Question

Q.2B Determine the reactions and draw the shear and bending moment diagrams for the beam 60KN 3 kN/m B 7 m 5 m El = constant 1 A BIELE c?

Answer 1

GOKN BAN/ А B * 7m 57 Degree of Static indeterminacy of the given beam Ds = Re -3 Re Total external reactions 1+3 = 4 so Ds = 4-3 Re=1 for Roller Support at A - => Ds Re=3 for fixed Support at c

so the We since Ds=1 given beam is indeterminate Structure, so to find out Reaction forces will apply compatibility Method ( force Method) and C Let Suppose vertical Reaction forcer at Support A are RA at Support and Rc Respectively and Moment с is Me Horizontal Reaction forces at support A and c will be because there is no horizontal loading Zero On the beam.

60KN 3 kN/m А. c B * 7m 5 m RA Ro Now Apply force equilibrium in Vertical direction EF, = 0 RA - 3 x 12 60 + Rc = 0 个 RA + RC = 3 6 -ci sign Convention Lepward force = +ve Downward force =-ve

So Moment about Since Support roller Support A is point A از زیبا be Kero. MA=0 → (3<12 x 12 ) + + 60x7 +Mc - RC XI2 = 0 636 + Mc ReX 12 (1) Sign Convention clockwise Moment = tve Anticlockwise Moment = tve' Now We will apbly compatibility Method. Since Support His a roller Support So Vertical deflection at Support I will be Zero,

uniformly distributed load of 3kN/m2 and point load 50 KN will produce down warel deflection Point A Reaction force RA produce upward deflection at A since net deflection at Support A=0 so Downward deflection due to uniformly distributed load and point load CookN) (3 kN/m) upward deflection due to Reaction RA

down word deflection due to Let Support load is si tool formely dis toebuted and due to is Sa 60KN ymy my * SKNA A C= C Sit62

Dhenwared deflection due to uniformly dis ti buted load (sv) - S, = 3 X 124 8 EI 7776 x W unitfm. S, (4) Downward E I AB = W.84 GET Downward de flection due to point load of Goki (52) 60 KN 5m my LB SB OB 62 Is* S2 Sots*

» Sy 60x53 + OB XAB 3 EI P А G L - ) 60X53 + 60852 x 2 EI 3 EI BB PL² REL AB = ЗЕІ S2 7750 = (1) Downward EL upward deflection due to Reaction fonce RA Let denote upward deflection due to RA is 83 12 m RA

RAX 123 11 3 EI => 576 RA 53 (1) upward EI So Since Zero at A net deflection i To tal Downward deflection upwaroi deflection =) Sit Szi S3 7776 7750 + 576 RA EI E I EI = > 155 26 EI 51 RA EI

RA 26. 955 KN from the eyevention () Rat Roa 96 RC = 96-RA =) => Rc = 96- 26.955 => RC = 69.045 KN

from the equation til 636 + Me - Rex12 = 0 69.045X12 = 0 636+ Met T => (Clockwise) KN-m 1 9 2.540 Me= Reactions RA 26.955 KN Rc 69.045 KN Mc = 192.540 KN-m (Clockiuise)

Shear force Diagram --> GOKN 3 kN/m A 2 3 - M = 192.540 Krom RA = 2¢,955 7 5 m I Rp = 69.045 kn 26.955 KN Linear K 5.955 kn B А -54.045 KN 7 Linear - 69.045 KN Calculation - Part A B take a section X, X, at a distance of x from the left support A Shear force VAB= 26.955-34 O<*$7m shear force at point A (x=0) VA = 26.955- 3x0 26.955

Shear force at VB = 26.955- 347 B (x=7m) VBE -5.955 KN Part take a Section X2 - X2 at a distance of a from point A. Shear force Vor = 26.955 - 3* - 60 Tx = 12m - VBC 33.045 - 3x 7 mart 12 at point B (x=7m) VB = -33.045 - 3x 7 = - 54.0u5 KN at point c Vc -33045- 3x12 - - 69.045 (x-12.4)

Bending Moment diagram → K →x GOKN 3kN/m 0 D Me: 192.540 kNong IX, 5M 7m Rc = 69.045 KN RA= 26.955 kN 1 Parabolic shape 115.185 kN-W Parabolic Shape ! A B 9.018 m - 1-192.540 kNm

Calculation a distouce of tako a Part AB Section X, X, at from the Support A. 26.955 x (3x) – Bending Moment MAB ㅕ MAB = 26.955 x - O <x 37m 3 x2 26.955 X0 MA= بدا م at point A Cx=0) хо2 O KN-m at point B (x= 7 m) MB = 26.955x7- 3x72 115.185 KN."

Part BC take Section X2 X2 at a distance of refroing Support A. Bending Moment MBC 26.955 X - 3 x 2 60 (x-7) 7-2X < 12- =) MBC –33.045 X. – 32 +420 at point с (K=/2 m) Mc = – 3, .045X12 – 3x12 +420 2 192: 540 KN-m point where Moment T020. Moc=0 -33.045 X- 3q2 +420 x = 9.018 m