cse the norm stress my be ccuted from (1.1) with sufficient ccurcy. Then the cross-section re A nd the stress σ depend on the oction ong the xis. If voume forces ct in the direction of the xis in ddition to the concentrted forces, then the norm force N so depends on the oction. Introducing the coordinte x in the direction of the xis we cn write: σ(x) = N(x) A(x). (1.4) Here it is so ssumed tht the stress is uniformy distributed over the cross section t fixed vue of x. In stticy determinte systems we cn determine the norm force N from equiibrium conditions one. If the cross-section re A is known, the stress σ cn be ccuted from (1.4). Stticy indeterminte systems wi be treted in Section 1.4. In engineering ppictions structures hve to be designed in such wy tht given mximum stressing is not exceeded. In the cse of br this requirement mens tht the bsoute vue of the stress σ must not exceed given owbe stress σ ow : σ σ ow. (Note tht the owbe stresses for tension nd for compression re different for some mteris.) The required cross section A req of br for given od nd thus known norm force N cn then be determined from σ = N/A: A req = N. (1.5) σ ow This is referred to s dimensioning of the br. Aterntivey, the owbe od cn be ccuted from N σ ow A in the cse of given cross-section re A.

7 1.1 Stress 11 Note tht sender br which is subjected to compression my fi due to bucking before the stress ttins n indmissiby rge vue. We wi investigte bucking probems in Chpter 7. Exmpe 1.1 A br (ength ) with circur cross section nd sight tper (inery vrying from rdius r 0 to 2 r 0 ) is subjected to the compressive forces s shown in ig Determine the norm stress σ in n rbitrry cross section perpendicur to the xis of the br. E1.1 r 0 2r 0 ig. 1.3 b x r(x) Soution We introduce the coordinte x, see ig. 1.3b. Then the rdius of n rbitrry cross section is given by r(x) =r 0 + r ( 0 x = r 0 1+ x ). Using (1.4) with the cross section A(x) =πr 2 (x) nd the constnt norm force N = yieds σ = N A(x) = πr 2 0 ( 1+ x ) 2. The minus sign indictes tht σ is compressive stress. Its vue t the eft end (x = 0) is four times the vue t the right end (x = ). Exmpe 1.2 Awtertower(heightH, densityϱ) withcross section in the form of circur ring crries tnk (weight W 0 ) s shown in ig The inner rdius r i of the ring is constnt. Determine the outer rdius r in such wy tht the norm stress σ 0 in the tower is constnt ong its height. The weight of the tower cnnot be negected. E1.2

8 12 1 Tension nd Compression in Brs W 0 H r i r x dx A A+dA dw r(x) r i b σ 0 σ 0 ig. 1.4 Soution We consider the tower to be sender br. The retionship between stress, norm force nd cross-section re is given by (1.4). In this exmpe the constnt compressive stress σ = σ 0 is given; the norm force (here counted positive s compressive force) nd the re A re unknown. The equiibrium condition furnishes second eqution. We introduce the coordinte x s shown in ig. 1.4b nd consider sice eement of ength dx. The cross-section re of the circur ring s function of x is A = π(r 2 r 2 i ) () where r = r(x) is the unknown outer rdius. The norm force t the oction x is given by N = σ 0 A (see 1.4). At the oction x +dx, the re nd the norm force re A +da nd N +dn = σ 0 (A +da). The weight of the eement is dw = ϱgdv where dv = A dx is the voume of the eement. Note tht terms of higher order re negected (compre Voume 1, Section 7.2.2). Equiibrium in the vertic direction yieds : σ 0 (A +da) ϱgdv σ 0 A =0 σ 0 da ϱgadx =0. Seprtion of vribes nd integrtion ed to da ϱg A = dx n A = ϱgx A = A 0 e ϱgx σ 0. (b) σ 0 A 0 σ 0

9 1.2 Strin 13 The constnt of integrtion A 0 foows from the condition tht the stress t the upper end of the tower (for x =0wehveN = W 0 ) so hs to be equ to σ 0 : W 0 A 0 = σ 0 A 0 = W 0 σ 0. (c) Equtions () to (c) yied the outer rdius: r 2 (x) =r 2 i + W 0 πσ 0 e ϱgx σ Strin We wi now investigte the deformtions of n estic br. Let us first consider br with constnt cross-section re which hs the undeformed ength. Under the ction of tensie forces (ig. 1.5) it gets sighty onger. The eongtion is denoted by Δ nd is ssumed to be much smer thn the origin ength. As mesure of the mount of deformtion, it is usefu to introduce, in ddition to the eongtion, the rtio between the eongtion nd the origin (undeformed) ength: ε = Δ. (1.6) The dimensioness quntity ε is ced strin. If, for exmpe, br of the ength = 1 m undergoes n eongtion of Δ =0.5 mm then we hve ε = This is strin of 0.05%. If the br gets onger (Δ >0) the strin is positive; it is negtive in the cse of shortening of the br. In wht foows we wi consider ony sm deformtions: Δ or ε 1, respectivey. The definition (1.6) for the strin is vid ony if ε is constnt over the entire ength of the br. If the cross-section re is not 1.2 Δ ig. 1.5

