341_pdfsam_1Mechanics of Materials(3 Ed)[Team Nanban]Tmrg

8/18/2019 341_pdfsam_1Mechanics of Materials(3 Ed)[Team Nanban]Tmrg

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or

(5.6)

Similarly, from Eq. 5.3,

(5.7)

Equations 5.6 and 5.7 can be stated in words as follows:

• The change in shear from Section 1 to Section 2 is equal to the area underthe load curve from 1 to 2. (The “area” that results from negative p( x) isnegative.)

• The change in moment from Section 1 to Section 2 is equal to the area underthe shear curve from 1 to 2. (The “area” that results from negative V ( x) isnegative.)

Equations 5.2 through 5.7 will be very useful to us in Sections 5.4 and 5.5, wherewe draw shear and moment diagrams. And we can employ modifications of Eqs. 5.6and 5.7 to determine expressions for V ( x) and M ( x). Thus,

(5.8)

and

(5.9)

Equations 5.2 through 5.9 will be used extensively in Section 5.5 in construct-ing shear and moment diagrams (Examples 5.7 through 5.9).Also, whenever shearand moment expressions must be obtained for a beam with a distributed load otherthan a simple uniform load or a triangular distributed load, it is much easier to useEqs. 5.2 through 5.9 than to use a finite free-body diagram, as is illustrated inExample 5.6.

M ( x) M 1 x

x1

V (j) dj

V ( x) V 1 x

x1

p(j) dj

M 2 M 1 x2

x1

V ( x) dx

V 2 V 1 x2

x1

p( x) dx

Shear-Force and Ben

Moment Diag

Equilibrium Me

In Example 5.2 we obtained expressions for V ( x) and M ( x) for a simply sup-ported beam with distributed loading. However, to design a beam (i.e., to select abeam of appropriate material and cross section) we need to ask questions like“What is the maximum value of the shear force, and where does it occur?” and“What is the maximum value of the bending moment, and where does it occur?”These questions are much more readily answered if we have a plot of V ( x) and aplot of M ( x). These plots are called the shear diagram and the moment diagram,respectively.

5.4 SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS:

EQUILIBRIUM METHOD


2/20

In Sections 5.4 and 5.5, two methods for constructing shear and moment digrams are described:

• Method 1—Equilibrium Method - (Section 5.4): Use finite free-body digrams or Eqs. 5.8 and 5.9 to obtain shear and moment functions, V ( x) anM ( x); then plot these expressions.

• Method 2—Graphical Method - (Section 5.5): Make use of Eqs. 5.2 throug5.7 to sketch V ( x) and M ( x) diagrams (see Table 5.1).

As you study the example problems in these two sections, observe that maxmum positive and negative bending moments can occur at any of the following cro

sections of a beam: (1) a cross section where the shear force is zero (Examples 5and 5.9); (2) a cross section where a concentrated couple is applied (Examples 5.5.6, and 5.8); (3) a cross section where a concentrated load is applied and where thshear force changes sign (Example 5.7); and (4) a point of support where there isreaction force and where the shear force changes sign (Example 5.9).

The following examples illustrate the two procedures. Examples 5.4 through 5illustrate the Equilibrium Method; Examples 5.7 through 5.9 illustrate the GraphicMethod.A third method, the Discontinuity-Function Method, is presented in Section 5.

Shear-Force and Bending-Moment Diagrams: Equilibrium MethodExamples 5.4 through 5.6 illustrate the Equilibrium Method for constructing sheand moment diagrams.

322

Equilibrium of Beams

E X A M P L E 5 . 4

Figure 1 shows the simply supported beam of Example 5.2, including thereactions.The expressions for V ( x) and M ( x) obtained in Example 5.2 are

(a) Using the above expressions for V ( x) and M ( x), plot shear andmoment diagrams for this simply supported beam. (b) Determine thelocation of the section of maximum bending moment, and calculate thevalue of the maximum moment.

6 x 6 12 ftM 2 [140(12 x) 109 (12 x)

3] lb ft,

0 6 x 6 ftM 1 (220 x 20 x2) lb ft,

6 x 6 12 ftV 2 [140 103 (12 x)

2

] lb,

0 6 x 6 ftV 1 (220 40 x)lb,

Plan the Solution It is straightforward to plot the shear and momentdiagrams from the given expressions (e.g., using a computer). From the

Fig. 1

40 lb/ft

6 ft

220 lb 140 lb

6 ft

A

x

C B

x


3/20

shear diagram, the location of the section where V ( x) 0 can be deter-mined.Then the appropriate moment equation can be used to determinethe value of the moment at this critical section. Since there is more loadover the left half of the beam than over the right half, the maximum mo-ment should occur to the left of x 6 ft.

Solution (a) Plot the shear diagram and the moment diagram. Figure 2shows the plots of shear and bending moment.

Fig. 2 Shear and moment diagrams.

V (lb)

M (lb·ft)

M max = 605 lb·ft

x (ft)

x (ft)

220

–140

126

5.5

126

The shear

diagram.

(a)

The momentdiagram.

