Structure Final Math Problem

Question 1
(a).Free Body Diagram.
18’AR2
40 kip2’
2’15kip
2’
25kip
2’
B6’
R1
(b).
Reaction 1;
Fy =0,therefore, R1 =40 cos 20 + 15 = 52.59 Kip.
Reaction 2;
‑ Fx =0, therefore, R2 = 40 sin 20 + 25 =38.68 KiP.
Question 2.
(a). Pallowable = Fallowable /Area of Cable
Area of Cable = πd2/4 ,={π×(3/4)}/4 = 122.2 KiP.
(b). ‑ Vertical Forces = 0
‑ Fy = 0.
Taking FV to represent vertical components of forces in the cables;
FV = 122.2 cos ᶱ
Sum of upward forces = 2 × 122.2 cosᶱ
=244.4 cosᶱ
Therefore; 244.4 cosᶱ = 30
cosᶱ = 30/244.4
ᶱ = cos-1 (0.1227)
= 82.94
=83°
Question 3.
(a).The tipping point is 8’ from the left end of the beam.
Tipping point
12’8’
(b).Balance of moments about the tipping point.
300lb150lb
10’2’x
300×2 = (x) × 150
Therefore, x = 4’ from the tipping point.
Question 4.
4(a).Force Polygon.
a1
b
2
c3
d
e
f9
(b).Reaction force; R = 4+8+8+8+8
= 36 Kip.
Cable Stay 5 Analysis
20’
ᶱ=Arctan(20/7)7’
=70.71°
Vertical component of force in cable stay =4 kip
Force in cable stay -4/sinᶱ = 4/sin 70.71
= 4.24 Kip.
Cable Stay 4 Analysis
ᶱ= Arctan (20/5) = 78.69°
Force in cable stay 4 = 8/sin 78.69
=8.15 kip.
Cable stay 3
ᶱ= Arctan(20/3) = 81.47°
Force in cable stay 3 =8/sin 81.47 =8.09 kip.
Cable stay 2