Question

Use Moment Area Theorems in the beam shown in figure 3 to determine the following:
a- Deflection at C.

b- Deflection at D.

c- Slope at C.

d- Slope at D.

Use E for modulus of elasticity and I for the moment of inertia for the whole beam.

20KN B k C pin Roller 4m 6m 4m Figure 3

Answer 1

x3 X 20KN JO KN *2 Alla A C B $D 1 4ml 16on 4 m 1 x 1X2 K3 Ay B + B By = 20 + 10 # For DB, Mug = - 1003 ; 'By *. from equilibriun, ZMA=0Byx 10 = 10x14 + 2004 = 22 KN (4) Efy = 0 7 А. 8 KN (4) Bending moment at different span (Mu)] # for AC, Mul, = +88; ; 054, 54m # for CB, May = +8(1+4) - 2012 * Bending moment diagram (BMD) :7 diagrom * * Sign Convention in Sagging BM = tve Hogging BM :-ve -1222 +32 ; 0 54₂ 56m os ng semi BM EI +32 ΕΙ + + E 4m D 4m -40 EI

120KN LOKN Act *. Sign Convention : 7 downward deflection+ +ve Upward Clockwise slope Antic lockwise slope tve > - ve → - ve By moment avea method nd 2 theorem. = 10 9 Ź (401) "(19)*( 1x (*)*(*)*(not the one +14)· (6+1) +1840 3 EI Now, OA DA- YB 10 184 JEI A &. Again, & Ź*(*23.)*(4)* -)* (* + 256 3 EI we have, Det Ye OA*4 => Ac - 184x4 ЗЕІ 256 3 EI A = 160 EI 160 W

* Agalm, 4 3 ( *) *4*(234) + Ź (***) (9(4+ ') (b X +*86.4++) + ** (22) (4) (30+ $ - Seo EI » AD+Y > = O *14 → AD 184x14 BEI 560 EI * DD 896 3 EI (+) © from moment area method - 1st theorem, C А A 2 So, OA 184 3 EI -0 = (+ 321)-(4) Here, Os is clockwise 0-(-31) 64 Hence, slope at 'C' is ſei and in = * + 8 3 EI с Anti-clockwise direction. @ Op-OA х 2 Again, (4+3)*(***) +Ix/4+ 1)*(-40) 00-(184- Oo - 304 Hence, slope at 'D' is clockwise direction. 40 EI 304 and in * JEI - 3 EI