Question

Elastic bodies

1. ) Determine the reactions at A and B for the steel bar and loading shown in the Figure (Prob. 1), assuming a close fit at both supports before the loads are applied (2) Draw axial force diagram. (3) Find displacement at point C

Answer 1

1. for the given body

let the elastic modulus be E

then

1. Reaction at A = Ra

Reaction at B = Rb

from force balance

Ra + Fd + Fk = Rb

Rb - Ra = Fd + Fk

Fd = 300 kN

Fk = 600 kN

as the object will lsoe contact on application of compressive force like given at A

hence

Ra = 0

Rb = 900 kN

now if we consider density of the material = rho in kg/mm^3

then

Rb = 900 kN + rho*(400*300 + 250*300)g

2. the following is an axial force diagram

axial force = F

For 0 < y < 150 mm

F = 900 kN + rho*(250*300 + 400*(300 - y))*g

for 150 < y < 300 mm

F = 300 kN + rho(250*300 + 400*(300 - y))*g

for 300 < y < 450 mm

F = 300 kN + rho(250(600 - y))*g

for 450 < y < 600 mm

F = rho*250*(600 - y)*g

3. displaceemtn at C = dy

stress = (300 kN + rho*300*350*g)/400

strain = dy/300

hence

(300 kN + rho*300*350*g)*300/400*dy = E

(300 kN + rho*300*350*g)*3/4*E = dy

where E is elasrtic modulii of the material

and rho is the density oif the material in kg/mm^3