Question

Q6 (12 marks) In the figure below, the steel beam ABC is simply supported at A and C. Its cross-section is the rectangular tube as shown where dimensions are in in. Force F varies such that the magnitude of deflection at the mid-span of the beam is between 0.01" and 0.05". Determine the mean stress and alternating stress developed at the top surface of cross-section B. Given, a = 7", b = 12". A ab — In Simple supports 4X2X1/4 Steel tube beam cross-section

Answer 1

given FINO re a=0.71 b= 1211 a tau At equilibrium toto - o - Cool I reatical forcee so og tris RA+ Rc-f=0 Rat Rear Moment at A - Rex Carty) - f(a)=0 BC Exa - 7CF) - 74 pe cayo & 2412 - 1a. RA-F- Fra a fb (atb) Catbl Conaides a gection N-N, at a distance & from point a Mn= RA (RO) & (n-a) = 12f (1-0) - f (xt)

Eldy - 12 F ca-o) - f (a-a da? = 14 (x-o) – f(a-A) Integrating е в са е смес ER (yl = 12f (x-03 f (x-7) Gate At no le at point A EI (O) = 0-0 to + ca At a = 1911 e at point a YATO 270 ditupe yea de E16) = 126 (19) - F (12) + 6(19) to 1986 Joom 722 240 - 280 & cala Ga – a2842 F +22.8421k en gang frjo -22, Guler 22 for ateden 294106 pel oito o abip 4

2 - 4 2 - 3 - 2.416667 ing EIF 10.0833416 pou Man man at xetin EL (Yatin = 12 (A) – 22.8421XAR El Ixt.in - - 123.7894f I mana – 10 7 66 317 xlog . f in y = 0.01 = 5661.490 P for y= 0.05 = 28307.489 P. Mmore = 89 Ob max = pin. eye y 2.416667 Oblomara = 1.829398f

for R=5661.494 Rand 6 - 10357, 1148 Peil for f= 28307.489 Pound alternatingetrage (Obl 51785.541 Pe mean atrege - Site - 31071.343 Pri mean = 31071.34315 Pan