STRUCTURAL ANAL.W/MOD MASTERING ACCE
STRUCTURAL ANAL.W/MOD MASTERING ACCE
18th Edition
ISBN: 9780134713649
Author: HIBBELER
Publisher: PEARSON
Question
100%
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Chapter 14, Problem 14.1P
To determine

The structure stiffness matrix for the truss.

Expert Solution & Answer
Check Mark

Answer to Problem 14.1P

The structure stiffness matrix for the truss is shown below.

k=AE[0.35360.35360.35360.353600000.35360.85360.35360.35360000.50.35360.35361.06080.35360.35360.35360.35360.35360.35360.35360.35361.06080.35360.35360.35360.3536000.35360.35360.85360.35360.50000.35360.35360.35360.353600000.35360.35360.500.85360.353600.50.35360.3536000.35360.8536]

Explanation of Solution

Concept Used:

Write the expression for direction cosine in x-direction.

λx=xFxNL          ...... (I)

Here, coordinate of x at near end of the member is xF and coordinate of x at far end of the member is xN.

Write the expression for direction cosine in x-direction.

λy=yFyNL          ...... (II)

Here, coordinate of y at near end of the member is yF and coordinate of y at far end of the member is yN.

Write the stiffness matrix for the member.

AEL[λx2λxλyλx2λxλyλxλyλy2λxλyλy2λx2λxλyλx2λxλyλxλyλy2λxλyλy2]

Here, cross-sectional area of the member is A, length of the member is L, modulus of elasticity is E and moment of inertia is I.

Write the expression for total structural stiffness matrix.

k=k1+k2+k3+k4+k5          ...... (III)

Here, the structural stiffness matrix is k.

Calculation:

The free body diagram of the truss is shown below.

STRUCTURAL ANAL.W/MOD MASTERING ACCE, Chapter 14, Problem 14.1P

          Figure (1)

Consider the member-1.

Calculate the length of member 1.

L=(1 m)2+(1 m)2=2 m

Calculate direction cosine in x-direction.

Substitute 1 m for xF, 0 m for xN and 2 m for L in Equation (I).

λx=1 m0 m2 m=0.70711

Calculate direction cosine in y-direction.

Substitute 1 m for yF, 2 m for yN and 2 m for L in Equation (II).

λy=1 m2 m2 m=0.70711

The stiffness matrix for member-1 is shown below.

Substitute 2m for L, 0.70711 for λx and 0.70711 for λy in the stiffness matrix.

   k1=AE2[(0.70711)2(0.70711)(0.70711)(0.70711)2(0.70711)(0.70711)(0.70711)(0.70711)(0.70711)2(0.70711)(0.70711)(0.70711)2(0.70711)2(0.70711)(0.70711)(0.70711)2(0.70711)(0.70711)(0.70711)(0.70711)(0.70711)2(0.70711)(0.70711)(0.70711)2]

k1=AE[0.35360.35360.35360.35360.35360.35360.35360.35360.35360.35360.35360.35360.35360.35360.35360.3536]1234          ...... (IV)

Consider the member- 2.

Calculate the length of member 2.

L=(1 m)2+(1 m)2=2 m

Calculate direction cosine in x-direction.

Substitute 2 m for xF, 1 m for xN and 2 m for L in Equation (I).

λx=2 m1 m2 m=0.70711

Calculate direction cosine in y-direction.

Substitute 0 m for yF, 1 m for yN and 2 m for L in Equation (II).

λy=0 m1 m2 m=0.70711

The stiffness matrix for member-2 is shown below.

Substitute 2m for L, 0.70711 for λx and 0.70711 for λy in stiffness matrix.

   k2=AE2[(0.70711)2(0.70711)(0.70711)(0.70711)2(0.70711)(0.70711)(0.70711)(0.70711)(0.70711)2(0.70711)(0.70711)(0.70711)2(0.70711)2(0.70711)(0.70711)(0.70711)2(0.70711)(0.70711)(0.70711)(0.70711)(0.70711)2(0.70711)(0.70711)(0.70711)2]

k2=AE[0.35360.35360.35360.35360.35360.35360.35360.35360.35360.35360.35360.35360.35360.35360.35360.3536]3456          ...... (V)

Consider the member- 3.