10 14 1 Tension nd Compression in Brs x dx u u+du undeformed br deformed br dx+(u+du) u ig. 1.6 constnt or if the br is subjected to voume forces cting ong its xis, the strin my depend on the oction. In this cse we hve to use oc strin which wi be defined s foows. We consider n eement of the br (ig. 1.6) insted of the whoe br. It hs the ength dx in the undeformed stte. Its eft end is octed t x, the right end t x +dx. If the br is eongted, the cross sections undergo dispcements in the x-direction which re denoted by u. They depend on the oction: u = u(x). Thus, the dispcements re u t the eft end of the eement nd u +du t the right end. The ength of the eongted eement is dx+(u+du) u =dx+du. Hence, the eongtion of the eement is given by du. Nowtheoc strin cn be defined s the rtio between the eongtion nd the undeformed ength of the eement: ε(x) = du dx. (1.7) If the dispcement u(x) is known, the strin ε(x) cn be determined through differentition. Reversey, if ε(x) is known, the dispcement u(x) is obtined through integrtion. The dispcement u(x) nd the strin ε(x) describe the geometry of the deformtion. Therefore they re ced kinemtic quntities. Eqution (1.7) is referred to s kinemtic retion Constitutive Lw Stresses re quntities derived from sttics; they re mesure for the stressing in the mteri of structure. On the other hnd, strins re kinemtic quntities; they mesure the deformtion

11 1.3 Constitutive Lw 15 of body. However, the deformtion depends on the od which cts on the body. Therefore, the stresses nd the strins re not independent. The physic retion tht connects these quntities is ced constitutive w. It describes the behviour of the mteri of the body under od. It depends on the mteri nd cn be obtined ony with the id of experiments. One of the most importnt experiments to find the retionship between stress nd strin is the tension or compression test. Here, sm specimen of the mteri is pced into testing mchine nd eongted or shortened. The force ppied by the mchine onto the specimen cn be red on the di of the mchine; it cuses the norm stress σ = /A. The chnge Δ of the ength of the specimen cn be mesured nd the strin ε =Δ/ cn be ccuted. The grph of the retionship between stress nd strin is shown schemticy (not to sce) for stee specimen in ig This grph is referred to s stress-strin digrm. One cn see tht for sm vues of the strin the retionship is iner (stright ine) nd the stress is proportion to the strin. This behviour is vid unti the stress reches the proportion imit σ P.Ifthe stress exceeds the proportion imit the strin begins to increse more rpidy nd the sope of the curve decreses. This continues unti the stress reches the yied stress σ Y. rom this point of the stress-strin digrm the strin increses t prcticy constnt stress: the mteri begins to yied. Note tht mny mteris do σ σ t = A σ = A σ Y unoding σ P ig. 1.7 ε p A ε

12 16 1 Tension nd Compression in Brs not exhibit pronounced yied point. At the end of the yieding the sope of the curve increses gin which shows tht the mteri cn sustin n ddition od. This phenomenon is ced strin hrdening. Experiments show tht n eongtion of the br eds to reduction of the cross-section re A. This phenomenon is referred to s ter contrction. Wheres the cross-section re decreses uniformy over the entire ength of the br in the cse of sm stresses, it begins to decrese ocy t very high stresses. This phenomenon is ced necking. Since the ctu cross section A my then be considerby smer thn the origin cross section A, the stress σ = /A does not describe the re stress ny more. It is therefore pproprite to introduce the stress σ t = /A which is ced true stress or physic stress. It represents the true stress in the region where necking tkes pce. The stress σ = /A is referred to s nomin or convention or engineering stress. ig.1.7 shows both stresses unti frcture occurs. Consider specimen being first oded by force which cuses the stress σ. Assume tht σ is smer thn the yied stress σ Y, i.e., σ<σ Y. Subsequenty, the od is gin removed. Then the specimen wi return to its origin ength: the strin returns to zero. In ddition, the curves during the oding nd the unoding coincide. This behviour of the mteri is ced estic; the behviour in the region σ σ P is referred to s inery estic. Now ssume tht the specimen is oded beyond the yied stress, i.e., unti stress σ>σ Y is reched. Then the curve during the unoding is stright ine which is pre to the stright ine in the iner-estic region, see ig If the od is competey removed the strin does not return to zero: pstic strin ε p remins fter the unoding. This mteri behviour is referred to s pstic. In the foowing we wi wys restrict ourseves to ineryestic mteri behviour. or the ske of simpicity we wi refer to this behviour shorty s estic, i.e., in wht foows estic wys stnds for inery estic. Then we hve the iner retionship

13 1.3 Constitutive Lw 17 Tbe 1.1 Mteri Constnts Mteri E in MP α T in 1/ C Stee 2, , Auminium 0, , Concrete 0, , Wood (in fibre direction) 0,7... 2, ,2... 3, Cst iron 1, , Copper 1, , Brss 1, , σ = Eε (1.8) between the stress nd the strin. The proportionity fctor E is ced moduus of esticity or Young s moduus (Thoms Young, ). The constitutive w (1.8) is ced Hooke s w fter Robert Hooke ( ). Note tht Robert Hooke coud not present this w in the form (1.8) since the notion of stress ws introduced ony in 1822 by Augustin Louis Cuchy ( ). The retion (1.8) is vid for tension nd for compression: the moduus of esticity hs the sme vue for tension nd compression. However, the stress must be ess thn the proportion imit σ P which my be different for tension or compression. The moduus of esticity E is constnt which depends on the mteri nd which cn be determined with the id of tension test. It hs the dimension of force/re (which is so the dimension of stress); it is given, for exmpe, in the unit MP. Tbe 1.1 shows the vues of E for sever mteris t room temperture. Note tht these vues re just guidnce since the moduus of esticity depends on the composition of the mteri nd on the temperture.

