Week 11 Challenge

1) vertical force = 125KN

Frictional force in longitudnal direction = 125* 0.3 =37.5 kN

Frictional Load in tranverse dirction = 10 % of the piping operational load

= 125*0.1 =12.5 kN

frictional load if teflon is placed =125*0.1 = 12.5 KN

2) Figure 1 - spring

Figure 2 - rigid strut

Figure 3 - stop+ rest

Figure 4 - Guyed + Rest + Stop

Figure 5 - guyed+ rest

Figure 6 - guyed

3) structural dead load =1KN/M²

operating dead load =1.9 KN/M²

empty dead load = 60% of operating dead load = 0.6*1.9 =1.14KN/M²

Test dead load =Dt =1.9 KN/M² (with water)