Question

This is a practice assignment, so that you can be comfortable scanning your calculation sheets and uploading to MyUni without the pressure of losing marks. Please make sure when submitting your assignment that: • You have scanned the correct side of the page, or both sides of the page if double sided. • The paper is in portrait orientation. • You have uploaded the correct assignment. Draw and annotate the shear force and bending moment diagrams for the beam shown in figure 1. Include your calculation sheets required to determine the shear force and bending moment values at critical points along the beam. 5 kN 12 kN 3 kN/m * im X * 3 m 2 m 3 m 1 m AB с D E Figure 1: Assignment 0 Submit your solution via MyUni by 3pm on Friday 31* of July.

Answer 1

RAX6-5x5-12x2 + 3x3 81.5 = A B с. AD € im 3m 2m 0 1 NDRight : 12 KN Ans :- 5KNM 3kN/m 3m EVEO, RA+RD = 12+5+9 = 26 KN taking moment about 'D' IRA=5.91 KN C S RO 20.08 KN calculation of shear Force span AB eo cse<1) T Na=RA RA 250, VA=5.g) kn spanBC (122 <4) Va=RA-5 span CD (452 <6m) Vx = RA-5-12, VC=-11.09 KN at se = 4 mi span De (6x<9) Na= RA-5-12 + Roccel X (x-6) :: X=6m, Noleft VDRight = 8.999 KN B x X Shay A Ap B NB = 0.91 kr x t B ine SKN 12 km - t A 3 kN/m € B 2 =-11-09 KN

Mx= RAXX-5X (2) - 12 x (x-4) Bending moment calculation span AB Myê RAXX x=0 m, MA=0 MA = O KN-m x=1m, Mo= 5.91 km Span Be anxa RAXX – 5X(X1) at x= 4 mi mo = 8.64 KN-M Mx= RAXX-5X (X-D) -12X (x-4) at x=6m, Mo= 5.91X6-SXS =1242 MD i -13.54 KN- span co span DE + RDX (X-6) at xa om, MD=-13.5 KH-m 2= gm, ME=0KN-m w X (X-672 5KM 12KN hl=3 3 Kot/m K B C TO e 3m 2m 3m 5.91 KN 5.9|KN 9 KN 0.9 KN 09IKN t A B S.FD e 11.09 KN 11.09 KN (8.64 KM-M 5.91kom + av D B.M.D tv 4.779m 13.5kNm

SKN 12KN t l RA x Distance of zero moment. Mx=0 =) RA XX-5(2-1) – 12X (X-4)=0 =) 5.91X - 5x+5-1222 +48=0 x=4.779 m Х