(b)

(b) Determine the maximum bending moment . The maximum momentoccurs where the shear vanishes. From the shear diagram and the equa-tion for V 1( x), it can be observed that V ( x) 0 in the interval 0 x 6 ft.Therefore,

V 1( xm) 220 40 xm 0S xm 5.50 ft

Then, the maximum moment is

M max M 1(5.50 ft) 605 lb ft Ans. (b)

Review the Solution The downward load on this simply supportedbeam bends it downward, so it is concave upward everywhere. This isconsistent with the fact that the bending moment is positive for the en-tire length of the beam. As expected, since there is more load over the

left half of the beam than over the right half, the maximum moment doesoccur to the left of x 6 ft. This is an example of a maximum momentthat occurs where V dM/dx 0.

Note how much easier it is to get a “feel” for the distribution of shear force andbending moment in a beam from shear-force and bending-moment diagrams than itis from the shear and moment functions alone.


4/20

Fig. 3 The shear diagram and themoment diagram.

Fig. 2 A free-body diagram.

E X A M P L E 5 . 5

Derive expressions for V ( x) and M ( x) for the cantilever beam with lin-early varying load shown in Fig. 1. Use these expressions to plot shearand moment diagrams for this beam.

Fig. 1

y

A B x

p( x ) p0

L

x

Plan the Solution We can use a finite free-body diagram to determinethe required expressions for V ( x) and M ( x).

Solution The triangle in the free-body diagram in Fig. 2 is similar to thetriangle in the problem statement; so, by similar triangles,

Ans

Ans

To plot these expressions for the shear force V ( x) and bending mo-ment M ( x), we first calculate V (L) and M (L). Figure 3 shows the plotsof V ( x) and M ( x).

Review the Solution The shear is positive everywhere, as we expecfrom the free-body diagram.The maximum shear occurs at the cantileversupport at B and is equal to the total area under the load curve in Fig. 1The upward load will bend the beam upward, so it will be concaveupward everywhere. This is consistent with the fact that the bendingmoment is positive for the entire length of the beam. The maximumbending moment occurs at B. Finally, the shear and moment have theproper dimensions, (F ) and (F L), respectively.

M (L) p0L

2

6V (L)

p0L

2,

M ( x) p0 x

3

6L

[ p( x)]a x

2b a x

3b

M ( x)

0aaM ba

0:

V ( x) p0 x

2

2L

a x2b p( x) V ( x) 0a F y 0:

p( x)

x

p0

LS p( x) a x

Lb p0

324

0.25

0.50

0.75Cubiccurve

(b) The moment diagram.

1.00

0.000.2 0.80.4 1.00.6

x / L

6 M ( x ) ––––– p0 L

2

0.25

0.50

0.75 Quadraticcurve

(a) The shear diagram.

1.00

0.000.2 0.80.4 1.00.6

x / L

2V ( x ) ––––– p0 L

y

x

p( x ) p( x )

M ( x )

V ( x ) x

2 x ––

3

x – 2

a


5/20

Fig. 1 An airplane wing.


E X A M P L E 5 . 6

A cantilever airplane wing (represented as a beam in Fig.1) has a distrib-uted load given by

p( x) p0 c1 a xL

b3 d

Derive expressions for V ( x) and M ( x) for the cantilever wing. Use these

expressions to plot shear and moment diagrams for this wing.

Plan the Solution It would be difficult to use a finite free-body diagramto determine the required expressions for V ( x) and M ( x), as was done inExample 5.4, since the load is not represented by a simple expression.Therefore, we will make use of Eqs. 5.8 and 5.9 to obtain expressions forV ( x) andM ( x). First,however,it is a good idea to draw a free-body diagramto define terms.

Solution Figure 2 shows a free-body diagram of the wing from cross-section x to the tip of the wing. All important terms are labeled on thisfree-body diagram. At the wingtip, x L, there is no shear or bendingmoment, so V (L) M (L) 0. From Eq. 5.8, we can write the following

expression for V ( x):

where p( ) is just the given load function with dummy variable substi-tuted for x. Since V (L) 0,

So,

Ans.V ( x) p0L c34

x

L

1

4 a xL

b4 d

p0L c 14

a jL

b4 jL

dL x

p0

L

x c 1

a j

Lb3

ddj

V ( x) L

x

p(j)dj

jj

V (L) V ( x) L

x

p(j)dj

x

y p(ξ)

V ( x ) M ( x )

L

ξ

x

x

p0 p( x )

L


6/20

In the previous section, Examples 5.4 through 5.6 illustrated how the equilibriu

method can be used to construct shear-force and bending-moment diagrams. In thsection, you will see how the graphical method facilitates the drawing of shear anmoment diagrams.

Shear-Force and Bending-Moment Diagrams: Graphical MethodExamples 5.7 through 5.9 illustrate the graphical method for constructing shear anmoment diagrams. Table 5.1 illustrates how Eqs. 5.2 through 5.7 are used in costructing (i.e., sketching) the diagrams, proceeding from the left end to the right enof the beam. If you study carefully the numbered steps that are given in the following examples and relate them to the entries in Table 5.1, you should be able

5.5 SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS:GRAPHICAL METHOD

Similarly, from Eq. 5.9,

Since M (L) 0,

So,

Ans

The shear and moment at the wing root, V (0) and M (0), are:

Finally, a computer was used to plot the above expressions for V ( x)and M ( x) that are shown in Fig. 3.

