Question

The simply supported beam AB in Figure 1 is subjected to a load variation given by w(x) = -kr". ܨܝ܂ Figure 1 (a) Determine the equation of the elastic curve in terms of El, x and L. (El is constant) (15 Points) (b) The beam has a length L of 1 m. Determine, in terms of k: (1) The reaction at the roller support. (3 Points) (ii) The bending moment at the section 0.2 m from end A, (that is, at x = 0.2 m). (4 Points)

Answer 1

Detailed solution has been given below. However, this problem has a lot principles applied into it. Though, I have explained almost everything, but still if any confusion or even a tiniest doubt arises, feel free to put that in the comments section, i'll be happy to help you out with it.

Page sal Let's first calculate the total load on the beam' D В. Let's take a element of dr_thickness, at a distance I from beint A. Then, dw: W(x) dx Kx3 da {w(x)= kx3 - Given) do Total load - Sdul 0 Koc da $ (34) K44 4 0 4 Now, let's find reaction Ra, and RB . on So, we know that Sum of moment of all forces beam should be zero , as the beam is in erubhum So, taking moment of all forces about boint A 1) Weight (load) L dM, = (Koc²dx) x = M, M. - Sami = {kx40x 11 I =145 (as) KL 5/5 (iv) Moment due to RAS RAXO: 0 Gill Moment due to RB = M2a - RBXL

Parze NC Som adding al..the moments, EMO KL Rip La=0 5 ܕܩ 5 Ro. - K44 5 Also, due to vertical W = RATRB equilibrium ftranslatonal) ..RA: W-RB = KLY KL4 (4 Ba : KL4 20 calculate Now, let's of beam the bending moment So, B.M beam at a distance A, can be written as: of Oc from beint M(C): +RB20 - Kocs 20 Moc) KL42- Kos S 20 an

Now, we koow EI d'y da M() K24x K5 Eldy dr2 5 20 Integrałng_above equahon on both sides, we get: EI dy da Kl4r? 10 Kx5 HG 120 Aqaio in tarahing equahon _ön both sides, we ad Ely = KL40c3 Κα* 840 1 x + 2 30 Now), also we kool that at subborts, there is no de flection in the beamo ał poin A (2C=0, x=0) At boint B (ocals Y=0 Putting this Dimits_from_eaxuahon _and B in Cavs Tin geti we [2=0 and Go -27 KLO 840

Putting values of Good (2.11 ev lewe get EIY = KL73 30 Kr? 840 27 KLCC 890 ANS Lei Ya 28 KL" 3. K527- 27 KLOK 640 EI bo_he_eruation of elastic_curveis._given_as. ANS Y@c)= K[282493-067-27407) 840 EI PART 2 r 0 Reaction at Roller supporte_RA RA KL4/20 {from ean 1} ANS Putting La Im we get Re. K/20 in the given units is (ii) BM-_ot given section, I m from A BM: KL42/5 - K 205/20 from each iv Putting L-1, 3-0.2, it becomes BM = K (0:2) -0.0399 K 20 0:04K in qiznesnits) ANS KX0.2 S