Question

Q2 The 10 m long simply supported beam is subjected to a uniformly distributed load w = 10 kN/m throughout and a point load P =10 kN at the midspan of the beam, as shown in Figure Q2 (a). The cross section of this beam is depicted in Figure Q2 (b), which consists of three equal rectangular steel members. Self-weight of the beam is neglected. 30 mm P= 10 KN W = 10 kN/m 200 mm 5 m 5 m Figure Q2 (a) Figure Q2 (b) I (a) Determine the reactions, bending moment equation along the beam and draw the corresponding diagram. [10] (b) Using the method of double integration, determine the deflection of the beam at the midspan of the beam. Assume E-200 GPA. [15]

Answer 1

Solution X P-10KN walo kN/m ve vo Vyv N → B. AZ ولد 5m 5 ml 6 х B VA X let =0 = I10 KN ñ Assuming A and B are vertical reactions at point a and a respectively Let Ha be the horizontal reaction at 1. C be the mid point of Beam. Skeel Applying equilibrium equation in vertical direction Efv 20 NA + VB (10 X10) - 10 VA + VB Now, applying equilibrium equation in horizontal direction EfH 20 THAO I stepNOW applying moment equibrium about a EMA -0 => (V8 x 10) - (10x10x10.) - (1045) Anticlockwise moment = positive Clockwise moment = negative V8 -55 KN from equation 0 NA - 55 KN (Step27

consider a section X - X at a distance & from a Here a varies from o to 5 m Moment at section non is Mut tv.x 10.3.2 o Mze = +55 X - 520? (Parabolic variation) LOU (i) when a=0 ĮMA=0 when u=5 ( Just left to C) Mc = +55x5 5x1511 Mc = 150 KN-m Now consider a Section x-x (u from c ) in CB portion, Moment at Section 2-24 is Mge = + VA: (5+") - (10x5)(+n) – 10%. – 10.*•* H, 3 + 55( 54 ) - 50(+5 + x ) - 10X - 5%? i when n=0 Mc +150 KN-M 5X5 M (in when a= 5m 8 = 55(10) - 5017.5) - 1095 IMB

TOKN TOKN/m to B A A 5 m 5'm 21 .55 KN 55 KN 150 KN-m) R parabola BMD Ai C

(6) IOKN X A 10 KN/m Vc B → → the 5m 5m 2 - 55 KN А X V8=55 KN consider a Section x-x at a distance a from A, Moment at Section x-X is Mx = yh.x -102.7 -10(2-5) According to double Integration method EI oly Mx da? -> 2 EI dy vx; -10. -1002-5) dae EI dey 55x4-107? -1011-5) dx Integrate above equation Erdy 5522+4,-5 -1012-5) 55 x'+G45 x 34 -10 (x-5)? -0 3 -5.nl ES 2 3 da 2 z) EI dy da

(5 Again Integrate equation 0 3 4 3 as EI y 55 x x + GX + , 3 G1 - 10 (2-5) 3 4 3 2) EI y 9.17x3 + 9+47-0.425 - 1,67(8-5)— © (auj Applying Boundry condition cü when 2-0, y = 0 23 0 ca (ii) when r= 10 , y=0 0.4211019 1.67(513 0 = 9.178(10)3 + C,(10) = 9170 + 109 4200 208.75 2) 0 = G: - 476.185 3 EI = y Put value of G and Ę in equation ® 9.1723 - 476.185% - 0.4224 -1.67/3-55% for deflection at mid point n=5 EL. = 9.17X651 - 476-185(5) - 0.42(5)" - 167(5-5)3 . 1146.25 -2380.625 - 262.5 EI -1496.875 ЕІЯ 1496.815 es EI

Here, E=Q00 G1Pa = 200 X 10 Mpa for the given moment tigure, % Thestia 3 I = bo 12 30 x(2001 I = QOXIO x10mmy 4 oʻx(1003) "my 3 EI - 200 X 10XÓKN X PO X10 X m2 EL : 200 x 10 x 20 x 10 4 EI - 200 x0 KN/m? Xm ΕΙΣ 4000 KN-M y = -1496.875 4000 y = -0.3742 m Negative sign shows that deflection is in downward direction