Review the Solution The shear is negative everywhere,as we expect fromthe free-body diagram.The maximum shear (magnitude) occurs at the wingroot A and is equal to the total area under the load curve in the problemstatement.The magnitude, (3/4) p0L, is reasonable compared with the value

p0L for a uniform load. The upward load will bend the wing upward, so iwill be concave upward everywhere.This is consistent with the fact that thebending moment is positive for the entire length of the wing.The maximumbending moment occurs at the wing root A. Finally, the shear and momenhave the proper dimensions, (F ) and (F L), respectively.

M (0) 3

10 p0L

2V (0) 3

4 p0L,

M ( x) p0L2 c 3

10

3

4 a x

Lb 1

2 a x

Lb2 1

20 a x

Lb5 d

r 0L2 c 3

4 a j

Lb 1

2 a j

Lb2 1

20 a j

Lb5 dL

x

p0LL

x

c34

j

L

1

4 a j

Lb4 ddj

M ( x)

L

x

V (j)dj

M (L) M ( x) L

x

V (j)dj

326


0.4

(b) The moment diagram

x / L

M ( x ) ––––– p0 L

2

–1.0

–0.5

0

(a) The shear diagram

0.5 1.0

0.5 1.0

x / L

0.2

0

V ( x ) –––– p0 L


7/20

T A B L E 5 . 1 Shear and Moment Diagram Features

Load Shear Moment

Diagram Diagram Diagram

Equation p V M

1. Slope of shear diagram equals value of load

(Eq. 5.2)dV

dx p( x)

p2

p1

V 1 V 2 M 1 M 2

M 1

M 2V 1V 2

Slope = p1

2. Jump in shear equals value of concentrated load

(Eq. 5.4)¢V P 0

V 1 V 2 M 1 M 2

P0

M 1

M 2V 1

V 2 P0

Positive V -jump

3. Change in shear equals area under distributed-load diagram

(Eq. 5.6)V 2 V 1 x2

x1

p( x)dx

V 1 V 2 M 1 M 2

(Area) p

M 1

M 2V 1

V 2

V 2 – V 1 = (Area) p

4. Slope of moment diagram equals value of shear

(Eq. 5.3)

dM

dx

V ( x) V 1 V 2 M 1 M 2

M 1

M 2

Slope = V 1

V 1 V 2

5. Jump in moment equals – (value of concentrated couple)

(Eq. 5.5)¢M M 0 M 1

M 2

Negative M -jump

M 0

V 1 V 2

6. Change in moment equals area under shear diagram

(Eq. 5.7)M 2 M 1 x2

x1

V ( x)dxV 1 V 2 M 1 M 2

M 1

M 2

M 2 – M 1 = (Area)V

V 1 V 2

(Area)V

V 1 V 2 M 1 M 2

M 0


8/20

construct shear and moment diagrams for any beam with simple loading, once yohave determined the loads and reactions acting on the beam.2 There are also severMDSolids examples that employ the graphical method.

328


E X A M P L E 5 . 7

Use Eqs. 5.2 through 5.7 to sketch shear and moment diagrams for the

simply-supported beam shown in Fig. 1.

Plan the Solution We can use a free-body diagram of the beam AC todetermine the reactions at A and C. Since there is no distributed load onthe beam, p( x) 0 everywhere. Because of the concentrated load at Bwe need to consider two spans, 0 x a and a x L.

Solution

Equilibrium—Reactions: To determine the reactions A y and C y, we firsdraw the free-body diagram of the entire beam AC (Fig. 2).


Shear Diagram: Equations 5.2, 5.4, and 5.6 involve the shear. Usingthese equations, we can sketch V ( x) progressively from x 0 to x LIt is convenient to sketch the shear and moment diagrams directly belowthe load diagram (Fig. 3a). Each step involved in sketching V ( x) isnumbered in Fig. 3b.

1. The shear at x 0 is zero.

2. The shear at x 0 is determined from Eq. 5.4, that is, V A

A y . Note that, because of the sign convention for shearan upward concentrated force causes an upward jump in the shear

diagram.

3. For 0 x a, p( x) 0. Therefore, from Eq. 5.2, dV/dx 0.

P (L a)

L

A y P (L a)L

A yL P (L a) 0,aaM bC 0:

C y P a aLbPa C yL 0,aaM b

A

0:


2The graphical method is most useful when the “areas” in Eqs. 5.6 and 5.7 are simple rectangles triangles, that is, when the loads on the beam are either concentrated loads or uniform distributed loa

The graphical method is also useful in interpreting the results of an equilibrium-method solution.

L

A

a

C B

P

x

y

V ( x )

(a) Load diagram.

(b) Shear diagram.

(c) Moment diagram.

M ( x )

Pa( L – a) ––––––––

L

P( L – a) –––––––

L

–Pa ––– L

P( L – a) –––––––

L

Pa ––– L

x

L – a

A

a

C

B

P

P

(1)

(1)

(2)

(2)

(3)

(3)

(4)

(4)

(5)

(5)

(6)

x

L – a

A

a

C

C y A y

B

P

x

y

Fig. 1


9/20

V & M Diagrams—Determinate Beams is a computer programmodule that may be used to plot shear-force and bending-moment diagrams forstatically determinate beams. The solutions of MDS Examples 5.1–5.6 illustrate the

graphical method for constructing V and M diagrams.

