Structural Analysis (10th Edition)
Structural Analysis (10th Edition)
10th Edition
ISBN: 9780134610672
Author: Russell C. Hibbeler
Publisher: PEARSON
Question
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Chapter 12, Problem 12.1P
To determine

The forces in each member of the truss.

Expert Solution & Answer
Check Mark

Answer to Problem 12.1P

The forces in each member of the truss are shown below.

The force in member AE is, 2.08k(Compression).

The force in member BF is, 2.08k(Tension).

The force in member EF is, 1.67k(Compression).

The force in member AB is, 1.67k(Tension).

The force in member CE is, 2.08(Compression).

The force in member BD is, 2.08k(Tension).

The force in member DE is, 1.67k(Compression).

The force in member BC is, 1.67k(Tension).

The force in member AF is, 4.25k(Compression).

The force in member BE is, 2.5k(Compression).

The force in member CD is, 5.25k(Compression).

Explanation of Solution

Check the determinacy of the truss as shown below.

   m+r2j=0

Here, number of members are m, number of reactions are r and number of joints are j.

Substitute 11 for m, 3 for r and 6 for j.

   11+32×6>02>0

Hence, the given truss is statically indeterminate by 2 degree.

The following diagram shows the free body diagram of the truss.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip  1

Figure-(1)

Write the equation for moment about A.

   ΣMA=0(Cy×16ft)(4k×16ft)(5k×8ft)=0Cy×16ft=104kftCy=6.5k

Here, vertical reaction at C is Cy.

Write the equation for sum of vertical forces.

   ΣFy=0Ay+Cy3k4k5k=0Ay+Cy=12k ...... (I)

Here, vertical reaction at A is Ay.

Substitute 6.5k for Cy in Equation (I).

   6.5k+Ay=12kAy=(126.5)kAy=5.5k

Consider the force in diagonal members FB is in tension and AE is in compression.

   FFB=FAE

Here, force in member AE and FB are FAE and FFB.

Use the method of section and cut the section as shown in figure below and calculate the forces in the members.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip  2

Figure-(2)

Calculate the angles as shown below.

   cosθ=35sinθ=45

Here, the angle between the members FA and FB is θ.

Consider the joint F.

Write the Equation for sum of vertical forces.

   ΣFy=05.5k3kFFBcosθFAEcosθ=0

Substitute 35 for cosθ.

   5.5k3k2FFB×35=02FFB×35=2.5kFFB=2.08k(Tension)FAE=2.08k(Compression)

Write the equation for the sum of horizontal forces.

   ΣFx=0FFE+FFBsinθ=0

Here, force in member FE is FFE.

Substitute 2.08k for FFB and 45 for sinθ.

   FFE+(2.08k×45)=0FFE=1.67k(Compression)

Consider joint A.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip  3

Figure-(3)

Calculate the angles as shown below.

   cosθ=45sinθ=35

Write the equation for sum of vertical forces.

   ΣFy=0FFA+FAEsinθ5.5k=0

Substitute 2.08k for FAE and 35 for sinθ.

   FFA+(2.08k×35)5.5k=0FFA=4.25k(Compression)

Here, force in member FA is FFA.

Write the Equation for sum of horizontal forces.

   ΣFx=0FABFAEcosθ=0

Substitute 2.08k for FAE and 45 for cosθ.

   FAB2.08k×45=0FAB=1.67k(Tension)

Here, force in member AB is FAB.

Consider the force in diagonal members BD is in tension and EC is in compression.

   FBD=FEC

Here, force in member BD and EC are FBD and FEC.

Use the method of section method and cut the section as shown in figure below and calculate the forces in the members.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip  4

Figure-(4)

Calculate the angles as shown below.

   cosθ=45sinθ=35

Write the Equation for sum of vertical forces.

   ΣFy=05.535+FBDsinθ+FECsinθ=02FBDsinθ=2.5

Substitute 35 for sinθ.

   2FBD×35=2.5FBD=2.08k(Tension)FCE=2.08(Compression)

Here, force in the member BD and CE are FBD and FCE.

Write the equation for moment about B.

   (FED×6+FECcosθ×6+4×86.5×8)kft=0

Here, force in the member ED is FED.

Substitute 45 for cosθ and 2.08k for FEC.

   (FED×6+2.08×45×6+4×86.5×8)kft=0FED×6ft=10.01kftFED=1.67k(Compression)

Write the Equation for moment about E.

   (FBC×6+FBDcosθ×66.5×8+4×8)=0

Here, force in the member BC is FBC.

Substitute 45 for cosθ and 2.08k for FBD.

   (FBC×6+2.08k×45×66.5×8+4×8)kft=0FBC×6ft=10.01kftFBC=1.67k(Tension)

Write the Equation for sum of vertical forces.

   ΣFy=0FBE+FFBsinθ+FBDsinθ=0

Substitute 35 for sinθ, 2.08k for FFB and 2.08k for FBD.

   FBE+2.08k×35+2.08k×35=0FBE+1.25k+1.25k=0FBE=2.5k(Compression)

Here, force in the member BE is FBE.

Consider the joint C.

Angles will be same as calculated at joint B.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip  5

Figure-(5)

Write the Equation for sum of vertical forces.

   ΣFy=0FCD+FECsinθ6.5k=0

Here, force in the member CD is FCD.

Substitute 35 for sinθ, and 2.08k for FEC.

   FCD+(2.08k×35)6.5k=0FCD=5.25k(Compression)

Conclusion:

Therefore, the forces in each member of the truss are shown below.

The force in member AE is, 2.08k(Compression).

The force in member BF is, 2.08k(Tension).

The force in member EF is, 1.67k(Compression).

The force in member AB is, 1.67k(Tension).

The force in member CE is, 2.08(Compression).

The force in member BD is, 2.08k(Tension).

The force in member DE is, 1.67k(Compression).

The force in member BC is, 1.67k(Tension).

The force in member AF is, 4.25k(Compression).

The force in member BE is, 2.5k(Compression).

The force in member CD is, 5.25k(Compression).

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