STRUCTURAL ANALYSIS W/MOD MAST
STRUCTURAL ANALYSIS W/MOD MAST
10th Edition
ISBN: 9780134863375
Author: HIBBELER
Publisher: PEARSON
Question
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Chapter 5, Problem 5.1P
To determine

The tension in each segment of the cable and the vertical distance between points A and D.

Expert Solution & Answer
Check Mark

Answer to Problem 5.1P

The tension in segment AB is 15.21kN.

The tension in segment BC is 8.49kN.

The tension in segment CD is 10.82kN.

The vertical distance between points A and D that is yD .is 4.33m.

Explanation of Solution

Concept Used:

Write the expression for the net moment about end A of the system.

   MA=0   ......... (I)

Here MA is the sum of all the moments taken about the end A.

Calculations:

The free body diagram for the system is shown below.

  STRUCTURAL ANALYSIS W/MOD MAST, Chapter 5, Problem 5.1P , additional homework tip  1

  Figure-(1)

Here, the unknown vertical reactions are Ay and Dy at points A and D respectively and the unknown horizontal reactions are Ax and Dx at points A and D respectively.

The slope of cable CD is known, therefore resolve the tension in cable CD, that is, TCD in perpendicular and horizontal directions and determine the moment about point A.

Write the moment about A from Equation (I).

   Dx×yD+Dy×9m3KN×5m20KN×2m=0Dx×yDkN+9DykN15kN-m40kN-m=0Dx×yDkN+9DykN55kN-m=0   ......... (II)

Consider ΔBCE.

Given BE=CE=3m

Given that BEC is equal to 90°

Therefore ΔBCE is a right angled isosceles triangle, therefore,

   BCE=CBE=45°

Consider ΔDCF.

Apply Pythagoras theorem in ΔDCF.

   DC2=DF2+CF2   ......... (III)

Here, the distance between points D and C is DC, the distance between points D and F is DF and the distance between points C and F is CF.

Substitute 6m for DF and 4m for CF in Equation (III).

   DC2=6m2+4m2=36m2+16m2=52m2

Take square root on both sides.

   DC= 52 m 2 =7.21m

Calculate DCF.

   DCF=sin1DFCF   ......... (IV)

Substitute 6m for DF and 7.21m for CF in Equation (IV).

   DCF=sin1 6m 7.21m=sin10.832=56.32°

Calculate DCE.

   DCE=DCF+FCE   ......... (V)

Substitute 56.32° for DCF and 90° for FCE in Equation (V).

   DCE=56.32°+90°=146.32°

Calculate DCB.

   DCB+DCE+BCE=360°   ......... (VI)

Substitute 146.32° for DCE and 45° for BCE in Equation (VI).

   DCB+146.32°+45°=360°DCB=360°146.32°45°=168.68°

The free body diagram at point C is shown below.

  STRUCTURAL ANALYSIS W/MOD MAST, Chapter 5, Problem 5.1P , additional homework tip  2

  Figure-(2)

Apply Sine Law at point C.

   3kN sin 168.68° = T BC sin 146.32° = T CD sin 45° 3kN 0.196= T BC 0.555= T CD 0.707   ......... (VII)

Here, the tension in cable BC is TBC and the tension in cable CD is TCD.

Determine TCD and TBC by taking two equations at a time from Equation (VII).

   3kN 0.196= T CD 0.70715.306kN= T CD 0.707 15.306kN×0.707=T CDTCD=10.821kN

Thus the tension in segment CD is equal to 10.82kN.

   3kN 0.196= T BC 0.55515.306kN= T BC 0.555 15.306kN×0.555=T BCTBC=8.49kN

Thus the tension in segment BC is equal to 8.49kN.

Analyze point D.

  STRUCTURAL ANALYSIS W/MOD MAST, Chapter 5, Problem 5.1P , additional homework tip  3

  Figure-(3)

Apply Angle Sum Property in ΔDCF.

   DCF+CDF+DFC=180°   ......... (VIII)

Substitute 56.32° for DCF and 90° for DFC in Equation (VIII).

   56.32°+CDF+90°=180°CDF=180°90°56.32°CDF=33.68°

Calculate Dy.

   Dy=TCD×cosCDF   ......... (XI)

Substitute 10.82kN for TCD and 33.68° for CDF in Equation (XI).

   Dy=10.82kN×cos33.68°=10.82kN×0.83=8.98kN

Calculate Dx.

   Dx=TCD×sinCDF

Substitute 10.82kN for TCD and 33.68° for CDF.

   Dx=10.82kN×sin33.68°=10.82kN×0.554=5.99kN

Calculate yD from equation (II).

Substitute 8.98kN for Dy and 5.99kN for Dx in equation (II).

   5.99kN×yD+9m×8.98kN55kN-m=05.99×yDkN+80.82kN-m55kN-m=05.99×yDkN+25.82kN-m=0yD=4.33m

Consider ΔABG.

  STRUCTURAL ANALYSIS W/MOD MAST, Chapter 5, Problem 5.1P , additional homework tip  4

  Figure-(4)

Calculate the length AG.

   AG=9myD   ......... (XII)

Substitute 4.33m for yD in Equation (XII).

   AG=9m4.33m=4.67m

Calculate ABG.

   ABG=tan1AGBG   ......... (XIII)

Substitute 4.67m for AG and 2m for BG in Equation (XIII).

   ABG=tan1 4.67m 2m=tan12.335=66.81°

Analyze point B.

The free body diagram for point B is shown below.

  STRUCTURAL ANALYSIS W/MOD MAST, Chapter 5, Problem 5.1P , additional homework tip  5

  Figure-(5)

   ABH=ABG+90°   ......... (XIV)

Substitute 66.81° for ABG in Equation (XIV).

   ABH=66.81°+90°=156.81°

Calculate CBH.

   CBH=CBE+EBH   ......... (XV)

Substitute 45° for CBE and 90° for EBH in Equation (XV).

   CBH=45°+90°=135°

Calculate ABC.

   ABC+ABH+CBH=360°   ......... (XVI)

Substitute 156.81° for ABH and 135° for CBH in Equation (XVI).

   ABC+156.81°+135°=360°ABC=360°156.81°135°=68.19°

Apply Sine Law at point B.

   20kN sin 68.19° = T BC sin 156.81° = T AB sin 135° 20kN 0.93= T BC 0.39= T AB 0.70721.51kN= T BC 0.39= T AB 0.707   ......... (XVII)

Here, the tension in cable BC is TBC and the tension in cable AB is TAB.

Determine TCD and TBC by taking two equations at a time from Equation (XVII).

   21.51kN= T AB 0.707TAB=21.51kN×0.707TAB=15.21kN

Thus the tension in segment AB is 15.21kN.

Conclusion:

The tension in segment AB is 15.21kN.

The tension in segment BC is 8.49kN.

The tension in segment CD is 10.82kN.

The vertical distance between points A and D that is yD .is 4.33m.

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