Calculate direction cosine in x-direction.

Substitute 0 m for xF, 2 m for xN and 2 m for L in Equation (I).

λx=0 m2 m2 m=1

Calculate direction cosine in y-direction.

Substitute 0 m for yF, 0 m for yN and 2 m for L in Equation (II).

λy=0 m0 m2 m=0

The stiffness matrix for member-3 is shown below.

Substitute 2m for L, 1 for λx and 0 for λy in stiffness matrix.

k3=AE2[(1)2(1)(0)(1)2(1)(0)(1)(0)(0)2(1)(0)(0)2(1)2(1)(0)(1)2(1)(0)(1)(0)(0)2(1)(0)(0)2]=AE[0.500.5000000.500.500000]5678          ...... (VI)

Consider the member- 4.

Calculate the length of member 4.

L=(1 m)2+(1 m)2=2 m

Calculate direction cosine in x-direction.

Substitute 0 m for xF, 1 m for xN and 2 m for L in Equation (I).

λx=0 m1 m2 m=0.70711

Calculate direction cosine in y-direction.

Substitute 0 m for yF, 1 m for yN and 2 m for L in Equation (II).

λy=0 m1 m2 m=0.70711

The stiffness matrix for member-4 is shown below.

Substitute 2m for L, 0.70711 for λx and 0.70711 for λy in stiffness matrix.

   k4=AE2[(0.70711)2(0.70711)(0.70711)(0.70711)2(0.70711)(0.70711)(0.70711)(0.70711)(0.70711)2(0.70711)(0.70711)(0.70711)2(0.70711)2(0.70711)(0.70711)(0.70711)2(0.70711)(0.70711)(0.70711)(0.70711)(0.70711)2(0.70711)(0.70711)(0.70711)2]3478

k4=AE[0.35360.35360.35360.35360.35360.35360.35360.35360.35360.35360.35360.35360.35360.35360.35360.3536]3478          ...... (VII)

Consider the member- 5.

Calculate direction cosine in x-direction.

Substitute 0 m for xF, 0 m for xN and 2 m for L in Equation (I).

λx=0 m0 m2 m=0

Calculate direction cosine in y-direction.

Substitute 0 m for yF, 0 m for yN and 2 m for L in Equation (II).

λy=0 m2 m2 m=1

The stiffness matrix for member-5 is shown below.

Substitute 2m for L, 0 for λx and 1 for λy in stiffness matrix.

k5=AE2[(0)2(1)(0)(0)2(1)(0)(1)(0)(1)2(1)(0)(1)2(0)2(1)(0)(0)2(1)(0)(1)(0)(1)2(1)(0)(1)2]=AE[000000.500.5000000.500.5]4278          ...... (VIII)

Calculate total structural stiffness matrix.

Substitute the values of k1, k2, k3, k4 and k5 in Equation (III).

k=AE[0.35360.35360.35360.353600000.35360.85360.35360.35360000.50.35360.35361.06080.35360.35360.35360.35360.35360.35360.35360.35361.06080.35360.35360.35360.3536000.35360.35360.85360.35360.50000.35360.35360.35360.353600000.35360.35360.500.85360.353600.50.35360.3536000.35360.8536]

Conclusion:

The structure stiffness matrix for the truss is shown below.

k=AE[0.35360.35360.35360.353600000.35360.85360.35360.35360000.50.35360.35361.06080.35360.35360.35360.35360.35360.35360.35360.35361.06080.35360.35360.35360.3536000.35360.35360.85360.35360.50000.35360.35360.35360.353600000.35360.35360.500.85360.353600.50.35360.3536000.35360.8536]

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