Deflection of Beams

The deformation of a beam is usually expressed in terms of its deflection from its original unloaded position. The deflection is measured from the original neutral surface of the beam to the neutral surface of the deformed beam. The configuration assumed by the deformed neutral surface is known as the elastic curve of the beam.

 

Methods of Determining Beam Deflections

Numerous methods are available for the determination of beam deflections. These methods include:

1. Double-integration method 2. Area-moment method 3. Conjugate-beam method 4. Strain-energy method (Castigliano's Theorem) 5. Method of superposition Of these methods, the first three are the ones that are commonly used.

Double Integration Method

The working equation for the double integration method is:

EI (d2y/dx2 ) = M

EI dy/dx = ∫ Mdx + C1

EI y = ∫∫ M(dx)(dx) + C1 x + C2

where x and y are the coordinates shown in the figure of the elastic curve of the beam under load, y is the deflection of the beam at any distance x.

E = the modulus of elasticity of the beam, I = represent the moment of inertia about the neutral axis, M = bending moment at a distance x from the end of the beam. EI is called the flexural rigidity of the beam.  The first integration dy/dx yields the slope of the elastic curve and the second integration y gives the deflection of the beam at any distance x. The resulting solution must contain two constants of integration, C1 and C2 , since EI d2y/dx2 = M is of second order. These two constants must be evaluated from known boundary conditions concerning the slope and deflection at certain points of the beam. For instance, in the case of a simply supported beam with rigid supports, at x = 0 and x = L, the deflection y = 0, and in locating the point of maximum deflection, we simply set the slope of the elastic curve dy/dx to zero.

The first important part of the method is the formulation of a single moment equation M which is applicable for any value of x. For this purpose, the pointed bracket < > is used such that the value of the function inside the pointed bracket is neglected or made zero if it is negative for a certain value of x.

Problem: Obtain the deflection at point B of the beam below. Assume EI is constant all throughout. the span of the beam. Figure:

Solution: 200N 450 N/m

Moment equation M: A B C D

0 ≤ x < 2 2m 1m 2m

M = 400x 400N 920N

2 ≤ x < 3: V M

M = 400x - 200 <x-2> 3 ≤ x < 5 x 200N M = 400x - 200 <x-2> - 450<x-3><x-3>/2 400 N

V M

Single Moment equation 2m x-2 M = 400x - 200 <x-2> - 450 <x-3><x-3>/2 x = 400x - 200 <x-2> - 450 <x-3>2/2 400N 200N 450N/m

V M

Note: the single moment equation is formulated by cutting at the last beam segment with possible change in moment. 2m 1m x-3

400N x

M = 400x - 200 <x-2> - 225<x-3>2

EI (d2y/dx2 ) = M

EI (d2y/dx2 ) = 400x - 200 <x-2> - 225 <x-3>2

EI dy/dx = 400x2 /2 - 200<x-2>2 /2 - 225<x-3> 3/3 + C1

= 200 x 2 - 100<x-2>2 - 75<x-3> 3 + C1

EI y = 200x3 /3 - 100<x-2>3 /3 - 75<x-3> 4/4 + C1 x + C2 eqn. 1 Boundary conditions:

At point A : At point D : x = 0 ; y = 0, x = 5 : y = 0 Thus, from eqn. 1 Thus, from eqn. 1 0 = 0 – 0 – 0 + 0 + C2 EI (0) = 200(5) 3/3 - 100<5-2>3 /3 - 75<5 - 3> 4/4 + C1 5 + 0

C2 = 0 C1 = - [8333.33 - 900 - 300] / 5 = - 1426.67 Thus: EI y = 200x 3/3 - 100<x-2>3 /3 - 75<x-3> 4/4 - 1426.67x

At point B , x = 2 m. negative EI y B = 200(2) 3/3 - 100<2-2>3 /3 - 75<2- 3> 4/4 - 1426.67(2) EI y B = 533.333 - 2853.33 y B = - 2320 /EI (answer) . The NEGATIVE SIGN means downward.

Problem: For the previous problem, obtain the maximum deflection of the beam.

