Question

QUESTION 1 [15] For the simply supported beam subjected to the loading shown in the figure, a) Derive equations for the shear force V and the bending moment M for any location in the beam. (Place the origin at point A.) b) Report the maximum positive bending moment, the maximum negative bending moment, and their respective locations. 36 KN 180 KN-m X B C D 4 m 5 m 3 m Figure 1

Answer 1

For this problem first we'll apply equation of equilibrium , so that magnitude of reaction forces can be obtained.

After this we'll make Shear Force Diagram (S.F.D.) and Bending Moment Diagram (B.M.D.) so that values of shear force and Bending Moment can be obtained at every point.

Calculations for the same is given in the following images ::

4 m sm de 3 36kN B AY 180 KN.M RA Rc By Equilibrium condition :- {fy so =) Rathe= 36 - 360 RATRE EMA=0 =) +180 + Re (9) IN 36118 18) Ref 28 KN lal grc = 36x12 -180 from o&o ir 8 KN RA = 36-hes 36 -ha E 36-285 4m sm 3m 36 KN * Shear force in VA-s = A & KN E Verne = 180 KNN 8 KN 28kN - Vero 28 Bending Moment! - 36 35 ran Si FiB), 180 konian 8 km A B Go 2SEN 8 kn AtoB!- o cre4 +82 KNM knih (sagging Bloc! (4<x<9 I 1 IT Manx = (+8% -180) Knim BLiNM B c Ctodi (ore <12) B.M.D. e in Mysore = 82-180+28(x-9) T 29 KNM -108 KNim - 148 Krim

Side of Positive Moment Maximum Bending +32 KNim [Jus Just Left Point B 1 Maximum Negative -148 KN.m Just Right Side Bending Moment of Point 'B'
Hope you got the concept well...!!!