MDS5.1 – 5.6

4. At x a there is a downward force P , so V B P .

5. For a x L, p( x) 0, so 0.

6. The reaction at C causes which closes the shear diagram

back to zero at x L.

Moment Diagram: Equations 5.3, 5.5, and 5.7 relate to M ( x) and can beused to sketch the moment diagram in Fig. 3c. Steps in the constructionof the moment diagram are explained and numbered.

1. The moment at x 0 is zero [simply supported beam].

2. For 0 x a, Eq. 5.3 gives

3. At x a, M (a) can be determined from Eq. 5.7 as the area of the

rectangle under the shear curve from x 0 to x a. Therefore

is the

maximum bending moment.

4. For a x L, Eq. 5.3 gives

5. Equation 5.7 gives M (L) M (a) L

a

V ( x)dx Pa

L(L a),

dM

dx V ( x)

Pa

L constant.

M (a) Pa(L a)

LM (a) 0

a

0

V ( x)dx P (L a)

L(a).

dM

dx V ( x)

P (L a)

L constant.

¢V C Pa

L,

dV

dx

which closes the moment diagram back to zero at x L, as it should[simple support at C ].

Review the Solution The dimensions on the shear diagram (F ) and themoment diagram (F L) are correct. If we draw finite free-body diagramsof the ends of the beam, we get Fig. 4. Therefore, the shear diagram in Fig.

3b has the correct signs according to the free-body sketches in Fig. 4 and thesign convention in Fig. 5.6. The downward force will bend the beam asshown in Fig. 5, which is consistent with the fact that the bending momentis positive everywhere.The maximum bending moment occurs at the crosssection where the force P is applied and where the shear force changes sign.

E X A M P L E 5 . 8

Use Eqs. 5.2 through 5.7 to sketch shear and moment diagrams forsimply supported beam shown in Fig. 1.

Plan the Solution We can use a free-body diagram of the beam AC inFig.2 to determine the reactions at A and C . Since there is no distributed

Fig. 5

Fig. 4

A

P( L – a) –––––––

L

Pa –––

L

C

+ Shear

– Shear

A C B

P


10/20

load on the beam, p( x) 0 everywhere. Because of the concentratedcouple at B, we need to consider two spans, 0 x a and a x L.

Solution

Equilibrium—Reactions: We first determine the reactions A y and C y.

Shear Diagram: Equations 5.2, 5.4, and 5.6 involve the shear. Usingthese equations and the load diagram in Fig. 3a, we can sketch V ( x) pro-gressively from x 0 to x L. It is convenient to sketch the shear andmoment diagrams directly below the load diagram.Each step involved insketching V ( x) is numbered in Fig. 3b.

1. The shear at x 0 is zero.

2. The shear at x 0 is determined from Eq. 5.4, that is, V A

A y M 0/L. Note that, because of the sign convention for shearan upward concentrated force causes an upward jump in the shear

diagram.

3. For 0 x L, p( x) 0. Therefore, from Eq. 5.2, dV /dx 0, V ( x) M 0/L constant.

4. The reaction at C causes V C M 0/L, which closes the sheardiagram back to zero at x L.

Moment Diagram: Equations 5.3, 5.5, and 5.7 relate to M ( x) and can beused to sketch the moment diagram in Fig. 3c. Steps in the constructionof the moment diagram are explained and numbered.

1. The moment at x 0 is zero (simply supported beam).2. For 0 x a, Eq. 5.3 gives dM/dx V ( x) M 0/L constant.

3. At x a, M (a) can be determined from Eq. 5.7 as the area of therectangle under the shear curve from x 0 to x a. Therefore

4. At x a there is a negative jump in moment given by Eq. 5.5. So

.

5. For a x L, Eq. 5.3 gives dM/dx V ( x) M 0/L constant.

6. Equation 5.7 gives

which closes the moment diagram back to zero at x L, as it should[simple support at C ].

Review the Solution The dimensions on the shear diagram (F ) and themoment diagram (F L) are correct. Note that both the maximummoment, M 0a/L, and the minimum bending moment, M 0(L a)/Loccur at the cross section where the concentrated couple acts. Comparethis example with the previous one, where there was a concentratedforce.

M (L) M (a) La

V ( x) dx (M 0/L)(L a)

M (a) M 0a

L M 0

M 0(L a)

L

M (a) 0 a0

V ( x)dx M 0a/L.

A yL M 0 0S A y M 0

LaaM b

C

0:

M 0 C yL 0SC y M 0

LaaM b

A

0:Fig. 1


L

A

a

C B

x

y

M 0

L – a

A

a

C

C y

M 0

A y

B

x

y

V ( x )

(a) Load diagram.

(b) Shear diagram.

(c) Moment diagram.

M ( x )

x

L – aa B

(1)

(1)

(2)

(2)

(3)

(3)

(4)

(6)

(5)

(4)

x

A C M 0

–M 0( L – a)________

L

M 0 a____

L

M 0___

L

M 0___

L

M 0___

L

330

Fig. 3 Load, shear, and momentdiagrams.


