Question

2- For the cantilever beam and loading shown, determine the following: (a) The equation of the Moment at A; (b) The equation of the Moment at B. P - w

Answer 1

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This question belongs to basics of Strength of materials generally taught in third semester in civil engineering.

Now,Let us focus on solution part,

ya P=2 wa - Let us first find →x-anis axls mentioned in the question). the support reaction. Pawak (let us assume for convenience, all forces in kN & distances in meter) WKN/m I w kN/mx 2 m. RA (Reaction at A). Let us write the equation of equbm in y arn. (assuming upward as tref downwand on » RA + w x 29-p = o( Equbm in y-direction). -ve). RA = P - QwA. - RA= zwa- 2wa = - 4 wa (Hence, Reaction of A is - opposite to our assumption - facts downward (since negative infront of magnitute) & han a magnitute of twa). Now, Let us consider rotational equbm. af A. which means - net moment at Ate EMA = 0 (Rotational Equbm). RA=4wa Consider counterclockwise 2 wax a moment on positive) Rwaxa. EMA = D = -3.wa* 3a + Zwaxa + MA=0:-D solution Epation can be attributed as estute of moment at A... external

Now,

MA=0 kNm. (from egun I). Clie no moment is generated due to arrangement of external loading of their application). Now, let us try to write the egun of moment at B.. considering only the right portion of IBD & then writing the equbm egun at B. Sign convention. Consider counterclockwise moment as po negative, (since it is causing / producing higging in the beam) PXA - MB = 0 (internal moment equam at B) Arower MB = Pa- Requrned Equation of part (2) Now, using same technique for moment at A & similar Convention also. - Internal EMA = 0 6. Moment Equbm Egh). PX34 - Wx29x9 - MA =o. Required equation for Frist part of question. MA = wa 3a - 2wa2 |

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