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Beam Analysis

IntroductionThis application solves for the deflection, slope, bending moment, shear forces of an elastic beam. A Beam Analysis program is provided which is very general and accomodates a wide variety of supportconditions. It uses the principles of MacCauley or singularity functions.

This worksheet includes:The beam_analyze programUsing the Beam Analysis ProgramDeflection & Shear Force Diagrams3D Stiffness AnalysisVariable Distributed Load Analysis3D Concentrated Load AnalysisFind Maximum Distributed Force allowing 1% DeflectionFind Maximum Width of Beam for Given Height and LoadsChange Beam Boundary conditions to Clamped-Clamped and resolve the previous design

restart:

Some Useful MacrosThe following maps common names to appropriate singularity functions and simplify the definition of boundary conditions.dist_force := (x) -> Heaviside(x): # distributed forceconc_force := (x) -> Dirac(x): # concentrated forceconc_moment := (x) -> Dirac(1,x): # concentrated momentD2 := (x) -> D(D(x)): # short form for 2nd derivativeD1 := (x) -> D(x): # short form for 1st derivative

The beam_analyze programThe following subroutine solves for the deflection, slope, bending moment, shear forces of an elastic beam. The algorithm formulates the governing 4th order boundary value problem with McCauley forcing functions as the loading conditions, and appropriate boundary conditions for the supports. The BVP is then solved analytically using the method of superposition.

Inputs: a set of loading conditions using the concept of MacCauley functionsa set of boundary conditions that specifies beam supports

Output:

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(4.2.1)(4.2.1)

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(4.2.2)(4.2.2)

a table containing the desired solutions.

beam_analyze := proc(loadset,bcset) local N,i,yy,th,m,v,q,de,outab; # Form differential equation de := EI*diff(y(x),x$4) = sum(loadset[i],i=1..nops(loadset)):

dsolve({de} union bcset,y(x)): # Solve boundary value problem yy := rhs(%): # Extract deflection th := diff(yy,x): # Extract slope m := diff(yy,x$2): # Extract moment v := diff(yy,x$3): # Extract shear print(`Boundary Value Problem to Solve is`); print(lhs(de)=op(loadset),bcset); outab[`deflection`]:= simplify(yy): outab[`slope`] := simplify(th): outab[`moment`] := simplify(m): outab[`shear`] := simplify(v): outab:end:

Using the Beam Analysis Program

Problem StatementThe beam is simply supported and is loaded with a uniform distributed load across the whole beam, a distributed load beginning at , a concentrated force at , and a concentrated moment at . Use the subroutine beam_analyze to solve this problem.

Loading and SupportsDefine loading and supports (simply supported).loads := {-w*dist_force(x-a), -v*conc_force(x-b), m*conc_moment(x-c)};

supports := {D2(y)(0) = 0, D2(y)(L)=0,y(L)=0,y(0)=0};

Parameter ValuesExternal Load Locations and MagnitutesDistributed force starts at x = a and spans beamConcentrated force at x = bConcentrated moment at x = cDistributed load: w N/mConcentrated load: v N

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(4.3.1.1)(4.3.1.1)

Concentrated moment: m N.mMaterial Stiffness: EI N.m^2L:=10:a:=0:b:=2:c:=4:v:=500:w:=500:m:=10000:EI:=10^6:

Solve the problem:sol := beam_analyze(loads,supports):

Boundary Value Problem to Solve is

newsol:=convert(sol,table):

Post-ProcessingDeflection Diagramplot(newsol[deflection],x=0..L,title=`Beam Deflection`,axes=BOXED);

> > Shear Force Diagramplot(newsol[shear],x=0..L,title=`Shear Force Diagram`,axes=BOXED);

> > Bending Moment Diagramplot(newsol[moment],x=0..L,title=`Bending Moment Diagram`,axes=BOXED);

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(4.4.4.1.2)(4.4.4.1.2)

(4.4.4.1.1)(4.4.4.1.1)

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3D Stiffness AnalysisDefine Stiffness Relationship and Parameter RangeL:=10:a:=0:b:=2:c:=4:v:=500:w:=500:m:=10000:unassign('EI'):sol := beam_analyze(loads,supports):

Boundary Value Problem to Solve is

stiff:=convert(sol,table):EIrange := 1e5 .. 2e6;

(4.4.4.1.2)(4.4.4.1.2)

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Deflection versus Stiffnessplot3d(stiff[deflection],x=0..10,EI=EIrange,axes=BOXED);

Bending Moments versus Stiffnessplot3d(stiff[moment],x=0..10,EI=EIrange,axes=BOXED);

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(4.4.4.1.2)(4.4.4.1.2)

