Question

Problems S.13 (a) Draw the free-body diagram of the beam. (b) Determine the reactions at the supports. A 6 m 40 kN В 8 m 12 m Problem 5.13

Answer 1

40 KN sm m - h 12m - (a) Free body dioqram for AB beam. *40KN - H 7B tom - lam

Ase at A, that support is volle Support that is cohy only one reaction is acting at A Support : At B, that support is hinge Support that's why two reactions are acting at B supports (b) Finding Reactions Equilibrium Equations The Beam in Equilibriurn. So . are three (1) Sum of Horizantal forces one zero (2) Som of vertical Porces are zero : (3) Moment about any point is zero. (1) Efx=0 Az-Be =0>) Ax=Bx. - 0 (1) Eky =. 40+Bv = Boss By = -40KN - 0 B) { MB 50 (about B) AqX6-408(12-)=0 Ax = 40x4 = 26.666 KN - substitue the value of Ae in eq 0 => Ax= B2 [Ax = 26,66 KN

-40 e the direction By = -40 KN ( we are getting Negative value. So our assuming direction is wrong so change So, By is downwoond direction Az = 26:66kN], @x = 26.6.6kN) By = 40 kn) A 440 KN 3. Bu By