Question

w(x) = 4W* (2 - x) M(x) = Wox' 2wx3 wo Lx A beam is supported by a pin at A and a roller at B. It is loaded with a non-uniform distributed load, w(x). An equation for the internal resultant moment in the beam in terms of x, M(x), is given. Determine the magnitude of deflection at the point where x equals a. Wo -9,500 N/m L = 4 m E - 180 GP 1 - 35 x 10-6 m4 a = 2 m • 2.31 mm 4.09 mm 10.8 mm 3.08 mm 7.59 mm 14.2 mm

Answer 1

20 Waqwo -2-(1-4) '3L dx2 J dx23_2 - 902 300 Me 2 hlo x4 - 2 ho x2 + blo .We know, Eidy - Mx-X $57.joy wo (35(26) hobl ltas EI - bolo xo - folo jät + Wal.medte: 0 again integrate both side. E EI Y = Wo x6 Wo xs + Wol x3 +60+C2 Now Boundry condition :- ☺ Q7 x=0:29=0 Putting in eq?> =) Ō = 0 - 0 +0+0+C = |C2=0] @ at D = L > y=0 = 0 = Words Wo l + Wall + CIL+O 90: Wo LX – 3W024 + 5 Wo 24+90 CL 5 9064 = -3 WOL4 Wol? = Wo 22 → 16 = Wo 137 901 90

How (2000) 9500 9500 X (2000) 2002 + 9500 x 4000x2000) - 7000*30*4000! k, 60 f 1000 x18 g . = "1000x90x4000) + tia que portano ya 2000 =-2.5755 x103) 180 x 183 x 35 x10-6 x(1000)4 = - 4008818 mm Do deflection ever be 400 88 mm downward 4009 mm downward.