Question

(TYPE B) SOLVING PROBLEMS: 5) Derive the equation and draw the diagram of bending moment for the following beam with given shear-force diagram (3 marks). 3 kipt

Answer 1

Solution Type B)

3kiplet te 887 a 4ft t. ka From the sheay force diagram we have RA = 16 kip ? Reaction forces Re=8 kip J a = +16 kip VAB = 16-3x sheay forces on beam VBC =-8kip Vc = -8kip

Shear force is o at port which is of xo distance from A VAR = 16-3% => 0 = 16-3360 Xo = 16 ft Noco bending moment calculation {moving from A to co } MA = 0 MAB = Rmx – 3XXXX 0<x38} - Mag = Rax 3x? - IMAR = 16x -1.5x² Ø { osas8} at x = 0 MAB = MA = 16X0-1.5*(0)?=0 at n =* = 16 = 5.33 Mas = Mo = 16 x – 105*(19)* = 42.667 Kip-ft at x=8 MAB = Mg = 16x8 - 105 x(8) = 32 kip-ft Now MBC = . RAX - 3x8x14+4-8) Mec = Rpx -24(6-4) - 166-24(x-4) * Moe = 16x-240496 → Moc =-+36 $ 83x {12} ... at x=8 MBC = M3 = -8x8 +96 = 32 at x = 12 MBC = Mc =-8*12+96 = 0

% Equation of bending moment is for beam MAB = 16x -1.50? → OS X58 Moc = -88 +968 S x 312 Draising bending moment diagram

Bending (kip-ft) Moment 42.667 ----- C€ ---- - - - - - - - - - - - - - - - - 1 5.3333 12 X (ft)