KAJIDAYA BAHAN: FIXED COLUMNS

MECHANICS - CASE STUDY

Introduction

Basic Steel Frame Structure

Center Column Joint at Top
Click to view movie (72k)

A simple, four bay office building is being considered for additional equipment storage facility at the construction company where you work. The company would like to use the steel I-beams currently in storage at the company for the center columns. The list of available I-beams are given in the table at the left.

What is known:

Young's modulus, E, is 200 GPa.
Dead load on the roof is 300 kg/m².
Live load on the roof is 250 kg/m².
A factor of safety of 2.5 is required for all loads.
Columns are rigidly fixed to the ground.
Columns are assumed to be fixed at the top due to the rigid beams.
Yield stress in compression is 250 MPa.

Question

What I-beam should be specified for the center column to minimize the total weight of the building? Chose only from the list at the left.

Approach

Section Number
W310 x 67	W250 x 58
x 39	x 45
x 33	x 28
x 24	x 22
x 21	x 18
W200 x 59	W150 x 37
x 46	x 30
x 36	x 22
x 22	x 24

Wide-Flange Beams Available
in the Storage Yard

Roof loads, dead and live, need to be combined for total design load.
Roof load needs to be distributed to each of the columns.
Assume failure will be buckling.
Use Euler's formula for fixed-fixed columns.

MECHANICS - THEORY

Column Buckling

Four Basic Column Configurations

Besides the basic pinned-pinned (simply supported) column, there are other types of column boundary conditions. A column can be fixed at one or both ends, and if one end is fixed, then the other can be free.
Each of these basic column types (shown at the left) can be solved in a similar fashion as the pinned-pinned column by developing a differential equation and then solving it using the end conditions. (The derivations are omitted since they are similar to the simply support column case presented previously.)
The basic column types are given below with their critical load equation.

Pinned - Pinned	Free - Fixed	Fixed - Fixed	Pinned - Fixed

Buckling Equations for Critical Load P_cr

Pinned - Pinned	Free - Fixed	Fixed - Fixed	Pinned - Fixed
L_e = L	L_e = 2 L	L_e= 0.5L	L_e= 0.7L

Equivalent (or Effective) Lengths

Alternate Buckling Equation Formulation

The four buckling equations given above can be summarized into one equation using the concept of equivalent length (sometimes referred to as effective length). Each of the four buckled column types have a similar buckling shape for a given part of their length. These similar shapes are noted in the diagram as Equivalent Length, L_e. Using this equivalent length, L_e, a single buckling equation can be used for all four column types.

Slenderness Ratio, L_e/r

For Short Column (Low Slenderness
Ratio) Failure May be Compression

As a column height is reduced, the critical buckling load increases. This relationship can be seen if the basic Euler formula, P_cr = π²EI/L², is graphed. To make it easier to plot this function, it can be rewritten using the definition of the radius of gyration, r, which is
r = (I/A)^0.5 [note, this is not the radius]
Substituting, gives

This function is only accurate in predicting buckling for slender columns. While each material is different, a slenderness ratio of 100 or greater will generally fail due to buckling. For columns with a slenderness ratio less than 100, the material may fail in compression and the yield stress must be check in addition to buckling.
As demonstrated with the simulation on the left, the value of E, stiffness, has little effect on the general shape of the Euler curve.

MECHANICS - CASE STUDY SOLUTION

Building Load

This case requires the center column for a basic building be designed to carry the roof load. The roof load has a dead and live roof load of, 300 and 250 kg/m², respectively. Generally, a dead load is the static, non-changing load such as roof weight and equipment on the roof. On the other hand, a live load is changing loads, such as snow or wind. The total (worst case) should be considered when designing the column.
The other major condition is that the selection of the center column needs to be specified from a group of I-beams that are in storage. The best column will be the lightest column that can withstand the roof load. Both buckling and compression failure should be checked.

Center Column Load

Roof Load Area
Carried by Center Column

The center column will need to carry the roof load that is half way to each of the other columns. The total roof area carried by the center column is 10 m by 8 m as shown in the diagram at the left. The total roof load over this area is
     F = (10 m)(8 m)(300 + 250 kg/m²)(9.81 m/s²)
        = 431.6 kN
The last term is the standard gravitational constant.
The column requires a factor of safety of 2.5, so the design load needs to be increased by a factor of 2.5, giving
     P_cr = 2.5 (431.6 kN) = 1.079 MN

Required Moment of Inertia

Center Column Load

Both Directions must be
Considered for Buckling

The minimum moment of inertia is needed so that a suitable wide-flange I-beam can be chosen. Since both ends are assumed to be fixed, the Euler buckling equation is

Substituting known values give,

Solving for the moment of inertia,
I = 6.696 × 10^-6 m⁴ = 6.696 × 10⁶ mm⁴
The I-beam must have this moment of inertia (or greater) in both direction. Generally, I-beams have a higher inertia around the x-axis, but buckling can occur about either axis.

