Question

5. A 25-mm-square aluminum strut is maintained in the position shown by a pin support at A and by sets of rollers at B and C that prevent rotation of the strut in the plane of the figure Knowing that LAB = 1.0 m, Lac = 1.25 m, and Lcp 0.5 m, determine the allowable load P using a factor of safety with respect to buckling of 2.8. Consider only buckling in the plane of the figure and use E = 75 GPa

Answer 1

Given Square :a = 25 mm LAB? 1.om LBC = 1.25m Lep:- 0.5m Faclóng of safety nt 2.8 ... E= 75 G Pa

allowable load Formula Pallow Por where in factor of safety Per > Cržlical load. = a*ET 20? So, we need to finced the critical length co effective length and moment of inertia.

Find the effective length (Les Take portion cp.it has freed at one end and free at one end. . So, 2e = alco k=2x0,5 - Le = lm Take portion Bc it has fired at both ends. ... so le = 0.5 LBC . . = 6.5 * 1.25 Le = 0.695 m Take portion AB, it has Priced at one end and pinned at one end. . - Le =0.7LAB : -07X1 . Le =0.7m Take maximum of the above effective length Le=im; it is the maximum Le for the L=1000mm

Now oinertia Find moment of I = ď - For square - T [ ] 5.2852 2 = 32552.08 mm Substituce the values & and be in per Per =*ELAx45x18) X (38552.08) B ioco) 2 | - ટીમ ૧૩. RN Por 5 04.095#N put the Per in Pallowable 2.8 rate. – 8.605 KN The allowable load for struct is = 86605 kn/