Question

8.(a) (a) Show that the horizontal thrust in the following two hinged arch shown is zero. CM Parabolic y L 2 Α, B L (b) Find H if the couple is acting at x = L 3 from A. [10 + 10 = 20 Marks] 8.(b) An inclined frame ACB carries a uniformly distributed load as shown in figure. Construct the thrust, shear force and bending moment diagrams.

please attempt both the parts otherwise you will be just wasting my question and I would be forced to give u thumbs down

Answer 1

Solution: Part-o(a) كه Yc L/ с را to A L Hoje B tfor ac & с ABM RA=M u L first two hinged arch is converted into simply Supported beam (SSB). vertical to for BC het Reaction at A AB be Raske EMA = 0 (0) M-REO Rg=M RB = M sv M {fy {fy=0 (4+) Now for Hosizontal thrust RA TRB=0 is given by Standard formula RA= - Rg=-M 4= 5 Mas y dz É Ic here y=4 h (L-X) $y2dx L 4-%dex4*, L² orra=M (1) + L for 2=42 E IC h=yc = { y = yet to also seen from figure

Moment table : CB Ас A IB Member Origin Limit Μας cóment in ScB) o to 4/2 o to 4/2 +Mx -M LX نار Nuinuator (viz 니다 11 from Heqh S. Mae y dx S-MX x Y e de + EIC E IC 닐 +Mx Ye da EIC +M Y c X L² x= x2 ELL o x dx = -M Yc x L² EIL CX L2 8 = 0 heuce H=0 for " Numulator is zelo the Condition

(! Now Moment acting at x = at x = 1/₂ from A. bna M y Yo CL Birl tox } first we draw SSB راز M с 쇠 RÄ (M/L R ㄴ EMA = 0 (at) 14/2 ti 3 M/L M-RB.L=0 RB = M 4 EFG =0( 447) RA TRB=0 so RA= M. Hozigontal thrust is given by L formula H = Mssy dx 14 At E IC S 4² dx E IC h = Yc 2 for farabola, y = 4hx (L-X) hue x = 1/3 Y = 44cXL/₃ X 21 / = eye Į L2

X (۲) Moment table for ssb Member DB 3 A Since origin Limit origin oto 4/3 -M X o to 26/3 from B Mss +M L O Nuvitator 2 L/3 M S. Mss y dx MX dx + E IC L १ EIC -Mx & Yc J X. L2 IBEI from: Heqh 243 S-M Xx8 Yc M xx840 L EIC + M2 x&ite * 24? I g 18E IC - 8 Mxzxyd = 8MLY 162 KEIČ Denominator from H equ 18 64°C XL Sy²dx 99 EIBI ELC 11 T62 ETC = ETC زر hence S Mery dx EIC H 8L M 8 ML YC 162 EIT 64 y CXL = Syadx 1296 y Huy EIC