11/20

E X A M P L E 5 . 9

Determine the reactions and sketch the shear and moment diagrams forthe beam shown in Fig. 1. (This beam is said to have an overhang BC.)Show all significant values (that is, maxima, minima, positions of maximaand minima, etc.) on the diagrams.

Fig. 1


Plan the Solution We can use a free-body diagram of the whole beamto compute the reactions. Then we can use Eqs. 5.2 through 5.7 to sketch

the V ( x) and M ( x) diagrams, as we did in Examples 5.7 and 5.8.

Solution

Equilibrium—Reactions: The reactions must be determined first. Figure 2shows the appropriate free-body diagram.

Ans.

Ans.

Check: Is Yes8 32 40 16 0?a F y 0?

A y 8 kN

A y(4 m) (8 kN/m)(4 m)(2 m) (16 kN)(2 m) 0

aaM bB

0:

B y 40 kN

(8 kN/m)(4 m)(2 m) (16 kN)(6 m) B y(4 m) 0

aaM b A

0:

A

4 m

16 kN8 kN/m

2 m

C B

A C

A y B y

B

y

x

16 kN

8(4) = 32 kN

2 m

4 m 2 m


12/20

It is convenient to sketch the V (Fig. 3b) and M (Fig. 3c) diagramsdirectly below a sketch of the beam that has all of the loads and reactionsshown (Fig. 3a).

Shear Diagram: The following steps are used in sketching the sheardiagram (Fig. 3b).

1. V (0

) 0 [no shear at end of beam].2. V (0) 8 kN [Eq. 5.4].

3. dV/dx 8 kN/m [Eq. 5.2].

4. V (4) V (0) (8 kN/m)(4 m) 8 32 24 kN [Eq. 5.6].

5. V (4) V (4) 40 kN [Eq. 5.4].

6. dV/dx 0 [Eq. 5.2].

7. V (6) V (6) 16 0.

8. Since dV/dx 8 for 0 x 4 m, and since V (0) 8 kN, V 0at xm 1 m [Eq. 5.6].

Moment Diagram: The steps employed in constructing the moment dia-gram (Fig. 3c) using Eqs. 5.2 through 5.7 will now be described:

1. M (0) 0 [no moment at end of beam].

2. From dM/dx V ( x) we have the slope of M ( x) going from 8kN m/m at x 0 to zero at x xm 1 m. Therefore, the moment diagram for 0 x 1 m must have the general shape

[Eq. 5.3].

3. M ( x) is maximum where V ( x) 0 [Eq. 5.3].

4. [Eq. 5.7; areaof triangle].

5. From x 1 m to x 4 m, V ( x) gets progressively more negativeTherefore, M ( x) must have the general shape [Eq. 5.3].

6.[Eq. 5.7; net of areas of triangles]

7. dM/dx V ( x) 16 kN [Eq. 5.3].

8. [Eq

5.7; no moment at end of beam].

The maximum shear occurs just to the left of the support at B andhas a magnitude of 24 kN. The maximum positive moment occurs whereV 0 at x 1 m and has a magnitude of 4 kN m; and the maximumnegative moment occurs at the support B, and it has a magnitude of32 kN m.

Review the Solution By imagining cuts just to the right of A (Fig. 4a) just to the left of B (Fig. 4b), and just to the right of B (Fig. 4c), we cancheck the sign of the shear at these points.

The moment diagram is best checked by seeing if the sign of themoment diagram corresponds to a reasonable deflected shape, that isconcave upward where M ( x) is positive and concave downward where

M (6) M (4) 6

4 V ( x)dx 32 kN m (16 kN)(2 m) 0

32 kN mM (4) M (0)

4

0 V ( x)dx 0 12(8 kN)(1 m)

12 (24 kN)(3 m)

M (1) M (0)1

0 V ( x)dx 12(8 kN)(1 m) 4 kN m

332


Fig. 4

V (kN)

(a) Load diagram.

(b) Shear diagram.

(c) Moment diagram.

M (kN·m)

x (m)

x m

x (m)

(1)

1

8

4

8

–24

–32

(1)

1 m

(2)

(8)

(2) (3) & (4)

(4)(3)

(5)(6)

(5)

(6)

(7)

16

1

(8)

(7)

A C B

16 kN

16

40 kN8 kN

8 kN/m

4 m 2 m

A

(a) (c)(b)

B C

+ Shear

8 kN

8 kN

16 kN

16 kN

– Shear + Shear

B C

24 kN

40 kN

16 kN


13/20

M ( x) is negative, according to the sign convention that is given Fig. 5.6c.

Where M ( x)

0, the beam is locally straight, that is, it is neither concaveupward nor concave downward. We are able to sketch (Fig. 5) a plausi-ble deflection curve that passes over the supports at A and B and that isconcave upward where M ( x) is positive and concave downward whereM is negative.The distributed load between A and B and the concentratedload at C could, indeed, cause the beam to deflect as sketched.

Fig. 5 A sketch showing the deflection of beam AC.

