Assignment 2: Problem Formulation

Sittipong Jarernprasert

Send me an e-mail (sj2@andrew.cmu.edu)

Due : September 16, 1999

Problem formulation for analyzing a frame structure.

Assumption:

1. Beams are always horizontal.

2. Beams-to-beam connections are always orthogonal.

3. All beam-to-beam and beam-to-column connections are supported type. No connected type of connection.

4. Loads are applied through the shear center of the beam.

5. Loads are vertical.

6. Columns are vertical.

7. No supporting displacement when calculating all component on each beam.

8. Young’s modulus (E) and Second moment of inertia (I) of each beam is constant.

9. No moment transfer between beam-to beam and beam-to column in any cases.

Requirement for the frame sturcture:

1. Equation for solving shear force at any point on the beam (V(x)).

2. Equation for solving bending moment at any point on the beam (M(x)).

3. Equation for solving slope at any point on the beam ((q(x)).

4. Equation for solving deflection or displacement at any point on the beam (w(x)).

5. Maximum internal shear force in each beam (V _MAX).

6. Maximum internal bending moment at any point in each beam (M _MAX).

7. Maximum slope of each beam (q_MAX).

8. Maximum deflection or displacement of each beam (w_MAX).

9. Axial force on each column.

Requirement data from user for each beam.(Input):

1. Order of Beam.

2. Total length of beam.

3. Total amount of loads.

4. Magnitude and position of each load.

5. Support: each support (left and right) need this information:

5.1 Position

5.2 Type of support (Beam or column)

5.3 Order of supported beam

5.4 Position from left end of supported beam.

5.5 E, and I of beam.

Definition

frame is combined by beam and column.

The characteristics of the frame structure:

• m is the number of beams in the structure.
• Each beam is labeled with a number j where j ranges from 1 to m.
• The external loads on each beam are described as outlined above.
• The local x axis of a beam corresponds to either the global X or Y axis of the overall problems. In the global coordinate system, Z is upward from the ground datum.
• Each beam is supported by two supports, either support may be another beam or column.
• If the support is a column:
• The load is transferred to the ground.
• The support is specified as a reaction, either A or B as described above.
• The location of the support is as specified above, measured from the left end of the beam in its local coordinate system.
• If the support is another beam:
• The beam-to-beam connection is orthogonal.
• The support is specified as a reaction, either A or B as described above.
• The location of the support is as specified above, measured from the left end of the beam in its local coordinate system.
• The number of the supporting beam is specified (k, where k goes from 1 to m, k not equal to j).
• The location of where beam j connects to the supporting beam k is specified as as a distance measured from the left end of beam k as x_j,k. This is the transfer of the reaction from beam j to beam k as a load.

Beams

A beam has only two supports (either other beams or columns), and supports multiple loads, either directly applied external loads or loads from other beams. The figure below illustrates the characteristics of a typical beam.

The specific characteristics of an individual beam are:

• The origin of a local right-handed cartesian coordinate system is located at the left end of the beam.
• x is the longitudinal axis of the beam, along it's neutral axis. x is positive along the length of the beam.
• y is the vertical axis, positive upwards (local beam coordinate system).
• L is the length of the beam.
• n is the number of point loads on the beam. Loads are positive in the direction of the positive y axis.
• Each load is denoted by a subscript i, where i ranges from 1 to n.
• Each load is described by its magnitude and position pair, denoted P_i and x_i.
• The beam is simply supported; the supports denoted A and B.
• The position and value of the reactions are x_a, R_a and x_b, R_b respectively.

Basic or governing equation:

Equation and solution procedure:

1. Find the reactions at support, use the force equilibrium in Y direction and bending moment equilibrium equations to solve for the reactions.

; We will get the equation for forces: When N is the total amount of external point load.

; We will get the equation for bending moment:

After we solve this system equation, we will get the equation for the reaction R_a and R_b. The equation for R_a and R_b can find at equation (1) and (2) in all equation part.

2. Find the internal shear force in any i^th section , we cut at point x from the left end and use use the force equilibrium in Y direction equation to find shear force. We will get an equation for shear(3).

The load P_j include both reaction R_a and R_b, M is N+2, and i is counted from the left end.

3. Find the internal bending moment in any section i, we cut at point x from the left end and use the bending moment equilibrium equation to find bending moment. We will get the equation for moment(4).

