Exercise 8 Determine the deflection at point C Assume the be

Exercise 8: Determine the deflection at point C. Assume the beam is rigid. The Modulus of Elasticity of the vertical steel rods is 200000 MPa Diameter of top rod 20 mm Diameter of bottom rod = 12 mm. Load at C 13 KN Distance AB = 1.00 metre. Distance BC 1.20 metres. Distance CD = 2.10 metres. 20 mm dia. 1.8 m Hinge 1.0 m 1.2 m 2.1 m 1.8 m 13 KN 12 mm dia.

Solution

Since C is a hinge,moment at C=0

this means vertical reaction at D=0 because R_D*2.1=0,which implies R_D=0

Considering moment equilibrium at A,we get

13*2.2-R_B*1=0

R_B = 28.6 kN(vertically upwards)

R_A = 28.6-13=15.6 kN(vertically downwards)

Therefore, both 12mm dia rod and 20 mm dia rod are in tension

tension in 12 mm dia rod = 15.6 kN=15600 N

tension in 20 mm dia rod = 28.6 kN=28600 N

Area of cross section of 12 mm dia rod = pi*12²/4=113.1 mm²

Area of cross section of 20 mm dia rod=pi*20²/4=314.1 mm²

Stress in 12 mm dia rod = 15600/113.1=137.9 N/mm²

Stress in 20 mm dia rod = 28600/314.1=91.05 N/mm²

length of 20 mm dia rod = 1.8m

length of 12 mm dia rod = 1.8m

Modulus of ealsticity of steel = 2*10⁵ N/mm²

strain in 20 mm rod = 91.05/2*10⁵=4.55*10^-4

strain in 12 mm rod = 137.9/2*10⁵ = 6.89*10^-4

vertical displacement of A=6.89*10^-4*1.8=0.00124 m=1.24 mm(upwards)

vertical displacement of B=4.55*10^-4*1.8=0.000819=0.819 mm(downwards)

since the beam is rigid, the displacemt at C can be determined with the help of displacements at A and B

displacement at C = 1.24-[1.24-(-0.819)]/(1) * 2.2]=-3.29 mm=3.29mm(vertically downwards)

Therefore, point C displaces 3.29mm downwards