Question

Let us consider the cantilevered balcony beam of Example 4.3 again and solve it using a single beam element. Recall that the beam is a wide-flange W1S X 35, with a crosssectional area of 10.3 in2 and a depth of 17.7 in. The second moment of area is 510 in 4. The beam is subjected to a uniformly distributed load of 1000 lb/ft. The modulus of 208 Chapter 4 Axial Members, Beams, and Frames elasticity of the beam E = 29 X 1061b/in2. We are interested in determining the deflection of the beam at the midpoint B and the endpoint C. Also, we will compute the maximum slope that will occur at point C. 1000 lb/ft 10 ft TABLE 4.2 Equivalent nodal loading of beams Londing Equivalent Nodal Loading M-PL

Answer 1

given alu A= 10.3102 d=17-tin a = sloine hl=1000lblet = 29x108 lb link 20 Consider a section Non at a diktance & from ce M = -W ) n -o) y W. - er – W (n-012 nog 2 m W ) Integrating El dy & - de ta EI (Y) = – lal (ro)" conte con to au 24 M 1

Atx=lott D L slope dano le or Ello) - hitta 2 At x=1 .e at point A MA 0 Ello) = -4/24 et web (w the 24 Eddy = - W (0-0) + h natten FIUE - 04we're Woo deflection at the mid point Biex=2 ES (Y) = a + bile u El Yo = – It alle 384

- to stars It 384 will El Wa 1000 lblet Lloft = 29x166 16/in2 1=510 in E = 14790 X10b eo fine del ta - 102.70 833 xlof et et2 GB - - 17 (1000) Close 384 102.708 33 Rob (y=-4.3103449 x0² Et Yo =-0.05192 in =) deflection at Hd = hehe end point a - Close Close 8x 102.708 33x106 lyc = – 0.0121t et y = – 0.14604 in

=) elope at point c at x=0 & 9 dYc = lall² = GEO Clooo) (18) 6x 102.8708354106 a ale = 1.62241 8 x103 rad