14 18 1 Tension nd Compression in Brs A tensie or compressive force, respectivey, cuses the strin ε = σ/e (1.9) in br, see (1.8). Chnges of the ength nd thus strins re not ony cused by forces but so by chnges of the temperture. Experiments show tht the therm strin ε T is proportion to the chnge ΔT of the temperture if the temperture of the br is chnged uniformy cross its section nd ong its ength: ε T = α T ΔT. (1.10) The proportionity fctor α T is ced coefficient of therm expnsion. It is mteri constnt nd is given in the unit 1/ C. Tbe 1.1 shows sever vues of α T. If the chnge of the temperture is not the sme ong the entire ength of the br (if it depends on the oction) then (1.10) represents the oc strin ε T (x) =α T ΔT (x). If br is subjected to stress σ s we s to chnge ΔT of the temperture, the totstrinε is obtined through superposition of (1.9) nd (1.10): ε = σ E + α T ΔT. (1.11) This retion cn so be written in the form σ = E(ε α T ΔT ). (1.12) Singe Br under Tension or Compression There re three different types of equtions tht ow us to determine the stresses nd the strins in br: the equiibrium condition, the kinemtic retion nd Hooke s w. Depending on the probem, the equiibrium condition my be formuted for the entire br, portion of the br (see Section 1.1) or for n eement of the br. We wi now derive the equiibrium condition for n

15 1.4 Singe Br under Tension or Compression 19 1 ig. 1.8 x dx n(x) 2 N b x n dx dx x+dx N +dn eement. or this purpose we consider br which is subjected to two forces 1 nd 2 t its ends nd to ine od n = n(x), see ig The forces re ssumed to be in equiibrium. We imgine sice eement of infinitesim ength dx seprted from the br s shown in ig. 1.8b. The free-body digrm shows the norm forces N nd N +dn, respectivey, t the ends of the eement; the ine od is repced by its resutnt ndx (note tht n my be considered to be constnt over the ength dx, compre Voume 1, Section 7.2.2). Equiibrium of the forces in the direction of the xis of the br : N +dn + n dx N =0 yieds the equiibrium condition dn dx + n =0. (1.13) In the speci cse of vnishing ine od (n 0) the norm force in the br is constnt. The kinemtic retion for the br is (see (1.7)) ε = du dx, nd Hooke s w is given by (1.11): ε = σ E + α T ΔT. If we insert the kinemtic retion nd σ = N/A into Hooke s w we obtin

16 20 1 Tension nd Compression in Brs du dx = N EA + α T ΔT. (1.14) This eqution retes the dispcements u(x) of the cross sections nd the norm force N(x). It my be ced the constitutive w for the br. ThequntityEA is known s xi rigidity. Equtions (1.13) nd (1.14) re the bsic equtions for br under tension or compression. The dispcement u of cross section is found through integrtion of the strin: ε = du dx du = ε dx u(x) u(0) = x 0 ε d x. The eongtion Δ foows s the difference of the dispcements t the ends x = nd x = 0 of the br: Δ = u() u(0) = With ε =du/dx nd (1.14) this yieds Δ = 0 0 ε dx. (1.15) ( ) N EA + α T ΔT dx. (1.16) In the speci cse of br (ength ) with constnt xi rigidity (EA = const) which is subjected ony to forces t its end (n 0,N = ) nd to uniform chnge of the temperture (ΔT = const), the eongtion is given by Δ = EA + α T ΔT. (1.17) If, in ddition, ΔT =0weobtin

17 1.4 Singe Br under Tension or Compression 21 Δ = EA, (1.18) nd if = 0, (1.17) reduces to Δ = α T ΔT. (1.19) If we wnt to ppy these equtions to specific probems, we hve to distinguish between stticy determinte nd stticy indeterminte probems. In stticy determinte system we cn wys ccute the norm force N(x) with the id of the equiibrium condition. Subsequenty, the strin ε(x) foows from σ = N/A nd Hooke s w ε = σ/e. iny, integrtion yieds the dispcement u(x) nd the eongtion Δ. A chnge of the temperture cuses ony therm strins (no stresses!) in stticy determinte system. In stticy indeterminte probem the norm force cnnot be ccuted from the equiibrium condition one. In such probems the bsic equtions (equiibrium condition, kinemtic retion nd Hooke s w) re system of couped equtions nd hve to be soved simutneousy. A chnge of the temperture in gener cuses ddition stresses; they re ced therm stresses. iny we wi reduce the bsic equtions to singe eqution for the dispcement u. If we sove (1.14) for N nd insert into (1.13) we obtin (EAu ) = n +(EAα T ΔT ). (1.20) Here, the primes denote derivtives with respect to x. Eqution (1.20) simpifies in the speci cse EA = const nd ΔT =const to EAu = n. (1.20b) If the functions EA(x), n(x) nd ΔT (x) re given, the dispcement u(x) of n rbitrry cross section cn be determined through integrtion of (1.20). The constnts of integrtion re ccuted from the boundry conditions. If, for exmpe, one end of the br

18 22 1 Tension nd Compression in Brs is fixed then u = 0 t this end. If, on the other hnd, one end of the br cn move nd is subjected to force 0, then ppying (1.14) nd N = 0 yieds the boundry condition u = 0 /EA + α T ΔT. This reduces to the boundry condition u = 0 in the speci cse of stress-free end ( 0 = 0) of br whose temperture is not chnged (ΔT =0). requenty, one or more of the quntities in (1.20) re given through different functions of x in different portions of the br (e.g., if there exists jump of the cross section). Then the br must be divided into sever regions nd the integrtion hs to be performed seprtey in ech of theses regions. In this cse the constnts of integrtion cn be ccuted from boundry conditions nd mtching conditions (compre Voume 1, Section 7.2.4) x N(x) W W = x W b ig. 1.9 As n iustrtive exmpe of stticy determinte system et us consider sender br (weight W, cross-section re A) tht is suspended from the ceiing (ig. 1.9). irst we determine the norm force cused by the weight of the br. We cut the br t n rbitrryposition x (ig. 1.9b). The norm force N is equ to the weight W of the portion of the br beow the imginry cut. Thus, it is given by N(x) =W (x) =W ( x)/. Eqution (1.4) now yieds the norm stress σ(x) = N(x) A = W A ( 1 x ).

