Question

3. (a) For the following stress function to be a valid Airy stress function, what should be the relationship between B and D? $ = Bx?y* – Dy6 (8 marks) (6) A thin plate of thickness t fixed along one edge is loaded by a shear force P on the opposite edge as shown in Figure Q3. The assumed Airy function is: 0 = A(a - x)(2 – 3bºy) (i) Show that this is a valid Airy stress function. (6 marks) Page 2 of 4 326MAE (ii) Find the stresses at the boundaries. (6 marks) P 2b a Figure Q3 MacBook Air 000 000 F4 " F5 F6 F7 F8

Answer 1

Given Data : Airy Stress Function

To find : Validity of given Airy Stress Function and relationship between B & D.

To find : Stress at the boundaries.

Concept Used : An Airy Stress Function satisfies the equation given by Equation 01.

Given ф = Bх” у рує To find Relation between B & D. Concept used: Fox Valid Airy Stress function we have → equation 1 әхәу” руч от др 2014 д) - 2871 а.о Әх. ә“ , 2 “Ф + әЧg = 0 dx4 а. Әх? 0 Әx3 Оф Э Ә ахға, әз (2 вуч) = 3 вуз де 22²2y² (зия") = 2484- and др. 12 Вх у Золу1 48xy - sys д әЯ ay? ду = 240x® - 120ру» Ә93 ә“ф 94Bх?- 3 600 ау Now Foom equation ® о+ 2* (24ву?) + чв.- 36оду = 0 49 рух 3 Чох* = 36opy* ЧB (2y=12) = 360 D8% в (29+x?) 360 Ду- 29 2 B 0 158 Relation Between B&D. *+ 28-

6 Given: 0= Ala-2) (423624) Ø= A ( ay3–3aby - zy? + 36ey) Bø - A(-y+3634) ²6 Now Axo =0 36 240- ах. de2 3x3 2x4 2 328 - A(-391367) -O a40 do əxay 22x²29 axay and 29 = Al 3ay?- 3a bP3zy?+ 3bx) ay - Al Gay-0-688 +o) aya 040 dy3 aya Now For Valid Aidy stress function 0 muss satisfy the 040 240 ax4 22 dy ayu Putting values => 0 +0+0 =0 Hence given & is a valid Airy stress function. -A (62-61) equation + 2040 + SO = -30 i Now we know At any point Geox - La Guya at and exy = -do ду- дхду We know Oxuc = Albay-6367) GAY (a-x) Pyy=0 escy= -A (-34²7 367) 3A( 9263) Now Given at a=a Shear force =P & and thickness=2 Then shear stress excy = Р Р Area 요 Ady. tay

2b >c so at x=a & y=-b to + b exy=p tdy 3A (y? 62) = p tdy Р b 3A Sly=6%) dy t -b =) 3A p [ b -b → 3A 3 Р + 3 t p. t → ЗА Р t BA ( that by the - (6)(b+b)] 2A-20) 3A1 -45°) -P (A 48t Now Calaelation of stresses At Boundaries x=0 y=-b to tb Goc= 6 Ay (a-x) = 6 ay (a-o) = 6Aay 50c= force In Area AB FAB Area of AB t dy tdy FAB 6A at Sydy FAB 6 Aat (42 6Art ( 16 as FAB=0 с * Boundary AB A face AB 6 Aay b -b 2 20 80 Soc = FAB =0 Arla AB Gax=0 1990

Ency = 3A (4:52) - FAB Ady b + FAB = 3A + f(y=6)dy -b y3_ - 3 A+ - b = 3A+ (( 49 50 ) -1°Cb+b)] => 3 At [ 25?.200] * 394-4 (5) FAB » e per ZAB=P so ZAB= P Area of AB (CAB 25+ For Boundary AB Oxx=0 ogy=o Z=P abt For Boundary BC ² x= o toa & y= b ryy=0 Brac = 6Ay (9-x) = 6 Ab (a->) FBC = FBC Area BC it da

a JoAb (a-x)dąc = Fet t 154b"***** » 6Ablača?] = fec t » 3 Aba² FBC t FB C = 3 Abat and Now Suc=FBC Area BC FBC at Sre = 3 Aba't at Oxx = 3 Aba 5xx= -3pba 4bit -3pa Gan= 4b²t Exy = 3Aly²6²) = 3 A(6²6) Excy= 089 =0

FAD t dx a 0 For Boundary AD. x=0 to a y=-b Tyy=0 Græc = 6 Ay [~-~) => -6Abla-x) = FAD Area of AD FADE -6abt ja-x) dhe - 6 Abt (ax-x) - GADU (a? a?] FAD = - 3 Abta² *3* ( ) bota? FAD 3pa² 462 and Oxa = FAD Area of AD Scx= 3 Pa² 3pa 4b²at 46²7 Say= 3 Aly?b² = 3A (163) FAD at Bay 0 egyzo CD X= a For y = -b to b Boundary Ency = Force on CD Р Arca of co abt Oser = 6 Ay Carm) – 6 Ay Caca)=0 ogyo P so for Boundary CD 2bt 5xxx=0 exy =0