Structural analysis methods and applications

Structural Analysis

Scenario

Objective of this project is to design an internal working structure a-B-C-D-E-F to support extractor fans as well as other equipment. The structure inside the building is a brickwork at 10 miles south of Derbyshire and Derbyshire. The proposed design is to extend beam AB and CD into the substantial bricks walls at a and D. Moreover, the BE and CF are to be located by pins at ground level. The arrangement as well as associated loading is to be repeated at 10 m intervals indicated within the plan view. (Lightfoot, 1961).

The project constructs a moment distribution table and determines the moments at the joints to an acceptable level of accuracy.

Moment Distribution Table

Moment distribution is an interactive method to solve an indeterminate structure. Moment distribution is also a mechanical process to deal with statically intermediary structure by the successive approximation within the moments in the structure. Typically, the moment distribution is a preferred calculation to reinforce a concrete structure. (Volokh, 2002). To prepare the moment distribution table for this project, it is critical to analyze every joint of the structure to develop the fix-end moments. Analysis of the fixed-end moment of the project reveals that the fixed-end joints are not equilibrium. The goal of the project is to ensure that the fixed-end moments are released and the accuracy and equilibrium are achieved. Thus, the moment distribution method is a process of solving the set of simultaneous equation using the iteration technique to arrive at the acceptable level of accuracy of the structure. To apply moment distribution to make the joints to arrive at an acceptable level accuracy, the paper constructs Moment Distribution Table as being revealed in Table 1:

Table 1: Moment Distribution Table

I

L (m)

k=I/L

D

LL

self w_u

carry over

3309,519

3062,208

1880,912

947,9519

7885,906

11179,54

12314,61

13410,36

6161,137

-0,5

0

0

balance

-3309,52

0

-2965,87

-1131,55

-3246,98

-7007,72

-14858,6

-12862,5

-10781,3

-3696,38

0,275671

0

carry over

0

-1482,94

-565,773

-1623,49

-3503,86

-7429,31

-6431,24

-5390,67

-1848,19

0,137836

0

0

balance

0

0

1229,225

875,7058

1884,615

4018,607

8765,949

5910,958

3987,674

1108,832

-0,07599

0

0

carry over

614,6127

437,8529

942,3074

2009,303

4382,974

2955,479

1993,837

554,4159

-0,038

0

0

0

balance

-614,613

0

-828,096

-1180,64

-2349,55

-2697,33

-3130,14

-1274,13

-305,39

0,022798

0

0

0

carry over

218,9264

471,1537

1004,652

2191,487

1477,739

996,9184

277,208

-0,019

0

0

0

0

balance

-218,926

0

-885,483

-1278,46

-1348,67

-909,588

-805,807

-138,594

0,010466

0

0

0

0

carry over

-442,742

-639,228

-674,333

-454,794

-402,903

-69,2972

0,005233

0

0

0

0

0

balance

442,7416

0

788,1363

451,6507

315,2563

173,5627

43,82296

-0,00262

0

0

0

0

0

carry over

394,0681

225,8253

157,6282

86,78133

21,91148

-0,00131

0

0

0

0

0

0

balance

-394,068

0

-230,072

-97,7638

-39,9513

-8,05333

0,000827

0

0

0

0

0

0

carry over

-1150,36

-488,819

-199,756

-40,2666

0,004137

0

0

0

0

0

0

0

balance

1150,361

0

413,1452

96,00917

14,79893

-0,00152

0

0

0

0

0

0

0

carry over

206,5726

48,00458

7,399466

-0,00076

0

0

0

0

0

0

0

0

balance

-206,573

0

-33,2424

-2,95948

0,000279

0

0

0

0

0

0

0

0

SUM

3535,136

-3535,14

5330,502

-3080,4

-3431,77

6508,857

18692,22

-2205,93

1664,013

-889,46

-13725,1

0,275671

0

Based on the moment distribution table, the paper presents bending moment diagram in the next section.

Bending Moment Diagram

A Bending Moment Diagram (BMD) is an analytical tool used in conjunction with structural analysis to assist in the structural design. The bending moment diagram is achieved by determining the value of the bending moment and shear force at given moment. The paper has been able to determine the size and type of a member of a given material. More importantly, the bending moment diagram is used to determine the conjugation beam method or moment share method. (Caprani, 2008). A Bending Moment Diagram is also used to create a moment variation with the length of the beam. The bending moment diagram is used to determine the deflection, shear stress as well as the slopes of the structure.

The beam sign convention is as follows:

A bending moment = M (x)

A shear force =V (x).

As being revealed in fig 1, the normal convention for a positive bending is designed in "U" form and spins clockwise to the left and spinning counterclockwise to the right.

Fig 1: Normal Bending Moment Convention

Thus, the next step is to claculate the moment diagram.

Calculation of Bending Moment Diagram

This step determines the value of the moment, and the normal sign convention is used for the bending moment diagram and the functions are expanded to reveal the effects of each loading on the bending functions.

The first step is to obtain the bending moment force equation to deterime support reaction.

Determining support reactions:

The beam has three reaction components and they are Ax, Ay, and Dy.