19 1.4 Singe Br under Tension or Compression 23 Accordingy, the norm stress in the br vries inery; it decreses from the vue σ(0) = W/A t the upper end to σ() =0t the free end. The eongtion Δ of the br due to its own weight is obtined from (1.16): N Δ = EA dx = W ( 1 x ) dx = 1 W EA 2 EA. 0 0 It is hf the eongtion of br with negigibe weight which is subjected to the force W tthefreeend. We my so sove the probem by ppying the differenti eqution (1.20b) for the dispcements u(x) of the cross sections of the br. Integrtion with the constnt ine od n = W/ yieds EAu = W, EAu = W x + C 1, EAu = W 2 x2 + C 1 x + C 2. The constnts of integrtion C 1 nd C 2 cn be determined from the boundry conditions. The dispcement of the cross section t the upper end of the br is equ to zero: u(0) = 0. Since the stress σ vnishes t the free end, we hve u () = 0. This eds to C 2 =0ndC 1 = W. Thus, the dispcement nd the norm force re given by u(x) = 1 W 2 EA (2 ) x ( x2 2, N(x) =EAu (x) =W 1 x ). Since u(0) = 0, the eongtion is equ to the dispcement of the free end: Δ = u() = 1 W 2 EA.

20 24 1 Tension nd Compression in Brs Thestressisobtineds σ(x) = N(x) A = W A ( 1 x ). As n iustrtive exmpe of stticy indeterminte system et us consider br which is pced stress-free between two rigid ws (ig. 1.10). It hs the cross-section res A 1 nd A 2, respectivey. We wnt to determine the support rections if the temperture of the br is rised uniformy by n mount ΔT in region. The free-body digrm (ig. 1.10b) shows the two support rections B nd C. They cnnot be ccuted from ony one equiibrium condition: : B C =0. B B ΔT ΔT c C 01 "0" System = C ΔT B b "1" System C X ig Therefore we hve to tke into ccount the deformtion of the br. The eongtions in the regions nd re given by (1.16) with N = B = C: Δ 1 = N EA 1 + α T ΔT, Δ 2 = N EA 2 (the temperture in region is not chnged). The br is pced between two rigid ws. Thus, its tot eongtion Δ hs to vnish: Δ =Δ 1 +Δ 2 =0. This eqution expresses the fct tht the geometry of the deformtion hs to be comptibe with the restrints imposed by the supports. Therefore it is ced comptibiity condition.

21 1.4 Singe Br under Tension or Compression 25 The equiibrium condition nd the comptibiity condition yied the unknown support rections: N EA 1 +α T ΔT+ N EA 2 =0 B = C = N = EA 1A 2 α T ΔT A 1 + A 2. The probem my so be soved in the foowing wy. In first step we generte stticy determinte system. This is chieved by removing one of the supports, for exmpe support C. The ction of this support on the br is repced by the ction of the force C = X which is s yet unknown. Note tht one of the supports, for exmpe B, is needed to hve stticy determinte system. The other support, C, is in excess of the necessry support. Therefore the rection C is referred to s being redundnt rection. Now we need to consider two different probems. irst, we investigte the stticy determinte system subjected to the given od (here: the chnge of the temperture in region )whichis referred to s 0 -system or primry system (ig. 1.10c). In this system the chnge of the temperture cuses the therm eongtion Δ (0) 1 (norm force N =0)inregion ; the eongtion in region is zero. Thus, the dispcement u (0) C of the right end point of the br is given by u (0) C =Δ(0) 1 = α T ΔT. Secondy we consider the stticy determinte system subjected ony to force X. It is ced 1 -system nd is so shown in ig. 1.10c. Here the dispcement u (1) C of the right end point is u (1) C =Δ(1) 1 +Δ (1) 2 = X X. EA 1 EA 2 Both the ppied od (here: ΔT )swestheforcex ct in the given probem (ig. 1.10). Therefore, the tot dispcement u C t point C foows through superposition: u C = u (0) C + u(1) C.

22 26 1 Tension nd Compression in Brs Since the rigid w in the origin system prevents dispcement t C, the geometric condition u C =0 hs to be stisfied. This eds to α T ΔT X EA 1 X EA 2 =0 X = C = EA 1 A 2 α T ΔT A 1 + A 2. Equiibrium t the free-body digrm (ig. 1.10b) yieds the second support rection B = C. E1.3 Exmpe 1.3 A soid circur stee cyinder (cross-section re A S, moduus of esticity E S,ength) is pced inside copper tube (cross-section re A C, moduus of esticity E C,ength). The ssemby is compressed between rigid pte nd the rigid foor by force (ig. 1.11). Determine the norm stresses in the cyinder nd in the tube. Ccute the shortening of the ssemby. C S S C S C b ig Soution We denote the compressive forces in the stee cyinder nd in the copper tube by S nd C, respectivey (ig. 1.11b). Equiibrium t the free-body digrm of the pte yieds C + S =. () Since equiibrium furnishes ony one eqution for the two unknown forces S nd C, the probem is stticy indeterminte. We

23 1.4 Singe Br under Tension or Compression 27 obtin second eqution by tking into ccount the deformtion of the system. The shortenings (here counted positive) of the two prts re given ccording to (1.18) by Δ C = C EA C, Δ S = S EA S where, for simpicity, we hve denoted the xi rigidity E C A C of the copper tube byea C nd the xi rigidity E S A S of the stee cyinder by EA S. The pte nd the foor re ssumed to be rigid. Therefore the geometry of the probem requires tht the shortenings of the copper tube nd of the stee cyinder coincide. This gives the comptibiity condition (b) Δ C =Δ S. (c) Soving the Equtions () to (c) yieds the forces C = EA C EA S, S =. EA C + EA S EA C + EA S The compressive stresses foow ccording to (1.2): (d) σ C = E C EA C + EA S, σ S = E S EA C + EA S. Inserting (d) into (b) eds to the shortening: Δ C =Δ S = EA C + EA S. Exmpe 1.4 Acoppertube is pced over threded stee bot of ength. The pitch of the threds is given by h. Anutfits snugy ginst the tube without generting stresses in the system (ig. 1.12). Subsequenty, the nut is given n fu turns nd the temperture of the entire ssemby is incresed by the mount ΔT. The xi rigidities nd the coefficients of therm expnsion ofthebotndthetuberegiven. E1.4

