Question

P10.023 The simply supported beam consists of a W410 x 85 structural steel wide-flange shape [E - 200 GPa? - 316 x 10 mm. For the loading shown, use discontinuity functions to computer (a) the slope of the beam at A, and (b) the deflection of the beam at midspan. Assume LA - LCD - 2.1 m, lec-3.0 m, w = 64 kN/m. Answers: (a) 0 L (b) ...The number of significant digits is set to estion Open Show Work

Answer 1

In order to calculate the slope and delflection, first of all we need to calculate all the reactions at the support A and D.

w= 64 kN/m, Freebody diagram for the beam is represented im figure 1

We will take a section X-X at a distance "x" from support A, and calculate moment Mx at the same section.

using the equation of equilibriums

Reaction at A and D = 134.4 kN;

Reffer figure for explanation

Evo; 2.100 +2.10 = PATRO . 4.20 = RATRA EMD=0; RA X 7.2 - 2.10 X 6:15 - 21.00 1.05 = 0 VA - (23/C + 2 26 )a - 7.2 TPA = 2:10] BA= 2:1864 = 134.4 kN/ Ry = 4.26-27.60 TRA= 2.10 ] - [ Bg = 134.4kN Pesa

now using discontinuity equation the slope at support A and deflection at mid point(at 3.6 m from A) can be calculated as follows:

6460/m A năm by KN/m figurer 1344 m m x 2.11 3.0 X 21 B4.4 My = 134.4 2 - 64 Mx = 134.48 -322 using Discontinuity ogn we get; ti d y = Mx. Integreeting the above egh both sides we get dice eidy - f Mx. dx. 3 ③ = f (134.4x - 32x²) dx = 134.4 x ² – 32x3 + 4 - £ I dy = 67.2x² - 10.66x3+ C, (slope egn) Integrating both sides £1 y = {(67.2x² + 10.66x3 + c) dx = 67.2 x3 + 10.6674 +6,x+ 6 - E14 - 22.4x3 + 2.665x" + ( x + ( 2 (Caplection I

Applying Boundary condition to ga at x=0; y=0 meget a - 0= ototot (2 Applying another Boundary concetion to ep" (6 we get at x = 7.2 ; g = 0 ..0 = 22:4 (7.2)3 + 2.665/7:274 + + oto terroristebatu firal slope & Deflection Eations! £ Idy = 67.2 x² - 10.66x² - 0 & ti y = 22.4x3+ 2.665x4 Now; QA = slope at A or (dy). cet A ;x=0 7 +(dy) - 0-0 dy). - OA = 0

पा » Virid a y mid e at Yaid ; x = 72 = 3.6 Putting values in egn ② £ I Gonid - 22.4 (366)3 + 2.665/3-6) = 1045.094 + 447.6/7 = 1492.711 7 Ymid = Umid = 1492.711 here t = 200 apa & I = 3/6x100mm 4 ar t= 200 x 100 kN/m² I = 316 8 10-6 m4 I Vanid = 1492.711 2000106 x 31681076 I mid- 0.02361m. 1492.70 63200

Note: C1 and C2 are integral constants

Answer 2

Slope at A will not be zero.