Question

Calculate: The signpost is made of 350 steel. The bottom of the sign is to be 4 m above ground level and the top is to be 6 m above ground. It is to be 3 m wide, from the side of the column. The forces acting on the column are as follows: The signs dead weight is 7.2KN 1.5m across the sign, Self-weight of the column, 1.7kPa wind load normal to sign and on column Calculate the following for the SHS below: reaction forces ii) the shear stress by torsion and transverse shear force iii) bending moment iv) torque sign acting on the signpost seen below: i) 50 1200 R=24 mm 50 250 100 0 100 200 250 column

Answer 1

the sbuchure is similar to a cantiliver beam TEN wind load 31 1.5m coind load = OX AXCd cross-section is green = 17x(250x6%) x 2x2 A-T KN brag coefficiant; ca =2 for plate A = 0.25x2 = 0.512. , coind load = 6.7 kw

7.2 KN 702 km 1.5 a) To find reaction forces; use equilibrium conditions. EFx=0 Ha = 17 KN A EFy=0 RA -1.2 KN EMA = 7.2x1.5 = 10.8 kNm b) Shoar stress T= Shear force Area 3 7.2x10 0.25X2 14400 N/m shear skiess based on toansveese Shear force 14.4 KNIG?

Shear Stress based on Torque : T y = I Ip T* Torque Ip → polar moment of Intetia . Torque = Force x t distance = 1,7X1 = 1.7 km a 250 mm Ip= 0,208 a 3 3 = 0.25 m 0.2880,25 -3 = 3.25 XIQ mm 4 In - a 11 0.25 6 6 -4 6.51810 3 = 651.0 4x10mon 1.7x106 Z 6 248163 651.04x10 2 T- 62.67 kNlm

(3) iii) Pending moment at bottom o sigo M= 7.2*1.5-1.7X/ - 9.1 KNM Torque ; T 1.7x ! 1.7 Knin