24 28 1 Tension nd Compression in Brs Determine the force in the bot. X nh nh X 1 X X b ig Soution After the nut hs been turned it exerts compressive force X on the tube which cuses shortening of the tube. According to Newton s third xiom (ction = rection) force of equ mgnitude nd opposite direction cts vi the nut on the bot which eongtes. The free-body digrms of bot nd tube re shown in ig. 1.12b. The probem is stticy indeterminte since force cnnot be determined from equiibrium one. Therefore we hve to tke into ccount the deformtions. The ength of the bot fter the nut hs been turned, see the free-body digrm in ig. 1.12b, is given by 1 = nh. Its eongtion Δ 1 foows from X( nh) Δ 1 = + α EA T 1 ΔT ( nh). 1 Since nh, this cn be reduced to Δ 1 = X EA 1 + α T 1 ΔT. The chnge of ength Δ 2 of the tube ( 2 = ) is obtined from Δ 2 = X + α EA T 2 ΔT. 2 The ength of the bot nd the ength of the tube hve to coincide fter the deformtion. This yieds the comptibiity condition 1 +Δ 1 = 2 +Δ 2 Δ 1 Δ 2 = 2 1 = nh.

25 1.5 Stticy Determinte Systems of Brs 29 Soving the equtions eds to the force in the bot: ( X + ) +(α EA 1 EA T 1 α T 2 )ΔT = nh 2 X = nh ( (α T 1 α T 2 )ΔT ). EA 1 EA Stticy Determinte Systems of Brs In the preceding section we ccuted the stresses nd deformtions of singe sender brs. We wi now extend the investigtion to trusses nd to structures which consist of brs nd rigid bodies. In this section we wi restrict ourseves to stticy determinte systems where we cn first ccute the forces in the brs with the id of the equiibrium conditions. Subsequenty, the stresses in the brs nd the eongtions re determined. iny, the dispcements of rbitrry points of the structure cn be found. Since it is ssumed tht the eongtions re sm s compred with the engths of the brs, we cn ppy the equiibrium conditions to the undeformed system. As n iustrtive exmpe et us consider the truss in ig Both brs hve the xi rigidity EA. We wnt to determine the dispcement of pin C due to the ppied force. irst we ccute the forces S 1 nd S 2 in the brs. The equiibrium conditions, ppied to the free-body digrm (ig. 1.13b), yied 1.5 : S 2 sin α =0 : S 1 + S 2 cos α =0 S 1 = tn α, S 2 = sin α. According to (1.17) the eongtions Δ i of the brs re given by Δ 1 = S 1 1 EA = 1 EA tn α, Δ 2 = S 2 2 EA = 1 EA sin α cos α.

26 30 1 Tension nd Compression in Brs Br 1 becomes shorter (compression) nd br 2 becomes onger (tension). The new position C of pin C cn be found s foows. We consider the brs to be disconnected t C. Then the system becomes movbe: br 1 cn rotte bout point A; br 2 cn rotte bout point B. The free end points of the brs then move ong circur pths with rdii 1 +Δ 1 nd 2 +Δ 2, respectivey. Point C is octed t the point of intersection of these rcs of circes (ig. 1.13c) B A 2 1 α C 2 S 2 α S 1 b C 2 1 C Δ 2 Δ 1 α 1 Δ 1 C Δ 2 α v c C d C u ig The eongtions re sm s compred with the engths of the brs. Therefore, within good pproximtion the rcs of the circes cn be repced by their tngents. This eds to the dispcement digrm s shown in ig. 1.13d. If this digrm is drwn to sce, the dispcement of pin C cn directy be tken from it. We wnt to ppy grphic-nytic soution. It suffices then to drw sketch of the digrm. Appying trigonometric retions we obtin the horizont nd the vertic components of the

27 dispcement: u = Δ 1 = 1 EA tn α, 1.5 Stticy Determinte Systems of Brs 31 v = Δ 2 sin α + u tn α = 1+cos 3 α EA sin 2 α cos α. (1.21) To determine the dispcement of pin of truss with the id of dispcement digrm is usuy quite cumbersome nd cn be recommended ony if the truss hs very few members. In the cse of trusses with mny members it is dvntgeous to ppy n energy method (see Chpter 6). The method described bove cn so be ppied to structures which consist of brs nd rigid bodies. Exmpe 1.5 A rigid bem (weight W ) is mounted on three estic brs (xi rigidity EA) s shown in ig Determine the nge of sope of the bem tht is cused by its weight fter the structure hs been ssembed. W W A B E1.5 1 αα 2 3 b S 1 S 2 αα S 3 EA A Δ 1 α Δ 2 1 v A A 2 c ig A B β d v A v B Soution irst we ccute the forces in the brs with the id of the equiibrium conditions (ig. 1.14b):