Solution;

For simply supported beams, the point of maximum deflection is where the slope of the tangent to the elastic curve is zero, or dy/dx = 0

Assume that the point of maximum deflection is within the range 0 ≤ x < 2

EI dy/dx = 200 x 2 - 100 <x-2>2 - 75 <x-3> 3 - 1426.67 0 = 200x2 - 100 <x-2>2 - 75 <x-3> 3 - 1426.67 x = 2.67 m which is outside the range 0 ≤ x < 2, hence, analysis not correct

Assume that the point of maximum deflection is within the range 2 < x < 3

EI dy/dx = 200 x 2 - 100 <x-2>2 - 75 <x-3> 3 - 1426.67 0 = 200x2 - 100 <x-2>2 - 75 <x-3> 3 - 1426.67 0 = 200x2 - 100 x2 + 400x - 400 – 1426.67 0 = 100 x2 + 400x – 1826.67

- 400 + √ (400)2 - 4(100)(-1826.67) x = ------------------------------------------------------------------ = 2.72 m which is within 2 ≤ x < 3, 2(100) therefore, analysis is correct.

Thus, the maximum deflection is at x = 2.72m and EI y = 200x 3/3 - 100<x-2>3 /3 - 75<x-3> 4/4 - 1426.67x

EI y max = 200(2.72)3 /3 - 100 <2.72-2>3 /3 - 1426.67(2.72) = 1341.576 – 12.442 - 3880.542

y max = - 2551.408/EI (answer) The negative sign means downward.

Problem: Obtain the deflection at the free end of the cantilever beam shown below. Assume that EI is constant all throughout the beam‘s span.

Solution: Figure: w P

∑MA = 0 A B C MA w(L/2)(L/4) + PL – MA = 0 L/2 L/2

MA = w(L 2/8) + PL R A

∑Fy = 0 w P RA – w(L/2) – P = 0 MA

RA = wL/2 + P

Single moment equation: (Note: extend w to end of w beam and counter the extended w so that RA

the net effect is zero.) L/2 L/2

M = -MA + RA (x) - wx2 /2 + w < x – L/2> < x – L/2>/2 wx = - MA + RA x - wx2 /2 + w <x – L/2>2 /2 w thus: V EI d2y/dx2 = - MA + RA x - wx2 /2 + w<x – L/2>2/2 M A M EI dy/dx = - MA x + RA x2 /2 - wx3/[(2)(3)] w + w<x – L/2>3/[(2)(3)] + C1 eqn. 1 R A

EIy = - MA x2 /2 + RA x3/[(2)(3)] - wx4/[(2)(3)(4)] L/2 x – L/2 + w<x – L/2>4/[(2)(3)(4)] + C1 x + C2 x w(x –L/2)

EIy = - MA x2/2 + RA x3/(6) - wx4/(24) + w<x – L/2>4/(24) + C1 x + C2 eqn. 2

Boundary conditions: at A : x = 0 ; y = 0, thus , from eqn. 2 , C2 = 0 at A : x = 0 ; dy/dx = 0 (slope of tangent to the elastic curve is horizontal at the fixed support) thus, from eqn. 1, C1 = 0

Thus: EIy = - MA x2/2 + RA x3/(6) - wx4/(24) + w<x – L/2>4/(24) eqn. 2

At free end (point C), x = L, and

EIyC = - MA L2/2 + RA L3/(6) - wL4/(24) + w<L – L/2>4/(24)

= - [w(L 2/8) + PL] L2/2 + [wL/2 + P] L3/(6) - wL4/(24) + w<L – L/2>4/[(24)]

= - wL4/16 - PL3/ 2 + wL4/12 + PL3/ 6 - wL4/(24) + wL4/(384)

= wL4 [ -1/16 + 1/12 – 1/24 + 1/384] + PL3 [ -1/2 + 1/6]

= wL4 [ -24 + 32 – 16 + 1]/384 + PL3 [ -3 +1]/6 yC = [ - 7wL4/384 - PL3/ 3 ] / EI (answer) The NEGATIVE SIGN means DOWNWARD..