M positive M negative

Deflection

exaggerated

The idea that a positive bending moment makes a beam concave toward the yside,whereas a negative bending moment causes the beam to be concave toward the y side, is in accord with the definition of positive bending moment in Fig. 5.6c.Thisfact was used in Examples 5.7 and 5.9 above to check the bending moment dia-grams. In the next chapter we derive a mathematical relationship between bendingmoment and curvature, and in Chapter 7 we use this relationship to obtain expres-sions for the deflection of beams.

Wherever there is a discontinuity in the loading on a beam or where there is a sup-port, there will be a discontinuity in integrals that involve the loads and reactions.Between these discontinuities the integrals will be continuous. For example, thebeam in Fig. 5.10 has four intervals— AB,BC ,CD, and DE. Consequently, four sep-arate expressions V i( x) and four expressions M i( x) would be required to specifyV ( x) and M ( x) for this beam. The introduction of discontinuity functions simplifiesthe process of determining expressions for shear and bending moment (Section 5.6),and it greatly simplifies the process of solving for the slope and deflection of a beam(Section 7.5). For example, for the beam in Fig. 5.10, the shear V ( x) can be writtenas a single compact expression, valid for 0 x L, rather than as four separateexpressions. Similarly, the load p( x) and the moment M ( x) can each be written as asingle expression that is valid for 0 x L.

*5.6 DISCONTINUITY FUNCTIONS TO REPRESENTLOADS, SHEAR, AND MOMENT


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Macaulay functions will be used to represent distributed loads on beams, an singularity functions will be used to represent concentrated external forces anconcentrated couples. Together, they are referred to as discontinuity functionand are symbolized by angle brackets, called Macaulay brackets,3 that have thform: x a n.

Macaulay Functions. Macaulay functions are useful in expressing functionthat are zero up to some particular value of the independent variable and nonzerfor larger values of the independent variable. For integer values of n 0, thMacaulay functions x a n are defined by the following expressions:

n 0, 1, 2, . . . (5.1

For example, Fig. 5.11 shows the unit step function x a 0 and the unit ramp funtion x a 1, where a is the value of the independent variable x at which the discotinuity occurs. As illustrated by the unit ramp function in Fig. 5.11b, these functionhave the value zero for x a and the value ( x a)n for x a. The units of x aare the units of xn (e.g., ftn, mn, etc.).

Singularity Functions. The two singularity functions of interest here are thunit doublet function, x a 2, which can be used to represent a concentratecouple, and the unit impulse function, x a 1, which can be used to representIHIH

HIH IH

H x aIn e0 for x 6 a( x a)n for x a

IH

IH

334


FIGURE 5.10 A beam with several applied loads.

w1

w2 M 0

L

B

C

A

P

D E

x

3The English mathematician W. H. Macaulay (1857–1936) introduced the use of special brackets to re

resent these discontinuity functions. It has been common practice to use angle brackets for this purpoand to refer to them as Macaulay brackets.

FIGURE 5.11 Examples of Macaulay functions.

(b) Unit ramp function

x

1

1

x – a 1

a x

(a) Unit step function

a

1

x – a 0


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concentrated force. These are illustrated in Fig. 5.12. Singularity functions are simi-lar in some respects to the Macaulay functions, but they are defined for negativevalues of n. They become singular (i.e., infinite) at x a, and they are zero for x a.The units of singularity functions are the same as the units of xn, even though n isnegative.

Integrals of Discontinuity Functions. The rules of integration for disconti-nuity functions are given in Eq. 5.11.

(5.11)

Since we will use discontinuity functions in conjunction with beams of finitelength with their origin at x 0, as illustrated in Figs. 5.10 and 5.13, we need not beconcerned with values of x less than x 0. The origin, x 0, will therefore be as-sumed to be located such that, for x 0, x a n 0 for all n.

Use of Discontinuity Functions to Represent Loads, Shear, and Moment.Figure 5.13 illustrates how Macaulay functions can be used to represent distributed

loads on beams. It must be remembered that Macaulay functions continue indefi-nitely for x a. Therefore, when a particular load pattern terminates at some valueof x, a new Macaulay function must be introduced to cancel out the effect of thatprevious Macaulay function. Note how the dimensions of each term in p( x) are pre-served even though the term x a 0 is dimensionless whereas the term x a 1 hasthe dimensions of length.

IHIH

IH

H x aIndx • H x aIn1

for n 0

1

n 1 H x aIn1 for n 7 0

Discontinuity Functio

Represent Loads, Shear

Mo

FIGURE 5.12 Examples of singularity functions.

a0 x a0 x

(a) Unit doublet function. (b) Unit impulse function.

x – a –2 x – a –1

FIGURE 5.13 Distributed loads represented by Macaulay functions.

a4

a3

a2

x

a1

p2 p( x )

p1

x

p( x ) = p10 + ( p1 + p2) x – a1

1 – x – a20 x – a4

1 – x – a3 p 2______

a3 – a

2( )

p 2______a

3 – a

2( )


16/20

The sign conventions for external loads and for internal shear force and bending moment are given in Figs.5.6, and in Section 5.3 relationships among loads, sheforce, and bending moment were presented. By way of review,

(5.repeate

(5.repeate

(5.repeate

(5.repeate

To apply the latter two when p( x) and V ( x) are represented by discontinuity funtions,we let x1 0

and let V 1(0)M 1(0

) 0. Therefore, Eqs. 5.8 and 5.9 can bwritten as

(5.1

The load function for the beam in Fig. 5.14 can be written in terms of disconnuity functions by referring to Table 5.2; and we can use Eqs. 5.11 to integrate the

M ( x) x

0

V (j)djV ( x) x

0

p(j)dj,

M ( x) M 1 x

x1

V (j)dj

V ( x) V 1 x

x1

p(j)dj

¢M M 0

¢V P 0

336


1

P0

P0

(a) Load diagram.

