Question

Untuk rasuk kayu dan muatan yang ditunjukkan dalam Rajah S2, gunakan fungsi singulariti untuk menentukan cerun di hujung A dan pesongan pada titik tengah C. Diberi ketegaran lenturan, El = 168.75 kN.m. For the timber beam and loading shown in Figure Q2, use singularity functions to determine the slope at end A and the deflection at the midpoint C. Given that the flexural rigidity, El = 168.75 kN.m. P= 4 KN w = 5 kN/m 50 mm B C A D 150 mm -1 m 0.5 m 0.5 m Rajah S2 / Figure Q2

Answer 1

GKN w=5kN/mt 11 TRB Imt 0.5 m 0.5 solving for loed distributed total The point load 1 the at reaction forces: uniformly assumed center of gravity of which is acting effective point load to replace distributed load = Sko/mexlnt acting span 26 on Total Uniformly 5KN. and center of gravity is at center. 4KN SKN. 1 RB RARE o.s. 0.5 moments through points A. Taking 4805 + 5x1.5 - RBX2 12 + 7.5 = RRX2. RB= 9.56 = 4.75kN

downward forces. Equating upward and RATRB = 445. RAT 4.75- 4+5 RA = 4.25k N. 4 KN 5 kN/mt. RB = 4.75. RA-4.25 osm m 0.5m that know day dx2 Bending moment ET EI d²y dx² Rana & (x-015) – 5(x-1)/2 = 0 kQ-13 = 4.25 x 5 4 (1-05) 5 .(x-1)2 d². = 4.250 4(2-015) – 5 EI dy dx = EI 4.252 - 4x+2 - 2.5 (x-2x+1)] [4.252 - 4x +2 25 - 4x +2 - 2.5x² + 5x - 2.5 6 (0.25*-0.5 - 3-s*****] & [ 5.252 -0.5-2.5 x²

dy [5.25 -0.5-2.5 x2 dx2 EI integrating 2 -0.5x - 2.5 +G 5.25 2 dy dx EL >0) constant. ci is integral integrating Again 74 +Cxfa 9 - 1 [ 5.258 -0.52 - 2.5 72). constant C2 sis integral the To find C, and a boundary conditions at x= oy=o x=2 ; y=0 Substituting in at *=0 =0 equation (2). [5.25 x 02 04 2.3 +0th 0-0.5 6 ЕТ (=o.

Substituting x=2 y=0 in ein 2 24 -2.5 ter 48+o És Cs 5.25 12 2 6 0 - 5.253 ) - 05@)- 2:5 +24 OS 7- 1-3.3 +aco G - 1.35 20 = - 2.7 dy ED [5.25-05-25 -0:3x-25-135 da y = Es [ 5.25-OS-251.354] EI = ked 168.75KN.m2 Given at A. Slope 3 O 0.5(0)-2.5 [5.258056-os On- x=0 168.75 - 1.35 rad, --0.008 168.75 -0.008 rad. OA

deffection at cu Es (sas 4-0 (9-2545 Y = 9/x=1 -1.350 50] exsas bas 168.75 - 1.35 (0.875 0.875 -0.25 -0,2083-1234 168.75 -0.00523mt = 5.23mm y = 5.23 mm