28 32 1 Tension nd Compression in Brs S 1 = S 2 = W 4cosα, S 3 = W 2. With 1 = 2 = / cos α nd 3 = we obtin the eongtions: Δ 1 =Δ 2 = S 1 1 EA = W 4EAcos 2 α, Δ 3 = S 3 3 EA = W 2 EA. Point B of the bem is dispced downwrd by v B = Δ 3.To determine the vertic dispcement v A of point A we sketch dispcement digrm (ig. 1.14c). irst we pot the chnges Δ 1 nd Δ 2 of the engths in the direction of the respective br. The ines perpendicur to these directions intersect t the dispced position A of point A. Thus, its vertic dispcement is given by v A = Δ 1 / cos α. Since the dispcements v A nd v B do not coincide, the bem does not sty horizont fter the structure hs been ssembed. The nge of sope β is obtined with the pproximtion tn β β (sm deformtions) nd = cot α s (see ig. 1.14d) β = v B v A = 2cos3 α 1 4cos 3 α W cot α EA. If cos 3 α> 1 2 (or cos3 α< 1 2 ), then the bem is incined to the right (eft). In the speci cse cos 3 α = 1 2, i.e. α =37.5,itstys horizont. E1.6 Exmpe 1.6 The truss in ig is subjected to force. Given: E = GP, =20kN. Determine the cross-section re of the three members so tht the stresses do not exceed the owbe stress σ ow = 150 MP nd the dispcement of support B is smer thn 0.5 of the ength of br 3. Soution irst we ccute the forces in the members. The equiibrium conditions for the free-body digrms of pin C nd support B (ig. 1.15b) yied

29 1.6 Stticy Indeterminte Systems of Brs 33 C C A B S 1 45 S 2 =S S 2 b S 3 B ig S 1 = S 2 = 2 2, S 3 = 2. The stresses do not exceed the owbe stress if σ 1 = S 1 A 1 σ ow, σ 2 = S 2 A 2 σ ow,σ 3 = S 3 A 3 σ ow. This eds to the cross-section res A 1 = A 2 = S 1 σ ow =94.3 mm 2,A 3 = S 3 σ ow =66.7 mm 2. In ddition, the dispcement of support B hs to be smer thn 0.5 of the ength of br 3. This dispcement is equ to the eongtion Δ 3 = S 3 3 /EA 3 of br 3 (support A is fixed). rom Δ 3 < we obtin Δ 3 3 = S 3 EA 3 < A 3 > 2 S 3 E 103 = E 103 = 100mm 2. Comprison with () yieds the required re A 3 = 100 mm 2. () 1.6 Stticy Indeterminte Systems of Brs We wi now investigte stticy indeterminte systems for which the forces in the brs cnnot be determined with the id of the equiibrium conditions one since the number of the unknown quntities exceeds the number of the equiibrium conditions. In such systems the bsic equtions (equiibrium conditions, kinem- 1.6

30 34 1 Tension nd Compression in Brs tic equtions (comptibiity) nd Hooke s w) re couped equtions. Let us consider the symmetric truss shown in ig It is stress-free before the od is ppied. The xi rigidities EA 1, EA 2, EA 3 = EA 1 re given; the forces in the members re unknown. The system is stticy indeterminte to the first degree (the decomposition of force into three directions cnnot be done uniquey in copnr probem, see Voume 1, Section 2.2). The two equiibrium conditions ppied to the free-body digrm of pin K (ig. 1.16b) yied : S 1 sin α + S 3 sin α =0 S 1 = S 3, : S 1 cos α + S 2 + S 3 cos α =0 S 1 = S 3 = S () 2 2cosα. The eongtions of the brs re given by Δ 1 =Δ 3 = S 1 1, Δ 2 = S 2. (b) EA 1 EA 2 To derive the comptibiity condition we sketch dispcement digrm (ig. 1.16c) from which we find α α K S 2 S 1 α α S 3 =S 1 K b c 1 3 Δ 3 α K Δ 1 K Δ "0" System "1" System = + X d X ig. 1.16

31 1.6 Stticy Indeterminte Systems of Brs 35 Δ 1 =Δ 2 cos α. (c) With (), (b) nd 1 = / cos α we obtin from (c) ( S 2 ) 2 EA 1 cos 2 α = S 2 cos α EA 2 which eds to S 2 = 1+2 EA 1 EA 2 cos 3 α. The remining two forces in the brs foow from (): S 1 = S 3 = EA 1 EA 2 cos 2 α 1+2 EA 1 EA 2 cos 3 α. Note tht now the vertic dispcement v of pin K cn so be written down: v =Δ 2 = S 2 EA 2 = EA EA 1 EA 2 cos 3 α The probem my so be soved using the method of superposition. In first step we remove br 2 to obtin stticy determinte system, the 0 -system. It consists of the two brs 1 nd 3 nd it is subjected to the given force (ig. 1.16d). The forces S (0) 1 nd S (0) 3 in these brs foow from the equiibrium conditions s S (0) 1 = S (0) 3 = 2cosα. The corresponding eongtions re obtined with 1 = / cos α: Δ (0) 1 =Δ (0) 3 = S(0) 1 1 EA 1 =. 2 EA 1 cos 2 α. (d)

32 36 1 Tension nd Compression in Brs In second step we consider the stticy determinte system under the ction of n unknown force X ( 1 -system, see so ig. 1.16d). Note tht this force cts in the opposite direction on br 2 (ctio = rectio). Now we get S (1) 1 = S (1) 3 = X 2cosα, Δ (1) 1 =Δ (1) 3 = S(1) 2 = X, X 2 EA 1 cos 2 α, Δ(1) 2 = X. EA 2 The tot eongtion of the brs is obtined through superposition of the systems 0 nd 1 : Δ 1 =Δ 3 =Δ (0) 1 +Δ (1) 1, Δ 2 =Δ (1) 2. (f) The comptibiity condition (c) is gin tken from the dispcement digrm (ig. 1.16c). It eds with (d) - (f) to the unknown force X = S (1) 2 = S 2 : 2 EA 1 cos 2 α X 2 EA 1 cos 2 α = X cos α EA 2 X = S 2 = 1+2 EA. 1 cos 3 α EA 2 The forces S 1 nd S 3 foow from superposition: S 1 = S 3 = S (0) 1 + S (1) 1 = EA 1 EA 2 cos 2 α 1+2 EA 1 EA 2 cos 3 α A system of brs is stticy indeterminte of degree n if the number of the unknowns exceeds the number of the equiibrium conditions by n. In order to determine the forces in the brs of such system, n comptibiity conditions re needed in ddition to the equiibrium conditions. Soving this system of equtions yieds the unknown forces in the brs. A stticy indeterminte system of degree n cn so be soved with the method of superposition. Then n brs re removed in order to obtin stticy determinte system. The ction of the. (e)

