Question

H a) Calculate the reactions forces at the supports A and D. b) Use Macaulay notation to find equations for the shear force and the bending moment (M) c) Draw the shear force diagram, either by inspection or by using equations from part b). d) Draw the bending moment diagram, either by inspection or by using equations from part b) e) Determine the maximum value of the bending moment (M), and where on the beam it occurs. The length of the beam is L = 4.5m. The force P is given in by P= (60 *n+700) N And the distributed force w is given in /m by w (200 + 3000) N/m. Wheren is the last digit of your student number Enter your values of n, P and win the ansder box 7 P1120 N W 4400 N/m Also, write your values of n P and w at the top of your calculations page. Enter the values you have calculated in parts a and d, for the support reactions and maximum bending moment in the answer box RO Max

Answer 1

. P=1120 W- 4400 solution :- L = 4.5m ; P = (60xm + Feno) N { min A D 43 @esing equilibrium equodions. EMA-TM, co EM =O = (RxL) – P(35) -(websites 2P Fli - + 18 1120) + ( 44002 4.5 RA = 1846 667N reaction at A Ify =o=RA + Rp - P - CL Ro 1120 + 14400 (4400*4,5) - - 1846.667 3 Ro = 5873:33N Isreaction foce at D. CS Scanned with CamScanner

, W 66 0 1) B a Lxx Marxe = Ride ; SV +RA Monte = + RAR P.(x-tg) – 1972-7j23) Moment Masera = + RAx-Plor-23) (x-3) shear force Ven=+Ry -P -052 }) 2 W CS Scanned with CamScanner

(By impection) Sif.D. 1846.66 74 726667 B -413 58 73.33N BM. D. (By inspection) 3920.006j N- 3860NM parabolic 2770 Nm A 0:1651m form max. B.M. (M. mase.) My = fix – Plz-hs) – (4-3)? Rap - Wfx-31) 1846-667 – 1120 – 4400 (R-32) = a Me =O= x = 3.1651 m CS Scanned with CamScanner

Nam Maximum value of B.M, (Mm) = 3920.0062 location is from pont A : x=3.1651 m from point o 1.3349 m left side, CScanned with CamScanner