Structural Analysis (10th Edition)
Structural Analysis (10th Edition)
10th Edition
ISBN: 9780134610672
Author: Russell C. Hibbeler
Publisher: PEARSON
Question
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Chapter 16, Problem 16.1P
To determine

The structure stiffness matrix for the frame.

Expert Solution & Answer
Check Mark

Answer to Problem 16.1P

The structure stiffness matrix for the frame is shown below.

   k=[4833.33004833.33000000130.97854.170130.97854.1700007854.17628333.307854.17314166.670004833.33004909.0105454.2875.75405454.280130.97854.1704338.697854.1704207.78007854.17314166.675454.287854.1711519144.415454.280261805.5600075.75405454.2875.75405454.2800004207.78004207.7800005454.280261805.565454.280523611.11]

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 global stiffness matrix for the member.

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

Calculation:

The free body diagram of the frame is shown below.

Structural Analysis (10th Edition), Chapter 16, Problem 16.1P

          Figure (1)

Consider the member-1.

Calculate direction cosine in x-direction.

Substitute 10 ft for xF, 0 ft for xN and 10 ft for L in Equation (I).

λx=10 ft0 ft10 ft=1

Calculate direction cosine in y-direction.

Substitute 0 ft for yF, 0 ft for yN and 10 ft for L in Equation (II).

λy=0 ft0 ft10 ft=0

The stiffness matrix for member-1 is shown below.

Substitute 1 for λx and 0 for λy in global stiffness matrix.

k1=[AEL00AEL00012EIL36EIL2012EIL36EIL206EIL24EIL06EIL22EILAEL00AEL00012EIL36EIL2012EIL36EIL206EIL22EIL06EIL24EIL]          ...... (III)

Calculate the value of AEL in stiffness matrix.

Substitute 20 in2 for A, 29×103 ksi for E and 10 ft for L.

AEL=(20 in2)×(29×103 ksi)(10×12)in=4833.33 kips/in

Calculate the value of 6EIL2 in stiffness matrix.

Substitute 29×103 ksi for E, 650 in4 for I and 10 ft for L.

6EIL2=6×(29×103 ksi)×(650 in4)(10×12)2in2=7854.17 kips

Calculate the value of 12EIL3 in stiffness matrix.

Substitute 29×103 ksi for E, 650 in4 for I and 10 ft for L.

12EIL3=12×(29×103 ksi)×(650 in4)(10×12)3in3=130.9 kips/in

Calculate the value of 4EIL in stiffness matrix.

Substitute 29×103 ksi for E, 650 in4 for I and 10 ft for L.

4EIL=4×(29×103 ksi)×(650 in4)(10×12)in=628333.3 kip-in

Calculate the value of 2EIL in stiffness matrix.

Substitute 29×103 ksi for E, 650 in4 for I and 10 ft for L.

2EIL=2×(29×103 ksi)×(650 in4)(10×12)in=314166.67 kip-in

Substitute 4833.33 kips/in for AEL, 7854.17 kips for 6EIL2, 130.9 kips/in for 12EIL3, 628333.3 kip-in for 4EIL and 314166.67 kip-in for 2EIL in matrix (III). k1=[4833.33004833.33000130.97854.170130.97854.1707854.17628333.3628333.37854.17314166.674833.33004833.33000130.97854.170130.97854.1707854.17314166.67314166.677854.17628333.3]          ...... (IV)

Consider the member-2.

Calculate direction cosine in x-direction.

Substitute 0 ft for xF, 0 ft for xN and 12 ft for L in Equation (I)

λx=0ft0ft12ft=0

Calculate direction cosine in y-direction.

Substitute 0 ft for yF, 12 ft for yN and 12 ft for L in Equation (II)

λy=0ft(12ft)12ft=1

The stiffness matrix for member-1 is shown below.

Substitute 0 for λx and 1 for λy in global stiffness matrix.

   k2=[12EIL306EIL212EIL306EIL20AEL00AEL06EIL204EIL6EIL202EIL12EIL306EIL212EIL306EIL20AEL00AEL06EIL202EIL6EIL204EIL]          ...... (V)

Calculate the value of AEL in stiffness matrix.

Substitute 20 in2 for A, 29×103 ksi for E and 12 ft for L.

AEL=(20 in2)×(29×103 ksi)(12×12)in=4027.78 kips/in

Calculate the value of 6EIL2 in stiffness matrix.

Substitute 29×103 ksi for E, 650 in4 for I and 12 ft for L.

6EIL2=6×(29×103 ksi)×(650 in4)(12×12)2in2=5454.28 kips

Calculate the value of 12EIL3 in stiffness matrix.

Substitute 29×103 ksi for E, 650 in4 for I and 12 ft for L.

12EIL3=12×(29×103 ksi)×(650 in4)(12×12)3in3=75.754 kips/in

Calculate the value of 4EIL in stiffness matrix.

Substitute 29×103 ksi for E, 650 in4 for I and 12 ft for L.

4EIL=4×(29×103 ksi)×(650 in4)(12×12)in=523611.11 kip-in

Calculate the value of 2EIL in stiffness matrix.

Substitute 29×103 ksi for E, 650 in4 for I and 12 ft for L.

2EIL=2×(29×103 ksi)×(650 in4)(12×12)in=261805.56 kip-in

Substitute 4027.78 kips/in for AEL, 5454.28 kips for 6EIL2, 75.754 kips/in for 12EIL3, 523611.11 kip-in for 4EIL and 261805.56 kip-in for 2EIL in matrix (III).

   k2=[75.75405454.2875.75405454.2804027.78004027.7805454.280523611.115454.280261805.5675.75405454.2875.75405454.2804027.78004027.7805454.280261805.565454.280523611.11]          ...... (VI)

The final matrix of the frame by assembling matrices (III) and (IV) is shown below.

   k=[4833.33004833.33000000130.97854.170130.97854.1700007854.17628333.307854.17314166.670004833.33004909.0105454.2875.75405454.280130.97854.1704338.697854.1704207.78007854.17314166.675454.287854.1711519144.415454.280261805.5600075.75405454.2875.75405454.2800004207.78004207.7800005454.280261805.565454.280523611.11]

Conclusion:

The structure stiffness matrix for the frame is shown below.

   k=[4833.33004833.33000000130.97854.170130.97854.1700007854.17628333.307854.17314166.670004833.33004909.0105454.2875.75405454.280130.97854.1704338.697854.1704207.78007854.17314166.675454.287854.1711519144.415454.280261805.5600075.75405454.2875.75405454.2800004207.78004207.7800005454.280261805.565454.280523611.11]

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