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Deformation of a Beam

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Deformation of a Beam 

? 2000 Harald Kammerer, GERB Schwingungsisolierungen, Germany
http://www.gerb.com 

Introduction 

This worksheet shows the calculation of the deformation of a simply supported beam 

The following Maple techniques are highlighted: 

 

  • Creating the model
 

 

 

  • Calculating the deformation of the beam
 

 

 

  • Creating an animation of the deformation
 

 

The Problem 

Consider an elastic both side simply supported beam. This beam can be for example a bridge. It is loaded by a moving load. This might be a car driving over the bridge. The situation is considered to be quasi static. So there are no dynamic effects respected. 

NOTE: Influence of shearing is not considered. 

 

The following variables are used: 

F:  Force 

xF: Position of the force related to the left end of the beam 

L:  Length of the beam 

x:  Position where the deflection of the beam is calculated 

EI: Product of the elastic modulus and the moment of inertia 

 

 

Further variables are defined at there first appearance. 

 

All units are m and N 

Set Parameters of the Problem 

 

Parameter 

Value 

Length of the beam 

`:=`(L, 20); -1 

 

Force 

`:=`(F, 10000); -1 

 

Number of steps for the load position 

`:=`(nx, 25); -1 

 

Length of steps between load positions 

`:=`(dL, `/`(`*`(L), `*`(nx))) = `/`(4, 5) 

Modulus of elasticity 

`:=`(Emod, `+`(`*`(3, `*`(`^`(10, 10))))); -1 

 

 

 

 

Dimension of the Cross Section 

The Cross section is for later calculations divided in three parts, the plate and the two webs.
 

`:=`(n, 3); -1 

`:=`(a, .5); -1 

`:=`(b, a); -1 

`:=`(d, 1); -1 

`:=`(e, d); -1 

`:=`(f, .5); -1 

 

Drawing-Canvas 


The dimensions of the  parts of the cross section are collected in sets.
 

`:=`(width, [`+`(`*`(2, `*`(a)), `*`(2, `*`(b)), d), b, b]); 1; `:=`(height, [f, e, e]); 1 

 

 

[3.0, .5, .5]
[.5, 1, 1] (3.1)
 

 

 

Boundaries of the cross section in z-direction 

 

`:=`(zmin, 0); -1; `:=`(zmax, `+`(e, f)); -1 

 

Collect the coordinates centers of gravity of every part in sets (arbitrary with respect to the left upper corner, other points are also possible) 

 

`:=`(cogy, [`+`(a, b, `*`(`/`(1, 2), `*`(d))), `+`(a, `*`(`/`(1, 2), `*`(b))), `+`(`+`(a, b, d), `+`(`*`(`/`(1, 2), `*`(b))))]) 

[1.500000000, .7500000000, 2.250000000] (3.2)
 

`:=`(cogz, [`+`(`*`(`/`(1, 2), `*`(f))), `+`(f, `*`(`/`(1, 2), `*`(e))), `+`(f, `*`(`/`(1, 2), `*`(e)))]) 

[.2500000000, 1.000000000, 1.000000000] (3.3)
 

 

Procedures 

All the main calculations and graphics are programmed in the form of procedures.   

To execute the code , left-click the icon. 

To view the code, right-click the icon, then select "View Code Edit Region" 

Procedures for the Calculation 

Area of the Cross Section 

Calculating the cross sectional area of a beam made up of three rectangular pieces. 

 

Button#crosssec 

 

Center of Gravity 

Calculating the center of gravity. 

 

Button#cog 

 

Moment of Inertia of a Beam 

Calculating the moment of inertia of the beam.  

 

Button#inertia 

 

Bending Moment of a Beam 

Simply supported beam at both ends with a single point load acting on it. 

 

Button#moment 

Button 

Deformation of a Beam 

Simply supported beam at both ends with a single point load acting on it (shear deformations are neglectable in this example). 

 

Button#deform 

 

Normal Stress of a Beam 

Simply supported beam at both ends with a single point load acting on it (shear deformations are neglectable in this example). 

Button#stress 

 

Procedures for the Plots 

Rotation and Following Translation of the Coordinate of a Point 

This procedure is used for the later graphical representation of the solution 

Button#transrot 

Drawing a Car in a x-y-Coordinate System 

This procedure is used to draw a car which symbolizes the moving load in the later graphical representation of the solution. 