Column Selection

There are currently 18 different wide-flange I-beams available for the construction of the building. They are list below.

Section Number	Weight	Area mm²	I_x 10⁶ mm⁴	I_y 10⁶ mm⁴
W310 x 67	67	8,530	145	20.7
x 39	39	4,930	84.8	7.23
x 33	33	4,180	65.0	1.92
x 24	24	3,040	42.8	1.16
x 21	21	2,680	37.0	0.986
W250 x 58	58	7,400	87.3	18.8
x 45	45	5,700	71.1	7.03
x 28	28	3,620	39.9	1.78
x 22	22	2,850	28.8	1.22
x 18	18	2,280	22.5	0.919
W200 x 59	59	7,580	61.2	20.4
x 46	46	5,890	45.5	15.3
x 36	36	4,570	34.4	7.64
x 22	22	2,860	20.0	1.42
W150 x 37	37	4,730	22.2	7.07
x 30	30	3,790	17.1	5.54
x 22	22	2,860	12.1	3.87
x 24	24	3,060	13.4	1.83

Wide-Flange Beams Available

Both I_x and I_y must be at least 6.696 × 10⁶ mm⁴ to satisfy the buckling requirements. The critical moment of inertia is I_y. There are several beams that have moment of inertia's greater than 6.696 in both directions. However, the lightest one is
W200 x 36

Compression Stress Check

Even though the column was designed assuming buckling, the compression stress should be checked to make sure it does not exceed the yield stress of the material. The compression stress is

σ = P/A = (1.079 MN)/(4,570 mm²)

= 236.1 MPa

This is less than the yield stress of 250 MPa for structural steel and will not yield in compression.

MECHANICS - EXAMPLE
Support Structure		Example
		A mechanical equipment support structure is made from three metal members, each 10 mm by 10 mm. The vertical members, CA and DB, are fixed to the ground. The horizontal member, AB, is pinned to the vertical members at A and B. If the equipment is expected to placed 10 cm from the the left, as shown, what is the maximum load F? Assume all columns are solid steel bars with a stiffness, E, of 200 GPa and a allowable stress of 400 MPa.

		Solution
Basic W8x31 I-Beam		With the slender vertical columns, 10 mm by 10 mm, it would be expect this structure to fail due to column buckling. However, it could also fail due to beam bending or column crushing. These other possible failure modes will be checked after load P is determined for just column buckling. Both columns, CA and DB, can be considered as fixed-pinned columns that have a critical buckling as P_cr = 2.046 π²EI / L² Moment of inertia for both columns is, I = ab³/ 12 = (0.01)(0.01)³ / 12 = 8.333x10^-10 m⁴ The allowable buckling load for each column is P_CA = 2.046 π² (200x10⁹) (8.333x10^-10)/0.25² = 53.85 kN P_BD = 2.046 π² (200x10⁹) (8.333x10^-10)/0.35² = 27.47 kN

Top Beam Section		Now, the critical buckling loads can be used to find F. First, the P_CA will be used to find F, and then P_BD. Taking moments about A gives, (10)F = P_BD(30) = 27.47(30) P = 82.41 kN Next, using P_BD and taking moments about B gives, (20)F = P_AC(30) = 53.85(30) F = 80.78 kN Thus the largest load F before either column will be buckle when F = 80.78 kN.

Maximum Crushing Load, 40 kN, in Column CA		Check column normal stress (crushing stress): σ = P/A = 80.78/0.01² = 807 MPa There is a problem. This is higher than the allowable of 400 MPa. It will fail due to crushing before buckling. The maximum column load is, P = (400 MPa) (0.01 m) (0.01 m) = 40 kN This means the largest F (moments about point B) will be 20(F) = 40(30) F = 60 kN

Maximum Bending Moment		Check beam bending stress: The largest bending moment due to a load of F = 60 kN (maximum due to column crushing) will be at the location of the load. M_max = 40(10) = 400 kN-m The bending stress will be σ = My/I = (400 kN-m) (0.005 m) / 8.333x10^-10 m⁴ = 2,400,000 MPa

		There is another serious problem. This is much higher than the allowable of 400 MPa. It will fail due to beam bending before buckling or column crushing. The maximum moment is, 400 MPa = M (0.005 m) / 8.333x10^-10 m⁴ M = 66.66 N-m This moment will be generated a reaction at A of P_A = (66.66 N-m)/ (0.1 m) = 666.66 N Thus, the maximum load F will be F (20) = 666.6 (30) F = 999.9 N = 1.0 kN Thus, the structure will fail due to beam bending before column buckling or column crushing. It is important to check all failure modes.

Thursday, 7 July 2011

FIXED COLUMNS