AREA – MOMENT METHOD

Theorem I : M/EI diagram Area = A The change in slope between tangents drawn to the elastic curve at any two points A and B is equal to the product of 1/EI multiplied by the area c.g. in the moment diagram between these two points (ӨAB = A) A B x Theorem II: elastic curve The deviation of any point B relative to a tangent drawn to the elastic curve at any other A B point A, in a direction perpendicular to the original position of the beam, is equal to the product of 1/EI multiplied by the moment of area about B ӨAB t B/A of that part of the moment diagram between points A and B. (t B/A = A x )

Sign Convention:

+’ve : A B ӨAB : counterclockwise t B/A : point B is above the tangent at A ӨAB t B/A

-’ve: ӨAB : clockwise t B/A : point B is below the tangent at A

ӨAB t B/A

A B

Problem: Using area-moment method, obtain the deflection at point C of the beam below. Assume EI is constant all throughout the beam’s span. P P

Solution: A B C D

Theorem II:

EItD/A = [(1/2)(PL/3)(L/3)(2L/9) P L/3 L/3 L/3 P

+ (PL/3)(L/3)(L/6 + L/3) +(1/2)(PL/3)(L/3)( L/9 + 2L/3)] V – diagram P

= PL3 [1/81 + 1/18 + 7/162] = PL3/9 EItC/A = [(PL/3)(L/3)(L/6) M- diagram -P

+ (1/2)((PL/3)(L/3)(L/3+L/9)] PL/3

= PL3 [ 1/54 + 4/162] = 7PL3/162] L/9 L/6 L/9 2L/9

By similar triangles; Elastic curve A C D tD/A /L = [ tC/A + δC ] / (2L/3) δc δC = (2/3) tD/A - tC/A ) tC/A tD/A = ( PL3 )[ (2/3)(1/9) – 7/162]/EI δC = 5PL3/(162EI) (Answer)

Problem: For the beam in the previous problem, find the maximum deflection.

Solution:

At the point of maximum deflection m, the M diagram: PL/3

tangent to the elastic curve is horizontal.

From the figure of the elastic curve, 2L/9 L/3 L/3 L/3

ӨA = ӨAm x

For small angles: A m D ӨA = t D/A / L = PL2/9EI (eqn. 1) ӨA δm

Өam t m/A t D/A

From theorem I: EIӨam = ½(PL /3)(L/3) + PL /3( x – L/3) (eqn. 2)

Thus, the location of maximum deflection is solved: PL2 / 9 = ½(PL /3)(L/3) + PL /3( x – L/3) x = L/2 (or at midspan) By similar triangle: (δm + t m/A )/ (L/2) = t D/A /L Theorem II: δm = t D/A /2 - t m/A EIt m/A = ½(PL/3)(L/3)(L/2 – 2L/9) = PL3 [1/18 – 5/648]/EI +(PL/3)(L/2 – L/3)(L/2 – L/3)/2 = 31PL3 /648EI

= 5PL3 / 648

MOMENT DIAGRAM BY PARTS:

1. The resultant bending moment at any section caused by any load system is the algebraic sum of the bending moments at that section caused by each load acting separately, or:

M = (∑M) L = (∑M)R

where (∑M) L = sum of moments caused by all forces to the left of the section.

x (∑M) R = sum of moments caused by all forces to y = kxn

the left of the section. h 2. The moment effect of any single specified loading is always some variation of the general equation y = k xn b

with area = [1/(n+1)] bh

and center x = [1/(n+2)] b

Moment diagrams of cantilever beam:

1. Couple or moment 3. Uniform wo

Area = -CL Area = -1/6woL3

x = L/2 x = L/4 degree of curve = 0 degree of curve = 2

2. Concentrated P 4. Triangular wo

Area = - PL2 Area = -1/24woL3

x = L/3 x = L/5 degree of curve = 1 degree of curve = 3

Problem: For the beam loaded as shown in the figure below, (a) Draw the moment diagram by parts with respect to left support A . (b) Compute the deviation of point A with respect to a tangent drawn at point B. (c) Compute the deflection at 3m from left support of the beam. Assume constant EI.

