Session Lecture note : Strength of Materials

Session

Lecture note :Pramudiyanto, M.Eng.

g{x V|ä|Ä tÇw cÄtÇÇ|Çz XÇz|ÇxxÜ|Çz Xwâvtà|ÉÇ WxÑtÜàÅxÇàFaculty of Engineering, State University of Yogyakarta

Strength of Materials

Deflection of Transversally Loaded Beam

05


we will be concerned with another aspect inthe design of beams, namely, the determinationof the deflection. Of particular interest is thedetermination of the maximum deflection of abeam under a given loading, since the designspecifications of a beam will generally includea maximum allowable value for its deflection.

A knowledge of the deflections is required toanalyze indeterminate beams. These are beamsin which the number of reactions at thesupports exceeds the number of equilibriumequations available to determine theseunknowns.


a prismatic beam subjected to purebending is bent into an arc of circleand that, within the elastic range, thecurvature of the neutral surface can beexpressed as

To determine the slope and deflection of the beam at any given point, we firstderive the following second-order linear differential equation, which governs theelastic curve characterizing the shape of the deformed beam :


Cantilever BeamThe cantilever beam AB is of uniformcross section and carries a load P at itsfree end A. Determine the equation ofthe elastic curve and the deflectionand slope at A.

SOLUTIONUsing the free-body diagram of the portion AC of the beam, where C is located at a distance x from end A, we find:

Substituting for M and multiplying both members by the constant EI, we write

Integrating x, we obtain :


We now observe that at the fixed end B wehave x = L and = dy/dx = 0. Substitutingthese values and solving for C1, we have:

Which we carry back into :

Integrating both members, we write :

At B we have x = L, y = 0. substituting thiswe have :

0

Carrying the value of C2, we obtain the equation of the elastic curve :

Or 3 2The deflection and slope at A :

3

2


Beam with distributed loadingThe simply supported prismatic beam AB carries a uniformly distributed load w per unit length. Determine the equation of the elastic curve and the maximum deflection of the beam.

SOLUTION

Drawing the free-body diagram of the portion AD of the beam and taking moments about D, we find that:

Substituting for M and multiplying both members of this equation by the constant EI, we write :

221

21 wxwLxM

wLxwxdx

ydEI 212

21

2

2


Integrating twice, we have:

Observing that y=0 at both ends of the beam, we first let x=0 and y=0 in first equation and obtain C2=0. We then make x=L and y=0 in the same equation and write :

Carrying the values of C1 and C2 back into previous equation, we obtain the equation of elastic curve :

2134

123

121

241

41

61

CxCwLxwxEIy

CwLxwxdxdyEI

31

144

241

121

2410

wLC

LCwLwL

xLLxxEI

wy

xwLwLxwxEIy

334

334

224

241

121

241


Substituting the value obtained for C1, we check that the slope of the beam is zero for x=L/2 and that the elastic curve has a minimum at the midpoint C of the beam. Letting x = L/2, we have :

The maximum deflection or, more precisely, the maximum absolute value of deflection, is thus :

EIwLLLLLL

EIwyC 384

528

21624

43

34

EIwLy

3845 4

max


Beam with point loading

For the prismatic beam and the loading shown, determine the slope and deflection at point D.

We must divide the beam into two portions, AD and DB, and determine the function y(x) which defines the elastic curve for each of these portions.

SOLUTIONFrom A to D (x < L/4). We draw the free-body diagram of a portion of beam AE of length x < L/4. Taking moments about E, we have

Pxdx

ydEIxPM43

43

21

2

1


where y1(x) is the function which defines the elastic curve for portion AD of the beam. Integrating in x, we write

213

1

121

1

81

83

CxCPxEIy

CPxdxdyEIEI

From D to B (x > L/4). We now draw the free-body diagram of a portion of beam AE of length x > L/4 and write

PLPxdx

ydEILxPxPM41

41

443

22

2

2


where y2(x) is the function which defines the elastic curve for portion DB of the beam. Integrating in x, we write

4323

2

322

2

81

241

41

81

CxCPLxPxEIy

CPLxPxdxdyEIEI

The conditions that must be satisfied by the constants of integration have been summarized. At the support A, we must have x = 0 and y1 = 0. At the support B, we must have x = L and y2 = 0. Also, the fact that there can be no sudden change in deflection or in slope at point D requires that y1 = y2 and 1 = 2 when x = L/4. We have therefore:


384,

12811,0,

1287

4153611

4512,

4

1287

1283,

4

12100,

00,0

3

4

2

32

2

1

43

3

1

3

21

32

12

21

433

2

21

PLCPLCCPLC

CLCPLLCPLyyLx

CPLCPLLx

CLCPLyLx

Cyx

Substituting for C1 and C2, we write that for x L/4 :

xPLPxEIy

PLPxEI

1287

81

1287

83

23

1

22

1


Letting x = L/4 in each of these equations, we find that the slope and deflection at point D are, respectively,

EIPLy

EIPL

DD 2563,

32

32

We note that, since 0, the deflection at D is not the maximum deflection of the beam.


THANK YOUThat’s for now

Session Lecture note : Strength of Materials

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