axial force-shear force & bending moment diagrams

Thin-Walled Structures

99

CHAPTER 4

Axial Force, Shear Force and Bending Moment Diagrams

The following three methods to construct equilibrium shear force and bending moment diagrams for beams are described

•

the method of sections

•

the differential equation method, and

•

the semi-graphical approach.

In actual practice the distinction between these methods becomes blurred, because they may be used simul-taneously by the analyst. In the last section of this chapter the buoyancy force distribution on ships is described so that the longitudinal bending response can be computed.

4.1 Method of sections

If an aerospace or ocean vehicle structure is assumed to behave as a slender bar built-up from many members, then in order to determine stresses in the members we follow the approach in mechanics of materials and first determine the distribution of the internal forces and moments along the length. The objective of this section is to review how to determine these internal actions and plot them along the length of the structure. The relationship between the internal forces and moments to the stresses will be discussed in subsequent chapters. The review is limited to statically determinate problems such that internal forces and moments may be obtained by the equa-tions of statics alone.

Consider a straight slender bar with uniform cross section, which has a length 3

c

, and is simply supported at two locations as shown in Fig. 4.1. In a right-handed Cartesian coordinate system (

x,y,z

), take the

z

-axis is paral-lel to the length, the

y

-axis in the plane of the figure, and the

x

-axis perpendicular to the plane of the figure. The origin is at the left end of the bar; . The bar is in equilibrium subject to the external loads

F

and

which act in the

z-y

plane. Neglect the weight of the bar relative to the loads

F

and , as is frequently done

in structures where the applied loads are much greater than the weight of the structure. The load

F

is a point force acting at the left end of the bar, and is replaced by its horizontal and vertical components

F

z

and

F

y

. The

0 z 3c≤ ≤ py z( )

py z( )


100

Thin-Walled Structures

load is a distributed load and has units of force per unit length. It is assumed positive if it acts vertically

upward (positive y-direction), and negative if it acts downward. At a typical value for

z

we want to find the inter-nal axial force

N(z)

, shear force

V

y

(

z

), and bending moment

M

x

(

z

). We write the internal actions as

N(z)

,

V

y

(

z

), and

M

x

(

z

), since they are mathematically functions of the coordinate

z

. The basis for their determination is equi-librium.

Free-body diagrams of the bar removed from its supports and by imagining the bar is cut at some value of

z

are shown in Fig. 4.2. The internal actions

N

,

V

y

, and

M

x

are shown as well as the unknown support reactions

A

y

,

B

z

, and

B

y

. Let us assume the distributed load vanishes in this discussion. Internal actions

N

,

V

y

, and

M

x

are shown in their assumed positive senses. A positive

z

-face has its outward normal in the positive

z

-direction, and a negative

z

-face has its outward normal in the negative

z

-direction. That is, on a positive

z

-face a positive axial force

N

acts in a positive

z

-direction, a positive shear force

V

y

acts in the positive

y

-direction, and the posi-tive moment

M

x

acts clockwise, or as a vector in the positive

x

-direction by the right-hand screw rule. By New-ton's third law on action/reaction, positive values of

N

,

V

y

, and

M

x

acting on a negative

z

-face have a sense opposite to their positive values on the positive

z

-face. The sign convention for the internal forces and moments needs be carefully followed.

The usual procedure to determine

N(z)

,

V

y

(

z

), and

M

x

(

z

), is to first draw an overall free-body diagram of the bar removed from its supports and find the unknown support reactions, and then section the bar at various

z

-loca-tions to find

N

,

V

y

, and

M

x

. The reader should verify that the support reactions in Fig. 4.2 are ,

, and , where

F

y

and

F

z

are the known applied loads. Note that the force

A

y

has a sense

opposite to what was originally assumed.

