Question

3. Consider the beam-bending problem shown below. Using one element and assuming that the beam length is , modulus of elasticity E, area A, and moment of inertial: a) Solve for unknown displacements. b) Find the displacement at x = 1/2 4. For problem 3 above assume that these is no truss axial effect or D.O.F Compute components of the element stiffness matrix using shape functions. K22 k23 k24 k33 k34 K44

finaite element

Answer 1

i tried my best to solve above problem as per the knowledge i have.

solution is as follows along with some basics:

basic for deformation of beam with different supports:

L'A), of becim 840 On-O Oc measured in the direct" of 2 rofath of beam (O< 90 salway Yo 디 96-Y max Pflastic Bi lor Defin (urvez final In cantilever beam, at fixed end ,o=y=0 at free end, = max & y=max parabolic Shape bending (: not pure posit of so not beam circular arc 1/ W 2/2 yo DA? B JOC = Omax & -))- emat =Ynox YATO e=0 >(parabola ) In SSB, wherever Imax → Slope= 8 =0(Zero Omax y=0 + support but in cantilever, at free end Imax& Omax, Imp to find Ymax & Omax under symmetrically loaded 55 B at mid span Yzmax, o tat supports to- max, y=0 slope at supports are equal abovej e unlike in nature

Double Integration method, ܢܙ d²Yx-x 2 --11) Rx-x d&2 a²yx Mx-x= CEINA), Px-x dx² :-(2) x-x 2 to difft Governingreqn in terms of m, y eſdy ax Slopes tane- dy where 0-2 dy where or angular dy da da deflect Forsmaller dern, o

finding support reactions:

Let, take 9(w/m) as w(Nm) for simplified notification Sque / MA B RB RA equivalent to - Ill (Mm) 9,- WA (only wan) EI a deflection of Curve Y = Rp 23 361 Rot

EX=0 RA+RB=WL -(1) & EMA=0= MA-WL(22) + R6 (L) = MA- WL2 2 -RBL -(2) 3 unknown, 2 equ so, wing compatible egn Yg=0 a (yo) due to wl() +(42) due to Re(t) = 0 * Yos-W24 RB L + 3 EI SEI O a RA=SWL (1) & @amar Os Rg = 3602 (1) (G) 8 LOL? 8

solution:

L 1/ re WL beam = 3WL 8 Curve WL? WL2 x+ WLX = & WLX --(2) 2 2 و مدل FO dre 16 dyx-* *0.-(+) (a) Now take a section x-x' at a distance of "x" from the hinge supporst line from left side) as shown below: Bto A (x=0 to L) Now, moment about ze-x from left end, Mx-* = (304) + WL(**) -(1) -deflection Ro 50, as per governing eqn at x-x, (ET, /BUL dx²) -15WL) x² WL? EI X+6, ar 8 :-(3) Now, at XL = Ox-x= OA 5w3. © C, = - WL? 16 2 50, • (EI) dyxx (C=-3 WL² -5L x+62336-323 da Now, at x=0 = Oxnx = 0 dx (slope) (because it is By integrating egr hinge support) (EI) (9xx] = FSL x 2 + 1819x?- 31°x] +62 -(5) 16 NOW, at x = 0 - Yx-x = YR=0 af Cz = 0 & at x = L = Yx-x= YA = 0 Cq= [- 54*+.42* 34] 2 W L4 & C₂ = 48 eq" 6 - (EI) [yx-x]= [-51x2 + 41?x? 3x] WL4 + 24 ice. WL4 Yx-x = MS ET [- Slx +41°x?31?x]+ 24 EI .. (6) Lo required egn for unknown displacement = B W F5L x3 WLA 24 16 3

(b) Now, at x = x= 1 / 2 3 4 x 4 = ? put x= 4² !! from equation 1. we get, Yemet [-52(4):41+4)=3()] WLA + 24 EI Y -5_4: 3 = @% + L -47 + WLI 16 EI 24 2 - 24E1 WL 24-EI WLA ( 174"] + w1" 16 EI Yey WLA 384-EI yern mot* [i6924+ 2] required displacement at x=42 (sign only indicate direction of displacement (as uppeor moment is taken as positive, so displacem- ent downward will get as negative sion) (downward) 4 So, WL yo @% - --- 38 4 EI
if i have done any mistake, i believe that the procedure given above will help you.

THANK YOU.