33 1.6 Stticy Indeterminte Systems of Brs 37 brs which re removed is repced by the ction of the sttic redundnts S i = X i.nextn + 1 different uxiiry systems re considered. The given od cts in the 0 -system, wheres the i -system (i =1, 2,..., n) is subjected ony to the force X i.in ech of the stticy determinte uxiiry probems the forces in the brs nd thus the eongtions cn be ccuted. Appying the n comptibiity conditions yieds system of equtions for the n unknown forces X i. The forces in the other brs cn subsequenty be determined through superposition. Exmpe 1.7 A rigid bem (weight negigibe) is suspended from three vertic brs (xi rigidity EA) s shown in ig Determine the forces in the originy stress-free brs if ) the bem is subjected to force (ΔT =0), b) the temperture of br 1 is chnged by ΔT ( =0). E ΔT S 1 S 2 S 3 A /2 /2 b c ig Δ 1 Δ 2 Δ 3 Soution The system is stticy indeterminte to the first degree: there re ony two equiibrium conditions for the three unknown forces S j (ig. 1.17b). ) If the structure is subjected to force the equiibrium conditions re : S 1 + S 2 + S 3 =0, A : 2 + S 2 +2S 3 =0. The eongtions of the brs re given by (ΔT =0) Δ 1 = S 1 EA, Δ 2 = S 2 EA, Δ 3 = S 3 EA. () (b)

34 38 1 Tension nd Compression in Brs We sketch dispcement digrm (ig. 1.17c) nd find the comptibiity condition Δ 2 = Δ 1 +Δ 3. (c) 2 Now we hve six equtions for the three forces S j nd the three eongtions Δ j. Soving for the forces yieds S 1 = 7 12, S 2 = 1 3, S 3 = b) If br 1 is heted ( = 0), the equiibrium conditions re : S 1 + S 2 + S 3 =0, A : S2 +2S 3 =0, ( ) nd the eongtions re given by Δ 1 = S 1 EA + α T ΔT, Δ 2 = S 2 EA, Δ 3 = S 3 EA. (b ) The comptibiity condition (c) is sti vid. Soving ( ), (b )nd (c) yieds S 1 = S 3 = 1 6 EAα T ΔT, S 2 = 1 3 EAα T ΔT. E1.8 Exmpe 1.8 To ssembe the truss in ig. 1.18, the free end of br 3 (ength δ, δ ) hs to be connected with pin C. ) Determine the necessry force cting t pin C (ig. 1.18b). b) Ccute the forces in the brs fter the truss hs been ssembed nd force hs been removed. δ δ v Δ C C b c ig. 1.18

35 1.6 Stticy Indeterminte Systems of Brs 39 Soution ) The force cuses dispcement of pin C. The horizont component v of this dispcement hs to be equ to δ to ow ssemby. The required force foows with α =45 from (1.21): v = 1+ 2/4 = δ = EA 2/4 EAδ (2 2+1). b) The force is removed fter the truss hs been ssembed. Then pin C undergoes nother dispcement. Since now force S 3 in br 3 is generted, pin C does not return to its origin position: it is dispced to position C (ig. 1.18c). The distnce between points C nd C is given by v = S /4. EA 2/4 The comptibiity condition v +Δ 3 = δ cn be tken from ig. 1.18c. With the eongtion Δ 3 = S 3( δ) EA of br 3 we rech S 3 EA S /4 + S 3 EA 2/4 EA = δ S 3 = EAδ 2( 2+1). The other two forces foow from the equiibrium condition t pin C: S 1 = 2 S 3, S 2 = S 3.

36 40 1 Tension nd Compression in Brs Suppementry Exmpes Detied soutions to the foowing exmpes re given in (A) D. Gross et. ormen und Aufgben zur Technischen Mechnik 2, Springer, Berin 2010, or (B) W. Huger et. Aufgben zur Technischen Mechnik 1-3, Springer, Berin E1.9 Exmpe 1.9 A sender br (density ρ, moduus of esticity E) is suspended from its upper end s shown in ig It hs rectngur cross section with constnt depth nd inery vrying width. The cross section t the upper end is A 0. Determine the stress σ(x) due to the force nd the weight of the br. Ccute the minimum stress σ min nd its oction x A 0 ρ ig Resuts: see (A) σ(x) = + ρg A 0 2 (x2 2 ),σ min = ρgx,x 2 = A 0 x ρga 2. 0 E1.10 Exmpe 1.10 Determine the eongtion Δ of the tpered circur shft (moduus of esticity E) shown in ig if it is subjected to tensie force. d D Resut: see (A) Δ = 4 πedd. ig. 1.20