The load P_j include both reaction R_a and R_b, M is N+2, and i is counted from the left end.

4. Find the slope at any point in i range on the beam, we have to use this equation: . When we solve this differential equation or integrate both sides of this equation by substituting Mi(x), we will get the equation for slope(5.).

5. Find the displacement of deflection at any point on the beam in i^th range, we have to use this equation: . When we substitute q_i(x) to this equation and solve this differential equation by integrate both sides of this equation, we will get the equation for deflection(6).

6. For C and D, in each interval (xi,xi-1) we will get different Ci and Di. However, when the piecewise function is a continuing piecewise function, we can find the relationship between C_i and C_i-1, Di and D_i-1 by using the continuity condition that at any i^th connect point, slope and deflection in interval (xi-1,xi) and (xi,xi+1) have to be equal. From this condition, we will find that

C_i-1 = C_i = C_i+1 = C and D_i-1 = D_i = D_i+1 = D ; i = 1,2,…,M+1

7. For C and D we can get by solve the equation system from boundary condition that deflection or displacement at support is equal to zero(0). From this boundary condition, we will get this equation system: ; ; After we solve this equation system we will get the equation for C(7) and D(8)

8. Find point which has maximum or minimum slope in i^th range on the beam, we can find this point by differentiating q(x) and set it to zero (), and then find all x that satisfy this equation. We will get the equation for x(9).

9. Find point which has maximum or minimum deflection or displacement in i^th range on the beam, we can find this point by differentiating w(x) and set it to 0 (), then find all x that satisfy this equation. We will get the equation for x(10).

Procedure of calculation:

For each beam:

1. Transfer the reaction from above beam which was supported by this beam by checking an order of the support beam of all beams that was calculated before this beam. If order of this beam is equal to order of which beam that is calculated before from current calculating order to the first calculating order, then assign P_N+1 = -R and x_N+1 = x of that support of the first supported beam, and P_N+2 and x_N+2 for the second. Do this until the current calculating order equal 1. Next, change the last order of P and x to the new N.

2. Set P₀ = 0 , x₀ = 0 , M₀ = 0

3. Find reaction R_A and R_B from the equation (1) and (2) in all equation part.

4. Assign order of left reaction to order N+1 and right reaction to order N+2

5. Sort order of load and position by position from left to right from order 1 to N+2, and assign the order from sorting to the new order of each load and position from 1 to M. So M=N+2

6. Find internal shear force in each range of x. Shear force in each length is a constant because of a point load. Use equation(11)

7. Minimum and maximum internal shear force is the maximum of internal shear force in each range of x. Find the minimum and maximum of internal shear force from the equation(12) and (13).:

Maximum magnitude of internal shear is the greater between V_MAX and V_MIN

8. Find internal bending moment at the end point in each range of x. Bending in each range is a linear function because of a constant internal shear force. Therefore, the maximum and minimum of bending moment occur at the end of each length. Use equation(14).

9. Minimum and Maximum internal bending moment is the maximum of internal bending moment at the end point in each range of x. Find the minimum and maximum of internal bending moment from the equation (15) and (16).

Maximum magnitude of internal shear is the greater between M_MAX and M_MIN

10. Calculate the constant C and D to find slope and displacement by this equation (7) and (8)

11. For each range of x, maximum and minimum of slope (q) can be found at the point that have maximum or minimum value and at the boundary point of each range of x. The maximum of q on the beam is the maximum magnitude of all maximum and minimum in all range of x on the beam. Use equation (17).

12. For each range of x, maximum and minimum of displacement (w) can be found at the point that have maximum or minimum value and at the boundary point of each range of x. The maximum of w on the beam is the maximum magnitude of all maximum and minimum in all range of x on the beam. Use equation (18).

For each column:

1. Find all beam that was supported by this column and then find which support of this beam supported by this column.

2. Axial force for this column is summation of all reactions of beam supported by this column.

All equation:

1.

2.

3. ; ;

4. ; ;

5. ; ;

6. ; ;

7.

8.

where k₁ = order of position at x=x_a in range 1 to M+1

k₂ = order of position at x=x_b in range 1 to M+1

9. ; ;

10. ; ;

11. ; ;

12. ; ;

13. ; ;

14. ; ;

15. ; ;

16. ; ;

17. ; ;

18. ; ;