Structural Analysis (10th Edition)
Structural Analysis (10th Edition)
10th Edition
ISBN: 9780134610672
Author: Russell C. Hibbeler
Publisher: PEARSON
bartleby

Concept explainers

Question
Book Icon
Chapter 15, Problem 15.1P
To determine

The reactions at supports

Expert Solution & Answer
Check Mark

Answer to Problem 15.1P

The support reaction Q4 is 17.495kN.

The support reaction Q5 is 7.5kN.

The support reaction Q6 is 5kN-m.

Explanation of Solution

Given:

The flexural rigidity, EI of the beam is constant.

Concept Used:

Write the expression for fixed end moments in a beam with uniformly distributed load.

   FEM=wl212 ...... (I).

Here, the fixed end moment is FEM, the load on the beam is w and the length of the beam is l.

Write the expression for fixed end moments in a beam with point load at mid-span.

   FEM=wl8 ...... (II)

Write the expression for load matrix.

   Q=kD ...... (III)

Here, Q is the load matrix, k is the stiffness and D is the displacement.

Calculations:

Calculate the fixed end moment at 1 using Equation (I).

Substitute 5kN/m for w and 2m for l in Equation (I).

   FEM1=(5 kN/m)×( 2m)212=(20kN-m)12=1.67kN-m

Write the force matrix from the known force.

   Q=[51.6671.667Q45Q5Q6]

Write the displacement matrix for the known displacement.

   D=[D1D2D3000]

Develop the stiffness matrix.

   k=EI[ 12 L 3 6 L 2 12 L 3 6 L 2 6 L 2 4L6 L 2 2L 12 L 3 6 L 2 12 L 3 6 L 2 6 L 2 2L6 L 2 4L]

Calculate the member stiffness matrix for member 1.

   k1=EI[ 12 L 3 6 L 2 12 L 3 6 L 2 6 L 2 4L6 L 2 2L 12 L 3 6 L 2 12 L 3 6 L 2 6 L 2 2L6 L 2 4L]1243 ...... (IV)

Substitute 2m for L in Equation (IV).

   k1=EI[ 12 2 3 6 2 2 12 2 3 6 2 2 6 2 2 4 2 6 2 2 2 2 12 2 3 6 2 2 12 2 3 6 2 2 6 2 2 2 2 6 2 2 4 2 ]1243=EI[ 1.5 1.5 1.5 1.5 1.52 1.51 1.5 1.5 1.5 1.5 1.51 1.52]1243

Calculate the member stiffness matrix for member 2.

   k2=EI[ 12 L 3 6 L 2 12 L 3 6 L 2 6 L 2 4L6 L 2 2L 12 L 3 6 L 2 12 L 3 6 L 2 6 L 2 2L6 L 2 4L]4356 ...... (V)

Substitute 2m for L in Equation (V).

   k2=EI[ 12 2 3 6 2 2 12 2 3 6 2 2 6 2 2 4 2 6 2 2 2 2 12 2 3 6 2 2 12 2 3 6 2 2 6 2 2 2 2 6 2 2 4 2 ]4356=EI[ 1.5 1.5 1.5 1.5 1.52 1.51 1.5 1.5 1.5 1.5 1.51 1.52]4356

Calculate the stiffness matrix for the beam.

   K=k1+k2 ...... (VII)

Substitute the values of k1 and k2 in Equation (VII).

   k=EI[ 1.5 1.5 1.5 1.5 1.52 1.51 1.5 1.5 1.5 1.5 1.51 1.52]1243+EI[ 1.5 1.5 1.5 1.5 1.52 1.51 1.5 1.5 1.5 1.5 1.51 1.52]4356k=EI[ 1.5 1.5 1.5 1.500 1.521 1.500 1.5140 1.51 1.5 1.503 1.5 1.500 1.5 1.5 1.5 1.5001 1.5 1.52]

Substitute the values of stiffness matrix and displacement matrix in Equation (III).

   [51.6671.667Q45Q5Q6]=EI[1.51.51.51.5001.5211.5001.51401.511.51.5031.51.5001.51.51.51.50011.51.52][D1D2D3000]

Solve the above matrix to get the values of displacement.

   5=EI(1.5D1+1.5D2+1.5D3) ...... (VIII)

   1.667=EI(1.5D1+2D2+D3) ...... (IX)

   1.667=EI(1.5D1+D2+4D3) ...... (X)

Solve Equations (VIII), (IX) and (X).

   D1=19.99EID2=11.6EID3=5EI

Calculate the value of Q4.

   Q45=EI(1.5D11.5D2) ...... (XI)

Substitute 19.99EI for D1 and 11.6EI for D2 in Equation (XI).

   Q45=EI((1.5× 19.99 EI)(1.5× 11.6 EI))Q4=(12.495+5)kNQ4=17.495kN

Calculate the value of Q5.

   Q5=EI(1.5D3) ...... (XII)

Substitute 5EI for D3 in Equation (XII).

   Q5=(EI(1.5× 5 EI))kNQ5=7.5kN

Calculate the value of Q6.

   Q6=EI(D3) ....... (XIII)

Substitute 5EI for D3 in Equation (XIII).

   Q6=(EI( 5 EI))kN-mQ6=5kN-m

Conclusion:

The support reaction Q4 is 17.495kN.

The support reaction Q5 is 7.5kN.

The support reaction Q6 is 5kN-m.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
For the truss shown in Fig 2, determine the nodal displacement and member forces using the stifness method for all elements of the truss. Assume for each member A = 0.0015 m2 and E = 200 GPa
5. Two 400 g blocks are connected by a rigid rod. The Blocks can rotate freely at the ends of the rod, so the rod does not apply any moments to the blocks. The blocks are in contact with the wall and floor, and can slide without friction. The system is released from rest when X=24cm and Y=18cm. Ignore the mass of the rod. What are the initial accelerations of Block A and Block B just after being released? Hint: See Assignment 4, Problem 2 for help getting a relationship between the acceleration of Block A and the acceleration of Block B. Y m = 400 g A | L g I 1 I I X B m = 400 g
The momentum of the force F = -100 i -70 j + 50 k around the point O is MO = 410 i- 300 j + 400 k, Determine the coordinates of the point through which the line of actionof F intercepts the yz plane.
Knowledge Booster
Background pattern image
Civil Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Structural Analysis
Civil Engineering
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:Cengage,
Text book image
Structural Analysis (10th Edition)
Civil Engineering
ISBN:9780134610672
Author:Russell C. Hibbeler
Publisher:PEARSON
Text book image
Principles of Foundation Engineering (MindTap Cou...
Civil Engineering
ISBN:9781337705028
Author:Braja M. Das, Nagaratnam Sivakugan
Publisher:Cengage Learning
Text book image
Fundamentals of Structural Analysis
Civil Engineering
ISBN:9780073398006
Author:Kenneth M. Leet Emeritus, Chia-Ming Uang, Joel Lanning
Publisher:McGraw-Hill Education
Text book image
Sustainable Energy
Civil Engineering
ISBN:9781337551663
Author:DUNLAP, Richard A.
Publisher:Cengage,
Text book image
Traffic and Highway Engineering
Civil Engineering
ISBN:9781305156241
Author:Garber, Nicholas J.
Publisher:Cengage Learning