Hello Please someone help me with this question Im stuck in

1 The steel bars AB and AC of the frame are pin-connected at the ends for buckling in the plane of the structure and Fixed-Fixed connected for buckling out of the plane of the structure. Determine the largest force F that can be applied to cause failure in buckling. Which member will buckle first? Take E 200 GPa and if oy 250 MPa, will the members fail by yielding? 4 m 6 m 50 mm x 60 mm 50 mm

Solution

solution:

1)here we have to analyse to member separately to calculate maximum force that system can take before buckling

2)here as per euler theory maximum load is given on member Ab is as follows

Pc1=2.05*pi^2*EI/L^2

where I=bd^3/12=100*50^3/12=1041666.667 mm4

E=200*1000 n/mm2

L=6000 mm

hence Pc1=117087.3208 N

3)maximum force in direction of member AC is

Pc2=2.05*pi^2*EI/L^2

here L=5000 mm

Pc2=168605.74 N

4)let divide Pc1 and Pc2 along x and y direction we get

Pc2x=101163.685 n

Pc2y=134884.41 N

Pc1y=117087.3208 N

Pc1x=0

hence resultant force in x and y direction is

Fx=101163.685 N

Fy=251971.73 N

5)here resulatnt force is R=(fx^2+Fy62)^.5=271521.35 N at angle of 21.875 degree to vertical

6) but resultant force acting on system in vertical direction only hence maximum force that system can take before buckling is Fy only

F=Fy=251971.73 N

7)stresses is given by

sy=Fy/(100*50)=50.39 MPa,

hence system will not subjected to yielding as Sy is less than yield strength

8)once the force F exceed the Fy value it will induce buckling in member AB first then when load exceed more than 271521.35 N the member Ac will buckle