Question

Analyze the given beam below and draw the shear and moment diagram for the entire span. 0.94 m 1.64 m 3.07 m 2kN/m F D E 1.38 m 13kN B 2.00 m D

Answer 1

onum coblem 3.0-19 H X & Knimt X Efx=0 f HF С. Hf=13 KN E 1:38m RE Efy=0 DB 13 KM RAT Rf = 2xlibre Rot Rf 23.28 KM 12.00 EMfo + A RA (13 X1.38)+ 2X 1.64X (3:07+ 1.6L) - R9X5-6550 Ra = 5.433 N1 Rfa-2.45 KN

* 1994m liftem X 3.0fm Kim Xi C V € 13KN 1.38m 2.เSEN vign convension 13KM B 1 SFD X2 209 X2 20 Vy-x-shear forceat.x-X moment at xx 5.413 KM mx-x - Bending Bmo part Ef (o sxc 3.07 220, x=3.07m Ve=VF = 2.15 kn Vx-X = 2.45 kn x=0 Mf=o mx-x = - 2- 2,15% 2= 3.07m me = ~6-601N KNm part ED (3.045&chifm) VE+2.15 KN 2= 3:07M Vzx==(2,15 + 2, (2-8:07) Voz+5.413 KM 2= 4.7 m ME =- =-6.601 KM-M X = 3:07m Max= - 2.15%- 2 C6-3.07% 2 mon-12.816 kNm X= Hoflm 565 palt DC ChoHS X 66-659M NC= VD 5.43 kn Vx-x2 + (2xl-64) + 2.15 = +5.43 kn, x=41fm, mor-12.816 kaum Mxax = -2.152 - 2x1.66X CX-389) 2= 6.65 5-65 mc:2535_KADO -17.926- m

PAKABE (OSXC 2m) VX-X = O KN mx-x-OKNam part BC (25263.38 m) Nx- xa-BKN 227223.38, Vg=Vc=-13KN mx-x= 13x6-2) 2= 200 MB=Oku-m 2= 3.38 Mca-17.94 KM-m 5.43 KN 2. 15kn 17.94 D KMMC С E 6.601 1° KNam ☺ 12.816 192 KNY KM-m В B. 13 kr BMD SFD &