Button#carplot 

Draw the Supports of the Beam 

 

Button#support 

Draw the Bridge Head on Both Sides 

 

Button#bridgeplot 

Plot the Deformation 

In this procedure the deflection of the beam will be plotted. To show the current position of the load there are a little car an an arrow drawn. 

 

Button#plotsol 

Plot the Normal Stress in the Beam 

In this procedure the normal forces in the beam will be plotted. To show the current position of the load there is an arrow drawn. 

 

Button#stressplot 

 

Execution 

Calculation 

Calculate the center of gravity by using the procedure cog. 

`:=`(yzc, cog(n, width, height, cogy, cogz)); 1 

[1.500000000, .5500000000] (5.1.1)
 

 

Boundaries of the cross section in z-direction in relation to the center of gravity 

`:=`(zmins, `+`(zmin, `-`(yzc[2]))); 1; `:=`(zmaxs, `+`(zmax, `-`(yzc[2]))); 1 

 

 

-.5500000000
.9500000000 (5.1.2)
 

 

Calculate the moment of inertia by using the procedure inertia. 

In this example only the moment of inertia with respect to the horizontal axis is necessary. 

 

`:=`(Iy, inertia(n, width, height, cogy, cogz, yzc)[1]); 1 

.4520833334 (5.1.3)
 

 

Area of the Cross section of the beam 

`:=`(A, crosssec(n, width, height)); 1 

2.50 (5.1.4)
 

 

For the further calculation the product of elastic modulus and moment of inertia is interesting. 

`:=`(EI, `*`(Emod, `*`(Iy))); 1 

0.1356250000e11 (5.1.5)
 

 

Calculate the bending moment depending on x and xF. 

`:=`(M, convert(moment(x, xF, L, F), Heaviside)); -1 

`+`(`*`(10000, `*`(x, `*`(Heaviside(`+`(xF, `-`(x)))))), `-`(`*`(500, `*`(x, `*`(xF)))), `*`(10000, `*`(xF)), `-`(`*`(10000, `*`(xF, `*`(Heaviside(`+`(xF, `-`(x)))))))) (5.1.6)
 

 

Calculate the deformation of the beam depending on x and xF 

`:=`(dfrm, deform(x, xF, L, EI, M)); -1 

 

 

Separate the deflection 

`:=`(w, eval(dfrm[1])); -1 

 

 

Separate the gradient 

`:=`(R, eval(dfrm[2])); -1 

 

 

Calculate the normal stress in the beam depending on x and xF. 

`:=`(sigma, stress(x, z, xF, L, Iy, M)); -1 

 

Create the Animation 

Calculate the line of deflection for every position of the load and return the solution in form of a plot by using the procedure plotsol. Thereby the load is moving over the bridge like a driving car. The deflection of the bridge is 50 000 times inflated. Note: there are no real dynamic effects regarded. When the car reaches the right end of the bridge, it drives back to the left end. The result is shown by an little animation. 

 

Button#line of deflection 

 

 

Calculate the normal stress for every position of the load and return the solution in form of a plot by using the procedure stressplot

 

Button#normal stress 

View the Result 

To view the animations, left-click the plot and press in the animation toolbar at the top of the document. 

 

Deformation 

display([seq(P[i], i = 0 .. `+`(`*`(2, `*`(nx))))], insequence = true, scaling = constrained, axes = none); 1 

Plot_2d  
 

 

The maximum deflection of the beam is reached when the load is positioned in the middle of the beam. The maximum deflected point is then the center of the beam. The maximum deflection is thus 

 

`:=`(fmax, evalf(subs(x = `+`(`*`(`/`(1, 2), `*`(L))), deform(x, `+`(`*`(`/`(1, 2), `*`(L))), L, EI, moment(x, `+`(`*`(`/`(1, 2), `*`(L))), L, F))[1]))); 1 

-0.1228878648e-3 (5.3.1)
 

 

Normal Stress 

display([seq(S[i], i = 0 .. nx)], insequence = true, scaling = unconstrained); 1 

Plot_2d  
 

 

The stress for the situation when the load is positioned in the middle of the beam at the center of the beam is 

`:=`(Smax, evalf(subs(x = `+`(`*`(`/`(1, 2), `*`(L))), stress(x, zmaxs, `+`(`*`(`/`(1, 2), `*`(L))), L, Iy, moment(x, `+`(`*`(`/`(1, 2), `*`(L))), L, F))))); 1 

 

105069.1244 (5.3.2)
 

 

 

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