Solution: Figure:

∑MA = 0 R2 (4) = 500(2) + 400(3)(1.5) c R2 = 700N

∑Fy = 0 R1+ R2 = 500 + 400(3) R1 = 500 + 1200 +- 700 R1 = 1000N

(a) The moment diagram by parts, with respect to point A, are drawn by considering a fixed support at A and drawing the moment diagrams due to each load.

- The moment diagram due to R2 of 700N is triangular with moment at A = 700(4) = 2800 N.-m. - The moment diagram due to 500 N is triangular with moment at A = -500(2) = -1000 N.-m. - The moment diagram due to 400N/m is parabolic with moment at A = -½(400)(3)2 = - 1800 N.-m.

(b) t A/B : By Theorem II

EI t A/B = ½(2800)(4)(4/3) + ½(-1000)(2)(2/3) + 1/3(-1800)(3)(3/4)

= 5450 N-m

(c) Deflection at point C, 3m from left support EI t C/B = ½(2800/4)((1)(1/3) = 116.67

by similar triangles:

tA/B /4 = (tC/B + δc)/1

δc = t A/B /4 - t C/B = 1/EI[5450/4 – 116.67]

= 1245.83/EI downward as assumed (Answer) A C B δc

tA/B tC/B

Problem : For the beam shown in figure below, compute the value of P that will cause the tangent to the elastic curve over support C to be horizontal. What will then be the value of EIδ under the 100-lb load? Figure: 100 lb. P

Solution: A B C D ∑MC = 0

RA (10) = (100)(6) - P(3) 4ft 6ft 3ft

RA = 60 - 0.3P RA = 60 - 0.3P RC

20/3 ft 10RA = 600 –3P

The moment diagram by parts with = 432 lb-ft respect to point c is shown at the left. 172.8. If the tangent to the elastic curve over support C is horizontal, then; 2ft 4ft - 3P EI t A/C = 0 0 = ½(600 – 3P)(10)(20/3) + ½(-600)(6)(8) -600

P = 56 lbs. A B Cthus: 600 – 3P = 432 δB = t B/C

and, under the 100-load: EI δB = EI t B/C = ½(432)(6)(4) + ½(172.8)(6)(2) – ½(600)(6)(4) = - 979.2 lbs. ft3 ( negative sign means that point B in the elastic curve is below the tangent drawn at C.)

CONJUGATE BEAM METHOD

Actual Beam Conjugate Beam

EIy = deflection EI d2 y/ dx2 = M

EI dy/dx = slope EI d3 y/ dx3 = dM/dx = V

EI d2 y/ dx2 = moment = M EI d4 y/ dx4 = dV/dx = load

Thus: 1. If M/EI of the actual beam is used as the fictitious load in the conjugate beam,

2. The fictitious or conjugate shear is equal to the actual slope, and

3. The fictitious or conjugate moment is equal to the actual deflection.

Actual beam Conjugate beam

y = 0 y = 0 M = 0 M = 0 slope ≠ 0 slope ≠ 0 V ≠ 0 V ≠ 0

y = 0 y ≠ 0 M = 0 M ≠ 0 slope = 0 slope ≠ 0 V = 0 V ≠ 0

y = 0 M= 0 slope ≠ 0 V ≠ 0

Problem: Using conjugate beam method, obtain the deflection at point C of the beam below. Assume EI is constant all throughout the beam’s span.

Solution: Figure: P P

The moment diagram of the beam is A B C D

shown in figure (b).

The conjugate beam in this case is L/3 L/3 L/3

the same as the actual beam, and it is P (a) P

loaded with M/EI diagram as shown in PL/3

figure (c) Analyzing the conjugate beam: (b)

∑MD = o PL/3EI

RA,con (L) = 1/EI [(1/2)(PL/3)(L/3)(2L/9) + (PL/3)(L/3)(L/6 + L/3) +(1/2)(PL/3)(L/3)( L/9 + 2L/3)] RA,con = PL2 /9EI L/9 L/6

2L/9

RA,con

Cutting at C, and obtaining moment (c) MC.con = RA,con(2/3)(L) – ½(PL/3EI)(L/3)(L/3 + L/9) – (PL/3EI)(L/3)(L/6)

= 5PL3 /162EI (answer)