Applying the equation of static equilbrium, the value is

FX = 0, Ax= 0 (eq. 1)

F y = 0, Ay + Dy -180 x8-350=0

Ay +Dy= 1790 kN (eq. 2)

Considering Z. axis that passes through and taking the moment of all the forces in the structure at z-axis i.e taking clockwise -- ve as well as anticlockwise +ve

MZ= 0; Dy x 20- 350x180 -- 180 x 12x7 =0 (eq. 3)

Solving eq. 3, we get 63000+ 15120 =78120/20=3906 kN

Thus, Dy = 3906 kN

Substituting the value of Dy in eq.2, we get

Ay +Dy= 1790 kN

Ay +3906= 1790 kN

Ay= 1790-3906

Ay=-2116 kN

Shear Force Calculation

FA left = 0

FA left = -2116 kN

FB = -2116 -160 x8=-3396 kN

FC left = -3396 kN

FC right = -2116- 1280+ 3906 kN =510 kN

FD left=510 kN

FD left=510-510=0

Bending Moment Calculation

MA =0

MB = -2116x8 -160 x12x7= 3488 kN

MC = -2116x20 -160 x12x19= -78800 kN

Maximim bending +ve beding moment will be calculated using the property of similar triangles using shear force diagrams as follows:

-2116 / x=-3396 / (8-x)

-2116 / x=-3396 / (8-x) =13.24m

Thus, the maximum +ve bending moment is 13.24m.

Mmax (+ve ) -2116 x 13.24 -180 x (13.24) x (13.24)/2=

=43792.62 kNm

Mmax (-ve) = -78800 kNm at point C

Thus, the point at which the bendingis at 0 is calculated as follows:

3488/a=-78800/(8-a)= -0.36m

Based on the moment bending diagram presented in Appendix 1, the paper discusses the appropriate factor of safety and the serial size of the structural section that should be used for beams, AB, BC and CD.

Safety Sized used for the Beam

The goal of this project is to enhance appropriate safety of the structure capable of resisting the anticipated loading to enhance adequate margin of safety. Typically, the structure is designed to anticipate that the joints possess rotational stiffness to ensure that the moment around the frame and distribution of forces are not significantly different from the calculation.

Elastic beam theory has been the basis of the structural steel design and analysis. Elastic beam theory is based on the steel structural element and analysis. The elastic beam theory reveals that maximum load that a structure could be able to support is assumed to be equal with the load and this could cause a stress somewhere within the structure. This assumed to be equal with the stress Fy of the materials. For example, the members should be designed so that the computed bending stress for the load is not to exceed the yield stress. Engineers have generally used the elastic beam theory to design structure and achieve satisfactory results. Elastic beam theory is an effective tool to enhance safety of the structure. It is essential to realize that non-application of elastic-beam theory in application of a structure could make a beam to deform under bending and shear. (Yaw. 2003).

For example, if a beam L >> d, the greatest deformation will cause the bending.

Equation of the Elastic beam theory is revealed as follows:

My=FyS (1)

From the equation, the yield My is equal to the yield stress Fy and multiply by the elastic modulus S.

where

S I?

c

moment of inertia c? is the distance from N.A. To outer cross section of fiber.

Figure 2

b d

As being revealed in Fig 2, the elastic modules of the rectangular section of b x d could be computed using:

Rectangular Cross Section:

The flexural formula

Using the Flexural Formula, the paper derives the rectangular cross section as follows:

Fy = MyC/1 = My1/c = My/s

My = FyS

1=bd3/12, c=d/2=S=1/c= bd3/12/d/2= bd2/6

My = FyS = Fy bd2/6

Based on the equation, the stress beam theory serves the purpose of equalizing the load in case of overload.

The beam is designed to resist a force .Typically, the stress diagram is revealed to have the ideal shape which is proportional to the limit of the same point of the steel. Thus, stress strain diagram is perfect and ideal to enhance safety. Conformity to the elastic beam theory will enhance the greater margin of beam AB, BC, and CD against collapse because the yield My is equal to yield stress Fy times of the elastic Modules of S. Theoretically, the strain member will permit still members to withstand the stress thereby prevent the structure against collapse.

By grouping the members, it is revealed that the members BE, and CF have the same L=8.4m, while the members BC and EF also have the same L=12m.

Thus, the concentrated load magnitude is P= 350 kN at a distance of 7m from support D

Uniform load intensity of B. 180 kN/m act on BC.

The calculation are as follows:

Using the Flexural stiffness of a member is represented as (EI/L).

Where ( E) is the modulus of elasticity,

Where (I) is the second moment of area

And divided by (L) which is the length of the member.

All the flexural stiffness of a member is represented as (EI/L).

Using elastic beam theory, the paper suggests the serial sizes for the beam AB, BC, and CD in Table 2.

Table 2: Serial Sizes for beams AB, BC and CD.

Beams

Serial Size (mm)

Weight

Depth (D) (mm)

Width (B) (mm)

Web Thk (t) (mm)

Flange Thk (T) (mm)

AB

178 x 102

19.0

4.8

7.9

BC

203 x 102

23.1

5.4

9.3

CD

203 x 133

30.0

6.4

9.6

List of References

Caprani, C. (2008). Structural Analysis III Moment Distribution. Colincaprani.UK.

Lightfoot, E. (1961). Moment Distribution: A Rapid Method of Analysis for Rigid- Jointed Structures. Taylor & Francis.

Volokh, K.Y. (2002). On foundations of the Hardy Cross method. International Journal of Solids and Structures. 39(16): 4197-4200.

Yaw, L.L. (2003). Moment Distribution the Real Explanation, and Why it Works. Walla University.

Appendix 1: Bending Moment Diagrams

UDL=w=1kN/m

Loading

Diagram

A

7.5kN

V (x) kN

Shear

Force

Diagram

+ve

-2.5

-ve

-22.5

M (x) kNm

quadratic

Bending

Moment

Diagram

43.792

+ve

10m

B

3.5m

C

3.5m

D

32.5kN

10m

10 kN

E

+ve

x

12.5

x

-ve