37 1.7 Suppementry Exmpes Exmpe 1.11 A sender br (weight W 0, moduus of esticity E, coefficient of therm expnsion α T ) is E x suspended from its upper end. It W 0 just touches the ground s shown in α T ig without generting contct force A ig Ccute the stress σ(x) if the temperture of the br is uniformy incresed by ΔT. Determine ΔT so tht there is compression in the whoe br. E1.11 Resuts: see (A) σ(x) = W 0 A ( 1 x ) Eα T ΔT, ΔT > W 0. EAα T Exmpe 1.12 The br (cross section re A) shown in ig is composed of stee nd uminium. It is pced stress-free between two rigid ws. Given: E st /E = 3,α st /α =1/2. A ig C stee uminium B E1.12 ) Ccute the support rections if the br is subjected to force t point C. b) Ccute the norm force in the whoe br if it is subjected ony to chnge of temperture ΔT ( =0). Resuts: see (A) ) N A = 3( ) 3 2, N B = b) N = E stα st AΔT. 3 2,

38 42 1 Tension nd Compression in Brs E1.13 Exmpe 1.13 The coumn in ig consists of reinforced concrete. It is subjected to tensie force. Given: E st /E c =6,A st /A c =1/9. cut cross section Determine the stresses in the stee nd in the concrete nd the eongtion Δ of the coumn if E st,a st E c,a c ig ) the bonding between stee nd concrete is perfect, b) the bonding is dmged so tht ony the stee crries the od. Resuts: see (A) ) σ st =4 A, σ c = 2 3 A, Δ = 2, 5 EA st b) σ st =10, Δ =. A EA st E1.14 Exmpe 1.14 A sender br (density ρ, moduus of esticity E, ength ) is suspended from its upper end s shown in ig It hs rectngur cross section with constnt depth. The width b vries inery from 2b 0 t he fixed end to b 0 t the free end. Determine the stresses σ(x) nd σ() nd the eongtion Δ of the br due to its own weight. Resuts: see (B) σ(x) = 1 2 ρg (2 + x)x + x, σ() =3ρg 4 2b b 0 cut x cross section b(x) ig ρg2, Δ = (3 2n2). 4E

39 1.7 Suppementry Exmpes 43 Exmpe 1.15 A rigid chir D (weight negigibe) is supported by three brs (xi rigidity EA) sshownin ig It is subjected to force t point B. ) Ccute the forces S i in the brs nd the eongtions Δ i of the brs. b) Determine the dispcement B C of point C. ig Resuts: see (A) ) S 1 =, S 2 =0, S 3 = 2, Δ 1 = EA, Δ 2 =0, Δ 3 = 2 EA, b) u C =0, v C =2 2 EA E1.15 Exmpe 1.16 Two brs (xi rigidity EA) re pin-connected nd supported t C (ig. 1.26). ) Ccute the support rection t C due to the force. b) Determine the dispcement of the support. C α ig E1.16 Resuts: see (A) )C = sin α cos2 α 1+cos 3 α, b) v C = 1 1+cos 3 α EA.

40 44 1 Tension nd Compression in Brs E1.17 Exmpe 1.17 Consider thin circur ring (moduus of esticity E, coefficient of therm expnsion α T,internrdiusr δ, δ r) with rectngur cross section (width b, thickness t r). The ring is heted in order to increse its rdius which mkes it possibe to pce it over rigid whee with rdius r. Determine the necessry chnge of temperture ΔT.Ccute the norm stress σ in the ring nd the pressure p onto the whee fter the temperture hs regined its origin vue. Resuts: see (B) ΔT = δ α T r, σ = E δ r, p = σ t r. E1.18 Exmpe 1.18 The two rods (xi rigidity EA) shownin ig re pin-connected t K. The system is subjected to vertic force. Ccute the dispcement of pin K. K 30 2 Resuts: see (B) u = 3 EA, v = EA. ig E1.19 Exmpe 1.19 The structure shown in ig consists of rigid bem BC nd two estic brs (xi rigidity EA). It is subjected to force. Ccute the dispcement of pin C. 1 B 2 30 C Resuts: see (B) u =0, v =3 3 EA. ig. 1.28

41 1.7 Suppementry Exmpes 45 Exmpe 1.20 ig shows winch K freight eevtor. The cbe 1 (ength, xi rigidity (EA) 1 ) of the winch psses over smooth pin K. Acrte(weight H W ) is suspended t the end of the cbe (see Exmpe 2.13 in W Voume 1). The xi rigidity (EA) 2 of the two brs 1 nd ig is given. Determine the dispcements of pin K ndoftheendofthe cbe (point H) due to the weight of the crte. Resuts: see (B) u =6.69 W, (EA) 2 v =3.86 W (EA) 2, f =2.83 W + W. (EA) 2 (EA) 1 E1.20 Exmpe 1.21 To ssembe the truss (xi rigidity EA of the three brs) in ig the end point P of br 2 hs to be connected with pin K. Assume δ h. Determine the forces in the brs fter the truss hs been ssembed. ig α α P K δ h E1.21 Resuts: see (B) S 1 = S 3 = EAδ cos2 α h (1 + 2 cos 3 α), S 2 = 2EAδ cos3 α h (1 + 2 cos 3 α).

42 46 1 Tension nd Compression in Brs Summry Norm stress in section perpendicur to the xis of br: σ = N/A, N norm force, A cross-section re. Strin: ε =du/dx, ε 1, u dispcement of cross section. Speci cse of uniform strin: ε =Δ/. Hooke s w: σ = Eε, E moduus of esticity. Eongtion: Δ = 0 ( ) N EA + α T ΔT dx, EA xi rigidity, α T coefficient of therm expnsion, ΔT chnge of temperture. Speci cses: N =, ΔT =0, EA =const Δ = EA, N =0, ΔT =const Δ = α T ΔT. Stticy determinte system of brs: norm forces, stresses, strins, eongtions nd dispcements cn be ccuted consecutivey from the equiibrium conditions, Hooke s w nd kinemtic equtions. A chnge of the temperture does not cuse stresses. Stticy indeterminte system: the equtions (equiibrium conditions, kinemtic equtions nd Hooke s w) re couped equtions. A chnge of the temperture in gener cuses therm stresses.

43