(b) Shear diagram.

(c) Moment diagram.

a3

a2

x

a1

p0

M 0

M 0

V ( x )

x

M ( x )

P0

x

x

– M 0

1

p0

FIGURE 5.14 Examples of the use of discontinuity functions to represent load, shear, anmoment.


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Case Load Shear Moment

1

2

3

4

5

6

T A B L E 5 . 2 A Summary of Loads, Shear, and Moment Represented by Discontinuity Functions

x

V

a

–M 0

–M 0

x – a –1

a

x

–M 0 x – a –2

M 0

x

a

P0

P0

x – a –1

p0

a

p0 x – a0

p1

a b

x

x – a 1 p1 –– b( )

x

2

a b

x – a 2 p2 –– b2( )

x

a b

pn

x – a n pn –– bn( )

x a

(n + 1)

V

bn x – a n + 1

pn –––––––– ( )

a

M

(n + 1)(n + 2)

bn x – a n + 2

pn –––––––––––– ( )

x a

V

x – a 3 p

2 ––– 3b2( )

a

M

x – a 4 p

2 –––– 12b2( )

x a

V

x – a 2 p

1 –– 2b( )

a

M

x – a 3 p1 –– 6b( )

x x a

p0 x – a1

V

a

M

x – a 2 p

0 –– 2

x a

P0

P0 x – a0

V

a

P0 x – a1

M

M

a

–M 0

–M 0 x – a0


18/20

discontinuity functions to get

(5.1

Note that the beam in Fig. 5.14 is shown to extend on the left to x 0. Singularifunctions are underscored in the above equations for p( x) and V ( x).

Table 5.2 summarizes load, shear, and moment relationships represented bdiscontinuity functions. The terms in this table that represent singularity functionare underscored to emphasize the fact that, strictly speaking, singularities have ifinite values at the point x ai and zero values everywhere else.Whereas it is common practice to represent concentrated couples and forces in the manner indicatein Case 1 and Case 2 of the “Load” column of Table 5.2, the moment M 0 in th“Shear” column of Case 1 is not a true concentrated shear force. Its effect is

cause the jump in moment at x a in the “Moment” column of Case 1, but othewise it can be ignored. Therefore, it is shown dashed in Table 5.2, Case 1. Note hothe amplitude of the load function is defined in Cases 4 through 6 to assure propedimensionality.

In this section, only statically determinate problems are considered. Thereforthe reactions are treated as known quantities that have been obtained by the use equilibrium equations.In Section 7.5, both statically determinate problems and staically indeterminate problems are examined.

M ( x) x

0

V (j)dj M 0H x a1I0

P 0 Ha a2I1

p0

2 H x a3I2

V ( x) x

0

p(j)dj M 0 H x a1I1 P 0 H x a2I

0 p0 H x a3I

1

p( x) M 0 H x a1I2 P 0 H x a2I

1 p0 H x a3I

0

338


E X A M P L E 5 . 1 0

For the beam in Fig. 1 of Example Problem 5.9, (a) use discontinuityfunctions to obtain expressions for p( x), V ( x), and M ( x), and (b) use thediscontinuity functions from Part (a) to construct shear and momentdiagrams for the beam, indicating the contribution of each term in thediscontinuity-function expressions.

The loads and reactions from Example Problem 5.9 are given inFig. 1.

Plan the Solution We can refer to Cases 2 and 3 of Table 5.2 to constructthe load function p( x) and then to perform the required integrations toget V ( x) and M ( x).

Solution (a) By referring to the Load column for Cases 2 and 3 in Table5.2, we can write

Ans. (1) (40 kN) H x 4 mI1 (16 kN) H x 6 mI1 p( x) (8 kN)H x 0 mI1 (8 kN/m)[ H x 0 mI0 H x 4 mI0]

Fig. 1

8 kN

A

40 kN

16 kN

B

8 kN/m

4 m 2 m

C


19/20

Observe that it is a very straightforward procedure to obtain the discontinuity-function expressions for V ( x) and M ( x). Each term in p( x), V ( x) and M ( x) can beevaluated separately, and the results summed to get the final discontinuity-functionexpressions. Likewise, graphs of p( x), V ( x), and M ( x) can be easily constructed fromthe discontinuity-function expressions.This makes the discontinuity-function methodan ideal one to serve as a basis for a computer program to evaluate and plot shear

diagrams and moment diagrams. In Section 7.5 discontinuity functions will be used tosolve beam deflection problems, including analysis of statically indeterminate beams.

