Question

A temporary support steel frame (E - 200 GPa) is made of 30-mm diameter solid circular steel rods (members turnbuckle rods (members BC and AD), all pin connected at the corners A, B, C, and D (see Figure below). The steel frame is braced by two cables AC and BD. A turnbuckle is fitted to cable AC and tightening it will induce tension in cable AC. Determine the largest allowable tension force in cable AC using a factor of safety of 2.75 with respect to buckling. (Note: you need to consider only buckling in the plane of the structure). 3.5m 2.25m

Answer 1

given data: AB and CD 30 mm dia BC and AD 22 mm dia factor of safety-2.75 calculation of length AC 3.52 +2.2524.16m we know force in the member can be calculated as L' Free body diagam for he member joining at C

BC TACSM AC CD

Taking summation of horizontal forces equal to zero

FBC-TACCOSA 2.25 FBc - TAC 4.16 (from traingle BCD) Fc 1.84 TAc

Taking summation of vertical forces equal to zero

F TacSİNo CD 3.5 FCD-Tacー- 4.16 (from traingle BCD)

From force formula

㎡*200* 10° *IBC 2.252 BC 4 * (兰)4-1 1 .499 * 10' mm" = 1 1 .499 * 1 0 9 ㎡ Inc = π2 *200* 10° * 1 1 .499 * 1 0-9 n14 FBCー 4.48 10 N 2.252 Allowable force force on CD 4.48* 10 factor of safety 2.75 = 1.62 * 10'N Therefore tension in the member is = 1.62 * 103 * 1.84 = 2.99" 10 N (from above equation)

LcD2 ㎡*200* 10° *1CD FCD = 3.52 CD 4 2 4 1CD-.* (-)' = 39.760 * 10,nm4-39.760 * 10 9㎡ ㎡*200* 109*39.760 * 10 9m4 6.40103 N 3.52 Allowable force-force on CD 6.40 10 factor of safety 2.75 2.32* 103 N Therefore tension in the member is Tc = 2.32 * 10, * 1.18 = 2.74 * 10 N from above equation

therefore small tension force i.e 2.74*10³ N is considered.