Question

Consider the frame and loading shown in the adjoining figure. Use the dimensions and load values in the table below. Analyse the frame and determine the reactions and internal actions required in the table below. Enter your answers in the space provided.

Dimensions:

h = 3.8m

a = 2.5m

b = 4.8m

c = 1.8m

w = 20kn/m

Consider the frame and loading shown in the adjoining figure. Use the dimensions and load values in the table below. Analyse the frame and determine the reactions and Internal actions required in the table below. Enter your answers in the space provided. w Dimensions: a 2.5 m 1.8 m Loading: w 20 kN/m Provide Answer for the followings to an accuracy of 2.5% Absolute Value Unit Direction Dy=1 Member Forces Shear force in menter CE Veel Me Maximum moment in member BC Mmax=1 Location of maximal moment Mmax From point

Answer 1

20 kn/m Aminama 39m Ex x 25m 1 yogm ( 1.8mt 207 using equilibrium equation - {fy=0 ; Dy + Ey - 200(2-574-8)=0 Dy + Ey= 146 KN - 0 { Fu=0; Dx-En=0 Dn= Ex E MEz0; 20x (Z-S+18)x (2-544.9+1-3) - Dy (48+19)=0 Dy = 120-56 KN (n) Ey= 25.43 kN (1) from ② Now, Taking equilibrium of LHS of hinge B. 20x 2.5² Dux 3.850 Du = 16:44 KN (E) [En= 16:44 KN (₃) from ②

Then shear force in member CE- t Ex cos (90-0) = 14.85KN tan 8 = 38 J 90-8=25.34" Eve=16.44 KN O= tan ! ( 38 ) 0=64.65° > Ey= 25.43 KN -- tyloso - 10-88 KN % VCE = Ey coso + Eu cos (90-0) [VLE = 25.73 kN (1)) By Calculating from above fig. Then, calculate moment at Me = Ey x1-8 ¢ Exx 3-8 Mc = 25-4381-8 + 16.44X 38 Me=108.246 kN-m 3 and calculating point of zero shear from 'C. Ey- 20x2=0 In=1-27m It is location of Mmax from point c. so, Mmax in BC = Ej (169+1-22) +E7X3-8 - 20x (1-9+123)? Substitute Ey and En values, we get Mmax = 46.29 KN-m g