V & M Diagrams—Determinate Beams is a computer programmodule that may be used to plot shear-force and bending-moment diagrams forstatically determinate beams. The discontinuity-function method, described inSection 5.6, was used in the development of the computer program module thatsolves the key equations and plots the shear and moment diagrams.The solutions of MDS Examples 5.7–5.9 illustrate this method.

MDS5.7–5.9

Integrating Eq. 1 by referring to Eqs. 5.11 or to the Shear column of Table 5.2, we get

(a) (b) (c)

(d) (e)

Ans. (2)

and, from the Moment column of Table 5.2, we get

(a) (b) (c)

(d

) (e

)Ans. (3)

(b) In Fig. 2, Eqs. (2) and (3) are plotted term-by-term, starting atthe left end of the beam; the separate terms are then summed so theresults can be compared with the shear diagram and the momentdiagram obtained in Example Problem 5.9.

Review the Solution Since the shear diagram in Fig. 2b and themoment diagram in Fig. 2c both close to zero at the right end, ourresults are probably correct.For this problem we could use Eqs. 5.2through 5.7 to check the shear and moment diagrams above. That

is, the procedure used to construct the shear diagram and the momentdiagram in Example Problem 5.9 can be used to check the resultsobtained by the discontinuity-function method.

(40 kN) H x 4 mI1 (16 kN) H x 6 mI1

M ( x) (8 kN) H xI1 (8 kN/m)[ 12 H xI2

12 H x 4 mI

2]

M ( x) x

0

V (j)dj:

(40 kN) H x 4 mI0 (16 kN) H x 6 mI0

V ( x) (8 kN) H xI0 (8 kN/m)[H xI1 H x 4 mI1]

V ( x) x

0

P (j)dj: B

8 kN/m16 kN

A

V(kN)

1 m

M (kN · m)

(a)

(a)

(b)

(c)

(e′)

(d )

(c)

x

x

(b)

(b′)

(a′)(c′)

(d ′)

(e)

40

16

8

80

16

48

4

–16

–32

–16

–24

–144

–48

40 kN8 kN

4 m 2 m

C

Fig. 2 Load, shear, and moment diagram


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Prob. 5.2-4. The shaft in Fig.P5.2-4 is supported by bearinat B and D that can only exert forces normal to the shaBelts that pass over pulleys at A and E exert parallel forcof 150 N and 300 N, respectively, as shown. Determine ttransverse shear force V C and the bending moment M C section C , midway between the two supports.

Prob. 5.2-1. For the simply supported beam AE in Fig.P5.2-1, (a) determine and , the internal resultants

just to the left of the 20-kN load at C , and (b) determineand , the internal resultants just to the left of the 20-kNload at D.

M D

V D

M C V C

SHEAR AND MOMENT IN BEAMS: EQUILIBRIUM

METHOD (METHOD OF SECTIONS)

5.7 PROBLEMS

In Problems 5.2-1 through 5.2.11 you are to determine the

internal resultants (transverse shear force V and bending

moment M) at specific cross sections of beams. Use the

sign conventions for V and M given in Fig. 5.6, p. 314,and always draw complete and correct free-body dia-

grams. The minus-sign superscript signifies “just to the

left of” the referenced point; the plus-sign superscript

means “just to the right of” the referenced point.

▼

B A C D E

10 kN 20 kN 20 kN

1 m 2 m1 m 1 m

P5.2-1, P5.4-5, P5.5-5, and P5.6-1

Prob. 5.2-2. For the beam AD in Fig. P5.2-2, (a) determinethe transverse shear force and the bending moment

at a section just to the left of the support at B, and (b)determine V E and M E at section E.M B

V B

Prob. 5.2-3. Transverse loads are applied to the beam inFig. P5.2-3 at A, C , and E, and a concentrated couple 3Pa isapplied to the beam at E. Determine expressions for (a) thetransverse shear force and bending moment at asection just to the left of the load at C , and (b) shearand moment just to the left of the support at D.Express your answers in terms of P and a

M D

V D

M C V C

2 kips1 kip

B

A D

C E

2 ft 2 ft 2 ft 2 ft

P5.2-2, P5.4-3, P5.5-3, and P5.6-2

B D

E A

C

3a 3a2a

P 2 P 2 P

3 Pa

2a x

P5.2-3, P5.4-2, P5.5-2, and P5.6-3

B A DC E

100 mm100 mm100 mm 150 mm

150 N

300 N

Prob. 5.2-5. Two transverse forces and a couple are applieas external loads to the cantilever beam AC in Fig. P5.2-Determine the transverse shear force V C and the bendinmoment M C at the fixed end C .

P5.2-4, P5.4-4, and P5.5-4

A B C12 kip . ft

6 kips

4 kips 4 ft 4 ft

P5.2-5, P5.4-8, P5.5-8, and P5.6-4

Prob. 5.2-6. For the cantilever beam AD in Fig. P5.2-6, dtermine the reactions at D; that is determine V D and MExpress your answers in terms of P and a.

a

2 P

A B C D

aa

2 P P

P5 2 6 P5 4 7 and P5 5 7

341_pdfsam_1Mechanics of Materials(3 Ed)[Team Nanban]Tmrg

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Transcript of 341_pdfsam_1Mechanics of Materials(3 Ed)[Team Nanban]Tmrg