One Dmensonal al Deformatons In ths secton, a specfc smple geometr s consdered, that of a long and thn straght component loaded n such a wa that t deforms n the aal drecton onl. The -as s taken as the longtudnal as, wth the cross-secton lng n the plane, g. 7... gure 7..: slender straght component; longtudnal as, cross-secton 7.. asc relatons for al Deformatons n statc analss of a structural component nvolves the followng three consderatons: () consttutve response () knematcs (3) equlbrum In ths hapter, t s taken for () that the materal responds as an sotropc lnear elastc sold. It s assumed that the onl sgnfcant stresses and strans occur n the aal drecton, and so the stress-stran relatons 6..8-9 reduce to the one-dmensonal equaton or, droppng the subscrpts, (7..) Knematcs (), the stud of deformaton, was the subject of hapter 4. In the theor developed here, known as aal deformaton, t s assumed that the as of the component remans straght and that cross-sectons that are ntall perpendcular to the as reman perpendcular after deformaton. Ths mples that, although the stran mght var along the as, t remans constant over an cross secton. The aal stran occurrng over an secton s defned b qn. 4.., 0 (7..) 0 Ths s llustrated n g. 7.., whch shows a (shaded) regon undergong a compressve (negatve) stran. ecall that ndvdual partcles/ponts undergo dsplacements whereas regons/lneelements undergo stran. In g. 7.., the partcle orgnall at has undergone a dsplacement u () whereas the partcle orgnall at has undergone a dsplacement u (). rom g. 7.., another wa of epressng the stran n the shaded regon s (see qn. 4..3) 67
Secton 7. ) ) (7..3) 0 0 ) ) gure 7..: aal stran; before deformaton, after deformaton oth dsplacements u () and u () of g. 7.. are postve, snce the partcles dsplace n the postve drecton f the moved to the left, for consstenc, one would sa the underwent negatve dsplacements. urther, postve stresses are as shown n g. 7..3a and negatve stresses are as shown n g. 7..3b. rom qn. 7.., a postve stress mples a postve stran (lengthenng) and a compressve stress mples a negatve stran (contractng) 0 0 0 0 gure 7..3: Stresses arsng n the slender component; postve (tensle) stress, negatve (compressve) stress qulbrum, (3), wll be consdered n the ndvdual eamples below. Note that, n the prevous hapter, problems were solved usng onl the stress-stran law (). Knematcs () and equlbrum (3) were not consdered, the reason beng the problems were so smple, wth unform (homogeneous) stress and stran (as ndeed also n the frst eample whch follows). Whenever more comple problems are encountered, wth non-unform stress and strans, (3) and perhaps () need to be consdered to solve for the stress and stran. 7.. Structures wth Unform Members unform aal member s one wth cross-secton and modulus constant along ts length, and loaded wth aal forces at ts ends onl. 68
Secton 7. ample onsder the bar of ntal length shown n g. 7..4, subjected to equal and opposte end-forces. The free-bod (equlbrum) dagram of a secton of the bar shown n g. 7..4b shows that the nternal force s also everwhere along the bar. The stress s thus everwhere / and the stran s everwhere and, from q. 7.., the bar etends n length b an amount (7..4) (7..5) Note that although the force actng on the left-hand end s negatve (actng n the drecton), the stress there s postve (see g. 7..3). D gure 7..4: unform aal member; subjected to aal forces, free-bod dagram Dsplacements need to be calculated relatve to some datum dsplacement. or eample, suppose that the dsplacement at the centre of the bar s ero, u ( ) 0, g. 7..4. Then, from qn. 7..3, ) ) ( ) D) ) ( D ) ) ) ( ) 4 (7..6) whch s another wa of sang that one can translate the bar left or rght as a rgd bod wthout affectng the stress or stran but t does affect the dsplacements 69
Secton 7. ample onsder the two-element structure shown n g. 7..5. The frst element s bult-n at end, s of length, cross-sectonal area and Young s modulus. The second element s attached at and has propertes,,. ternal loads and P are appled at and as shown. n unknown reacton force acts at. Ths can be determned from the force equlbrum equaton for the structure: P 0 (7..7) s usual, the reacton s frst assumed to act n the postve () drecton. Wth known, () the stress n the frst element can be evaluated usng the free-bod dagram 7..5b, () and usng g. 7..5c: P P () (), (7..8) and so the stran s P () (), (7..9) P Note that the stress and stran are dscontnuous at. P P P (c) P P gure 7..5: two-element structure subjected to aal forces and P, (b,c) free-bod dagrams or each element, the total elongatons are ths result, whch can be vewed as a volaton of equlbrum at, s a result of the one-dmensonal appromaton of what s reall a two-dmensonal problem 70
Secton 7. ) ) ) ) P P (7..0) If P, then 0 as epected, wth 0 and 0. Thus far, the stress and stran (and elongatons) have been obtaned. If one wants to evaluate the dsplacements, then one needs to ensure that the strans n each of the two elements are compatble, that s, that the elements ft together after deformaton just lke the dd before deformaton. In ths eample, the dsplacements at and are u ( ) ) (7..), ) ) compatblt condton, brngng together the separate relatons n 7.., s then ) P P ) (7..) ensurng that u () s unque. s n the prevous eample, the dsplacements can now be calculated f the dsplacement at an one (datum) pont s known. Indeed, t s known that u ( ) 0. ample onsder net the smlar stuaton shown n g. 7..6. Here, both ends of the twoelement structure are bult-n and there s onl one appled force,, at. There are now two reacton forces, at ends and, but there s onl one equlbrum equaton to determne them: 0 (7..3) n structure for whch there are more unknowns than equatons of equlbrum, so that the stresses cannot be determned wthout consderng the deformaton of the structure, s called a statcall ndetermnate structure 3. 3 See the end of.3.3 7