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(4.5.1.1)(4.5.1.1)

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Example Analysis ProblemsVariable Distributed Load AnalysisChange the external distributed load to be linearly dependent upon the location with the following criteria: Distributed load: LHS=200 and RHS=1000Concentrated load: 5500 at x=2 mConcentrated moment: none

PLOT the deflection and shear force diagrams:L:=10: a:=0: b:=2: c:=0: v:=5500: w:=200+80*x: m:=0: EI:=10^6:sol:=beam_analyze(loads,supports):

Boundary Value Problem to Solve is

newsol:=convert(sol,table):

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(4.4.4.1.2)(4.4.4.1.2)

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> > plot(newsol[deflection],x=0..L,title=`Beam Shear`,axes=BOXED);

plot(newsol[shear],x=0..L,title=`Beam Deflection`,axes=BOXED);

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(4.4.4.1.2)(4.4.4.1.2)

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(4.5.2.1)(4.5.2.1)

3D Concentrated Load AnalysisObserve the variation in deflection if the concentrated load varies from 0 to 10000. Remove the distributed load i.e. w=0.newsol:=convert(eval(op(subs(L=10,a=0,b=2,c=0,w=0,m=0,EI=10^6,op(sol)))),table):L:=10: a:=0: b:=2: c:=0: w:=0: m:=0: EI:=10^6: unassign('v'):sol:=beam_analyze(loads,supports):

Boundary Value Problem to Solve is

newsol:=convert(sol,table):plot3d(newsol[deflection],x=0..10,v=0..10000,axes=BOXED);

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(4.6.1.2)(4.6.1.2)

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(4.4.4.1.2)(4.4.4.1.2)

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(4.6.1.1)(4.6.1.1)

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Example Design ProblemsFind Maximum Distributed Force allowing 1% DeflectionGiven the following loading conditions:Distributed Load : starting at x = 0 to x = 10

defl:=convert(eval(op(subs(L=10,a=0,b=0,c=0,v=0,m=0,EI=10^6,op(sol)))),table):L:=10: a:=0: b:=0: c:=0: v:=0: m:=0: EI:=10^6: unassign('w'):sol:=beam_analyze(loads,supports):

Boundary Value Problem to Solve is

defl:=convert(sol,table):maxdefl:=defl[slope]=0;

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(4.6.1.2)(4.6.1.2)

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(4.6.1.4)(4.6.1.4)

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(4.6.2.2)(4.6.2.2)

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(4.6.3.2)(4.6.3.2)

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(4.6.3.1)(4.6.3.1)

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(4.4.4.1.2)(4.4.4.1.2)

(4.6.1.3)(4.6.1.3)

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(4.6.2.1)(4.6.2.1)

Max deflection will occur at mid span x = 5.newdefl:=simplify(subs(x=5,defl[deflection])=-0.01*L);

maxforce:=eval(solve(newdefl,w)) * N;

Find Maximum Width of Beam for Given Height and LoadsGiven the following loading conditions:Concentrated Load: 500 N at mid-spanMax deflection: 1% of length occurs at mid-spanheight: 0.1 m

Find the Cross Sectional Square Area for an aluminum beam that will support a 1% deflection

cross:=convert(eval(op(subs(L=10,a=0,b=5,c=0,v=500,m=0,w=0,EI=Youngs*(base*height^3)/12,op(sol)))),table):L:=10: a:=0: b:=5: c:=0: v:=500: m:=0: w:=0: EI:=Youngs*(base*height^3)/12:sol:=beam_analyze(loads,supports):

Boundary Value Problem to Solve is

cross:=convert(sol,table):maxhgt:=solve(simplify(subs(x=5,height=.1,Youngs=69E9,cross[deflection])=-.01*L),base);

Change Beam Boundary conditions to Clamped-Clamped and resolve the previous designRe-run the Beam_Analyze macro in order to determine the new general solution.New support conditions are indicated as follows.

supports := {D(y)(0) = 0, D(y)(L)=0,y(L)=0,y(0)=0};

L:=10: a:=0: b:=5: c:=0: v:=500: m:=0: w:=0: EI:=Youngs*(base*height^3)/12:

Solve the problemclamped := beam_analyze(loads,supports):

(4.4.4.1.2)(4.4.4.1.2)

(4.6.1.2)(4.6.1.2)

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(4.6.3.3)(4.6.3.3)

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(4.6.3.2)(4.6.3.2)

Boundary Value Problem to Solve is

cross:=convert(clamped,table):maxhgt:=solve(simplify(subs(x=5,height=.1,Youngs=69E9,cross[deflection])=-.01*L),base);