The Cantilevered Beam Problem:
Concentrated Moment Case

Van Warren
Warren Design Vision
Little Rock, Arkansas
7/19/95

Problem

A cantilevered beam is subjected to a concentrated moment somewhere along its span:

Determine the reaction and moment at the support, the deflection along the span, the slope at the endpoints, the shear and the bending moment.

Variables

L  = beam length
	a  = x coordinate of applied force
	Ra = the reaction at support A
	Ma = the moment at support A
	M  = the applied moment
	y(x)  = deflection as a function of x
	V(x)  = shear as a function of x
	M(x)  = bending moment as a function of x

	A subscript of '1', refers to interval 0 ≤ x ≤ a
	A subscript of '2', refers to interval a ≤ x ≤ L

Note that the point x = a is shared by intervals 1 & 2.

Initializations

x =.
Ra =.
Ma =.

strip2[u_] := First[First[u]] ;
ruleToEquality[u_] :=  u /. Rule -> Equal ;
equalityToRule[u_] :=  u /. Equal -> Rule ;
ruleToSet[u_] :=  u /. Rule -> Set ;
ruleToFunction[u_] :=  u /. Rule -> SetDelayed ;
rePresent[u_] := strip2[u] ;
enSconce[u_] := ruleToEquality[u] ;

srf[s_, l_]:=StringReplace[s,Join[{"["->"(", "]"->")"},l]];

numFix[s_]:=
StringReplace[s,
{
	"EI"->"(E*I)",
	"1.*"->"",
	"0.3333333333333333333"->"(1/3)",
	"(-x^2"->"(-(x^2)",
	"(-a^2"->"(-(a^2)",
	"(-b^2"->"(-(b^2)",
	"(-x^4"->"(-(x^4)",
	"(-a^4"->"(-(a^4)",
	"(-b^4"->"(-(b^4)"
}
];

tsf[s_] := ToString[s, FormatType->InputForm];

fix[e_, l_] := srf[tsf[e], l];

outPut[e_] := Print[numFix[numFix[ToString[InputForm[e]]]]];

Sum Forces & Moments

First we sum forces and moments to deduce the values of the reactions at support A.  Since this is an automatic proof we dispense with manual notation.  In Mathematica™ notation:

forcesEqn = Ra     == 0;
momentEqn = Ma - M == 0;

reactions =
	First	[
	Simplify	[
				Solve[{forcesEqn, momentEqn}, {Ra, Ma}]
				]
			];
reactions =
{
Ra -> Collect[Last[First[reactions]], {L}],
Ma -> Collect[Last[Last [reactions]], {L}]
};

The following expressions can be output to a spreadsheet or program for repetitive evaluation under varying conditions:

outPut[First[reactions]]
outPut[Last [reactions]]

We now have the reactions at each support in terms of the load, placement and beam length.  Now we need to compute the shear, bending moment, beam deflection and end slopes.  We use the sign conventions discussed previously in the derivation of the Cantilevered Beam, Concentrated Force problem.

Shear and Bending

Consider the following beam with imaginary cutting sections labeled A' and B'

Consider two beam segments, the segment A-A' and the segment A'-B'.

The external moment impressed on the structure does not produce any reactive force at the support A..  Therefore we record the shear as zero in the interval 0 ≤ x ≤ L and no sign convention is even necessary.  For further discussion of this issue see Beer & Johnston, Vector Mechanics For Engineers 1977, Statics & Dynamics, p 273.

Mathematically we have:

and no plot of a zero shearing force is provided.

We now consider the bending moment as a function of position.  Returning to the segment A-B', we note that the external moment acting on the structure tend to bow the cutting section so that it is negative in sense, i.e. the external forces bow the cutting section up.  This corresponds to a negative sign for the bending moment for the interval 0 ≤ x ≤ a.

There are no external loads impressed on the structure to the right of the cutting line marked B'.  Therefore the moment is zero for the interval a < x ≤ L.  This gives us a complete result for bending, symbolically:

and graphically:

The maximum magnitude of M(x) is just M.  For consistency with our other automatic proofs we repeat this argument using the symbolic manipulation facilities of Mathematica™:

Solve For Shear and Bending 0 <= x <= a

shear1   = V1[x] ->  0;
bending1 = M1[x] -> -Ma;

outPut[shear1]
outPut[bending1]

Solve For Shear and Bending a < x <= L

shear2   = V2[x] -> 0;
bending2 = M2[x] -> 0;

outPut[shear2]
outPut[bending2]

Slope and Displacement

As in previous editions we assign y deflection coordinates to two intervals along the beam.  The y1 coordinate refers to the interval 0 ≤ x ≤ a, and the y2 coordinate refers to the interval a < x ≤ L. To compute spanwise deflection and slope we refer to linear bending theory which (under the assumption of constant stiffness) states that

We immediately jump to the services of our symbolic expression muncher:

Solve For Slope and Displacement 0 <= x <= a

dy1dx'[x] == M1[x]/(EI) /. bending1;
slope1 = rePresent[DSolve[%, dy1dx[x], x]] /. C[1] -> C1;

y1'[x] == Last[slope1];
displacement1 = rePresent[DSolve[%, y1[x], x]] /. C[1] -> C2;

outPut[slope1]
outPut[displacement1]

Solve For Slope and Displacement a < x <= L

dy2dx'[x] == M2[x]/(EI) /. bending2;
slope2 = rePresent[DSolve[%, dy2dx[x], x]] /. C[1] -> C1p;
slope2 = dy2dx[x] ->
	Collect[Collect[Last[slope2], {Ra, Ma, EI}], EI];


y2'[x] == Last[slope2];
displacement2 =
	rePresent[DSolve[%, y2[x], x]] /. C[1] -> C2p;
displacement2 = y2[x] ->
	Collect[Collect[Last[displacement2], {Ra, Ma, EI}], EI];


outPut[slope2]
outPut[displacement2]

Boundary Condition Simultaneous Equations

We have four equations, and four unknown constants C1, C2, C1' and C2'.  The value of the constants can be determined by applying the four boundary conditions:

Evaluating these automatically we have:

bc1 = enSconce	[
				Reverse	[
						displacement1 /. x->0 /. y1[0]->0
						]
				];
bc2 = enSconce	[
				Reverse	[
						slope1 /. x->0 /. dy1dx[0]->0
						]
				];
bc3 =
(
	(
		(Last[displacement1] /. x-> a) -
		(Last[displacement2] /. x-> a)
	) == 0
);
bc4 =
(
	(
		(Last[slope1] /. x -> a) -
		(Last[slope2] /. x -> a)
	) == 0
);
constants = First	[
			Simplify[
			Solve	[
						{bc1, bc2, bc3, bc4},
						{C1, C2, C1p, C2p}
					]
					]
					];

collector[u_] :=
	First[u] ->
	Simplify[
		Collect[
			Collect[Last[u],
			{EI, Ra, Ma, MP}], EI
			   ]
			];
			
outPut[foo1 = collector[constants[[1]]]]
outPut[foo2 = collector[constants[[2]]]]
outPut[foo3 = collector[constants[[3]]]]
outPut[foo4 = collector[constants[[4]]]]

As a check we evaluate the constants in terms of the reactions:

constants /. reactions

This is as expected and requires little effort compared to checking by hand. We proceed to check our deflection and slope equations in a similar effortless manner:

displacement1 /. constants /. reactions
slope1        /. constants /. reactions
displacement2 /. constants /. reactions
slope2        /. constants /. reactions

End Slopes

Our only remaining errand is to evaluate the slope expression at the endpoints and to remember that end slopes come out in radians:

leftEndSlope = Simplify[slope1 /. x->0];
riteEndSlope = Simplify[slope2 /. x->L];

outPut[leftEndSlope]
outPut[riteEndSlope]

Output

Output useable deflection expressions.

deflExpr =
	IF[AND[0<=x,x<=a],A1, A2] /.
	A1 -> Last[displacement1] /.
	A2 -> Last[displacement2];

outPut[fix[deflExpr,{"C" -> "a_C","EI" -> "(E*I_a)"}]]
outPut[fix[deflExpr,{"C" -> "b_C","EI" -> "(E*I_b)"}]]
outPut[fix[deflExpr,{"C" -> "c_C","EI" -> "(E*I_c)"}]]

Output useable constant expressions.

outPut[fix[foo4,{"C" -> "a_C","EI" -> "(E*I_a)"}]]
outPut[fix[foo3,{"C" -> "a_C","EI" -> "(E*I_a)"}]]
outPut[fix[foo2,{"C" -> "a_C","EI" -> "(E*I_a)"}]]
outPut[fix[foo1,{"C" -> "a_C","EI" -> "(E*I_a)"}]]

Testing

Convert displacement functions.

ruleToFunction[displacement1]
ruleToFunction[displacement2]

?y1

?y2

Input parameters for a realistic example.

inputs =
{
	L->36, a->12, M->(300*12), EI->(402311.4063)
};

nconstants = constants /. reactions /. inputs

tmp1 = y1[x] /. nconstants /. reactions /. inputs;
tmp2 = y2[x] /. nconstants /. reactions /. inputs;

yA[x_] := tmp1;
yB[x_] := tmp2;

y[x_] := Which[And[ 0<=x, x<  12], yA[x],
               And[12<=x, x<= 36], yB[x]]

reactions /. inputs

Plot displacement functions.

Plot[Evaluate[y[x]], {x, 0, 36}]

Cross verification example problems.

Example Problem 1:

We run a BeamCALCulator™ that handles a concentrated moment:

Load Parameters

Moment M:  300*12 in-lbs;	Concentrated Moment
Length:        36 in;

Material Parameters

Material:     Aluminum 6061 T-6
Elastic Modulus: 10,100,000 psi
Yield Strength:      35,000 psi (Tensile)
Density:              0.098 pci

Cross Sectional Parameters

Outer Width  W: 1.00 inches
Inner Width  w: 0.85 inches
Outer Height H: 1.00 inches
Inner Height h: 0.85 inches