Secton 7. (c) gure 7..6: two-element structure bult-n and both ends; subjected to an aal force, (b,c) free-bod dagrams In terms of the unknown reactons, the strans are () () () (), (7..4) and, for each element, the total elongatons are, (7..5) nall, compatblt of both elements mples that the total elongaton 0. Usng ths relaton wth qn. 7..3-4 then gves, (7..6) The dsplacements can now be evaluated, for eample, ) (7..7) / / so that a postve dsplaces to the rght and a negatve dsplaces to the left. Note the general soluton procedure n ths last eample, known as the basc force method: qulbrum + ompatblt of Stran n terms of unknown orces Solve equatons for unknown orces 7
Secton 7. The Stffness Method The stffness method (also known as the dsplacement method) s a slght modfcaton of the above soluton procedure, where the fnal equatons to be solved nvolve known forces and unknown dsplacements onl: qulbrum n terms of Dsplacement Solve equatons for unknown Dsplacements If one deals n dsplacements, one does not need to ensure compatblt (t wll automatcall be satsfed); compatblt onl needs to be consdered when dealng n strans (as n the prevous eample). ample (The Stffness Method) onsder a seres of three bars of cross-sectonal areas,, 3, Young s modul,, 3 and lengths,, 3, g. 7..7. The frst and thrd bars are bult-n at ponts and D, bars one and two meet at and bars two and three meet at. orces P and P act at and respectvel. The force s constant n each bar, and for each bar there s a relaton between the force, and elongaton,, qn. 7..5: where k k (7..8) Here, k s the effectve stffness of each bar. The elongatons are related to the dsplacements, u u etc., so that, wth u u 0, D u u k u ku, k, 3 3 (7..9) There are two degrees of freedom n ths problem, that s, two nodes are free to move. One therefore needs two equlbrum equatons. One could use an two of P P, P 0, P 0 (7..0) 3 0 3 In the stffness method, one uses the second and thrd of these; the second s the node equaton and the thrd s the node equaton. Substtutng qns. 7..9 nto 7..0 leads to the sstem of two equatons u ku P k k u P k k (7..) k u whch can be solved for the two unknown nodal dsplacements 3 73
Secton 7. D P P P (c) P 3 (d) 3 3 gure 7..7: three bars n seres; subjected to eternal loads, (b,c,d) free-bod dagrams Note that t was not necessar to evaluate the reactons to obtan a soluton. Once the forces have been found, the reactons can be found usng the free-bod dagram of g. 7..7d. The stffness method s a ver sstematc procedure. It can be used to solve for structures wth man elements, wth the two equatons 7.. replaced b a large sstem of equatons whch can be solved numercall usng a computer. 7..3 Structures wth Non-unform Members onsder the structure shown n g. 7..8, an aal bar consstng of two separate components bonded together. The components have Young s modul, and crosssectonal areas,. The bar s subjected to equal and opposte forces as shown, n such a wa that aal deformatons occur, that s, the cross-sectons reman perpendcular to the as throughout the deformaton. Snce there are onl aal deformatons, the stran s constant over a cross-secton. However, the stress s not unform, wth and ; on an cross-secton, the stress s hgher n the stffer component. The resultant force actng on each component s and. Snce, the total elongaton s (7..) 74
Secton 7. gure 7..8: bar consstng of two separate materals bonded together 7..4 esultant orce and Moment onsder the force and moments actng over an cross-secton, g. 7..9. The resultant force s the ntegral of the stress tmes elemental area over the cross secton, qn. 3.., There are two moments; the moment d (7..3) M about the as and M about the as, M d, M d (7..4) Postve moments are defned through the rght hand rule,.e. wth the thumb of the rght hand pontng n the postve drecton, the closng of the fngers ndcates the postve M ; the negatve sgn n qn. 7..4b s due to the fact that a postve stress wth 0 would lead to a negatve moment M. d M M gure 7..9: esultants on a cross-secton; resultant force, resultant moments onsder now the case where the stress s constant over a cross-secton. Snce t s assumed that the stran s constant over the cross-secton, from qn. 7.. ths wll occur when the Young s modulus s constant. In that case, qns. 7..3-4 can be re-wrtten as, M d, M d (7..5) 75
Secton 7. The quanttes d and d are the frst moments of area about, respectvel, the and aes. These are equal to and, where (, ) are the coordnates of the centrod of the secton (see qn. 3..). Takng the as to run through the centrod, 0 results n M M 0. Thus, a resultant aal force whch acts through the centrod of the cross-secton ensures that there s no moment/rotaton of that cross-secton, the man assumpton of ths secton. or the non-unform member of g. 7..8, snce the resultant of a constant stress over an area s a force actng through the centrod of that area, the forces, act through the centrods of the respectve areas,. The precse locaton of the total resultant force can be determned b takng the moments of the forces, about the and aes, and equatng ths to the moment of the force about these aes. 7..5 Problems. onsder the rgd beam supported b two deformable bars shown below. The bars have propertes, and, and have the same Young s modulus. The are separated b a dstance. The beam supports an arbtrar load at poston, as shown. What s f the beam s to reman horontal after deformaton. 76