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Math Problem

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# Structure Final Math Problem (Math Problem Sample)

Instructions:

these are 15 structures problems, i need them to be completed no matter what, i am not looking for accuracy i just want them to be completed to the best of your ability

source..Content:

Question 1

(a).Free Body Diagram.

18â€™AR2

40 kip2â€™

2â€™15kip

2â€™

25kip

2â€™

B6â€™

R1

(b).

Reaction 1;

Fy =0,therefore, R1 =40 cos 20 + 15 = 52.59 Kip.

Reaction 2;

â€‘ Fx =0, therefore, R2 = 40 sin 20 + 25 =38.68 KiP.

Question 2.

(a). Pallowable = Fallowable /Area of Cable

Area of Cable = Ï€d2/4 ,={Ï€Ã—(3/4)}/4 = 122.2 KiP.

(b). â€‘ Vertical Forces = 0

â€‘ Fy = 0.

Taking FV to represent vertical components of forces in the cables;

FV = 122.2 cos á¶±

Sum of upward forces = 2 Ã— 122.2 cosá¶±

=244.4 cosá¶±

Therefore; 244.4 cosá¶± = 30

cosá¶± = 30/244.4

á¶± = cos-1 (0.1227)

= 82.94

=83Â°

Question 3.

(a).The tipping point is 8â€™ from the left end of the beam.

Tipping point

12â€™8â€™

(b).Balance of moments about the tipping point.

300lb150lb

10â€™2â€™x

300Ã—2 = (x) Ã— 150

Therefore, x = 4â€™ 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

20â€™

á¶±=Arctan(20/7)7â...

(a).Free Body Diagram.

18â€™AR2

40 kip2â€™

2â€™15kip

2â€™

25kip

2â€™

B6â€™

R1

(b).

Reaction 1;

Fy =0,therefore, R1 =40 cos 20 + 15 = 52.59 Kip.

Reaction 2;

â€‘ Fx =0, therefore, R2 = 40 sin 20 + 25 =38.68 KiP.

Question 2.

(a). Pallowable = Fallowable /Area of Cable

Area of Cable = Ï€d2/4 ,={Ï€Ã—(3/4)}/4 = 122.2 KiP.

(b). â€‘ Vertical Forces = 0

â€‘ Fy = 0.

Taking FV to represent vertical components of forces in the cables;

FV = 122.2 cos á¶±

Sum of upward forces = 2 Ã— 122.2 cosá¶±

=244.4 cosá¶±

Therefore; 244.4 cosá¶± = 30

cosá¶± = 30/244.4

á¶± = cos-1 (0.1227)

= 82.94

=83Â°

Question 3.

(a).The tipping point is 8â€™ from the left end of the beam.

Tipping point

12â€™8â€™

(b).Balance of moments about the tipping point.

300lb150lb

10â€™2â€™x

300Ã—2 = (x) Ã— 150

Therefore, x = 4â€™ 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

20â€™

á¶±=Arctan(20/7)7â...

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