Question

QUESTION 2 [30 Marks) The framed structure shown in Figure 2 has simple supports (i.e. is free to rotate) at joints A and G. Members AC and GE are vertical, while CE and EH are horizontal. There is an internal moment release at joint C. A uniformly distributed load of 5 kN/m is applied between D and E, and another uniformly distributed load of 4 kN/m is applied between F and G as shown. There is also a horizontal point load of 5kN at B and'a vertical point load of 3 kN at H. a) Determine all the reactions at the supports. b) Draw the (i) bending moment and (ii) shear force diagrams for the frame. 3k 5kNlm ITITITITI 5 kn * 3m 3m thuhuhuht 4xnim - 3m 3m zm Figure 2

Answer 1

C 5Koolm 3m 3m Become 13 kw 3m 3m 2m SKN Skub & B u kom 3m HA A Be He ATRA TRB 2 fy=0; RA TRG = (6+3)+3 a 18kN & fxzo HA + 5 = MB + 4x3) HA-HB = 7 KN (Span ABCY { Me =0 (Span ABC) & Mczo KAX 6 =-543 MAX 6 = -5x3 HA = -2.5 THA=2.5 kN (K HB = - 9.5 JNA = -2.5 kN ( HB = X-70 2.5 = a.5 (4) He = 9.5 (4) He = 9.50(–) { MB=0 RAX6 + 5x3 - (5x3x105) +(3+2) - (4x3x1.5) RA= 3.25 KN RB = 14.75 KN

Span AB (u bom A) - so = HA Ar: Mavr = 2.5xx - 0, SA=2.5 AN H=0, Maro x=3, SB=2.5 kN K= 3, MB= 705 Span BC (x from A) Sx= HA-5 My = HAXK - 5 (x-3) = 2.5-5 = 2.5k-5 (x-3) x=3, SB = -2.5 KN n=3 x= 6, SB=-2.5 KN MB=705 k=6, Mc= 0 Span CD (from • Sx = RA My = RAXX . = 3.25 = 3.25 x • no, Sc= 3.25 kn K=o, Mc=0 - x=3, So=3.25 KN K=3, Mo= 9.75 - Span DE (x from e) Sx= RA-5(x-3) Mx = RAXK - 5 (x-3) (x 3) = 3.25-5(x-3) = 3.25k - 5 (x-3) (4-3) - x=3 So= 3.25 kn x=3, Mo= 9.75 KN-m K=6 SE= - 11.75 kN x=6, ME= - 3 kN-m S 6 Atx 5.25 = 5(x x = 3.65 d=3.65, Mx = 10.806 kN-m - Span En (k from H) 3 Sx=3 My = - 3xK no, SH = 3 kN R=0, Mu=0 3 n=2, SE= BIN U=2, E =-6 kN-m Span GF (x from G) Sp= - Hah+ 4x ц, + Нахк Фцк Puzo, SG=-9.5 x=0, MG=0 2 x=3, SE-2.5 x=3 = MF=+10.5 ku-m o Sfo at value of x K=2.375 Mx + 11-2B1k Non 0 4x=9.5 K= 2.375m (from G)

Span FE (u from br) Sx= – MG + 4x3 - 9.5+ 12 = 2.5kN = 3, SF = 2.5 4-6, SE= 215 Me= + HG x x - 4x3 (4-105) as x=3, Mr +10.5 KN-m 1=6, ME + + 3kN-m 3 KN 5 kN/m 5XN- MN 3 4 kN/m 2.5kN e la GB 9.5kN 3.25kN 14.75 KN

11:757 S.F.D 9.5

10.5 (+) 9) W, 3 В. М D