The overall equilibrium free-body diagram of the bar is shown at the top of Fig. 4.3. The diagrams shown from top to bottom directly below the overall free-body diagram are the axial force diagram, the shear force dia-gram, and the bending moment diagram, respectively. The axial force diagram is obtained by sectioning the bar at any value of

z

, 0 <

z

< 3c, placing a positive internal force

N

on the cut faces, and then summing forces in the

z

-direction for one of the two free-body diagrams. Thus,

N

(

z

) is a constant, for all

z

, 0 <

z

< 3c. In a similar man-ner one obtains

V

y

(

z

) and

M

x

(

z

). Note that the shear force has a discontinuity at

z

= c since a point force acts there. It is necessary to consider free-body diagrams in two separate ranges of

z

to draw the shear force diagram; 0 <

z

< c, and c <

z

< 3c. The magnitude of the jump in the shear force is equal to the magnitude of the point

z

y

F Fy

Fz

c 2c

Fig. 4.1 A slender bar simply supported at two locations and subjected to force F and distributed load .py z( )

py z( )

py z( )

py z( )

Ay 3 2⁄–( )Fy=

Bz Fz= By Fy 2⁄=

Thin-Walled Structures 101

Differential equation method

force: . The shear force is a piecewise constant for this problem. The bending

moment is piecewise linear; i.e., Mx(z) = z Fy, 0 < z < c, and Mx(z) = (3c - z)(Fy/2), c < z < 3c. The bending

moment is continuous at z = c, but has a discontinuity in slope , where

. As illustrated in Fig. 4.2, there are two free-body diagrams for each cut. Either one may be

used to find N, Vy, and Mx. If the left free-body diagram is used to find N, Vy, and Mx, then equilibrium of the right free-body diagram will give the same values of N, Vy, and Mx. If it does not, there is either a math error, the sign convention is violated, or overall equilibrium is in error. Sketching the axial force, shear force, and bending moment diagrams by the method illustrated in this section is called the method of sections.

4.2 Differential equation method

The distributed load intensity , the shear force Vy(z), and bending moment Mx(z) are related by simple dif-

ferential equations at z, if no point forces or concentrated couples act at z. These differential equations are useful in the construction of the shear force and bending moment diagrams.

Fy

Fz N

Vy

Mxz

Vy

Mx

z

c

Ay

2c

By

Bz

Left and right FBD’s for cuts in the range 0 < z < c

Fy

Fz

Ay By

Bz

c 2c

Fig. 4.2 Free body diagrams (FBDs) of the slender bar with = 0.py z( )

overall FBD

V y c+( ) V y c-( )– 3 2⁄( )Fy=

M'x c+( ) M'x c-( )– 3 2⁄( )Fy=

M'x dMx dz⁄=

py z( )


102 Thin-Walled Structures

Consider a portion of a straight beam subjected to distributed load whose intensity is as shown in Fig.

4.4. By convention is positive upwards and negative downwards. A free-body diagram of a portion ∆z-

long of the beam is shown in Fig. 4.5. The shear force and bending moment change with z, and so their values at z + ∆z are different from their values at z. The distributed load acting on the segment ∆z is replaced by single force of magnitude acting long a line of action given by z = z*, where z < z* < z + ∆z. Mathematically this

is permissible by the mean value theorem (for integrals) for continuous functions of a single variable, and from this theorem we have

Fz

Fy

3/2 Fy

Fy/2

z

z

z

z

0

N

Fz

Vy

0

- Fy

Fy/2

Mx

0

- cFy

c 2c

+ N

Vy

Mx

N

VyMx

Fig. 4.3 Axial force N, shear force Vy, and bending moment Mx diagrams for the slender bar with = 0,py z( )

py z( )

py z( )

py∆z



Note that in the limit as , , and .

In the free-body diagram of Fig. 4.6 we sum forces vertically, divide by ∆z, and take the limit as to get

(4.1)

Often it is more convenient to use an integrated form of eq. (4.1). Integrating it from z1 to z we obtain

py1∆z------ py ζ( ) ζd

z

z ∆z+( )

∫=

∆z 0→ z* z→ py py→

z

y

Fig. 4.4 Distributed load intensity shown acting in the positive sense

py z( )

Mx + ∆Mx

Mx

Vy + ∆Vy

Vyz

z + ∆z

Fig. 4.5

py z( )

Mx + ∆Mx

Vy + ∆Vy

Vy

Mx

py z*( )∆z

z + ∆zz*z

Fig. 4.6

∆z 0→

zd

dV y py–=



(4.2)

The integrated form is valid only if there are no point loads between z1 and z. If we sum moments at z in the free-

body diagram, divide by ∆z, and take the limit as we get

(4.3)

The integrated form of eq. (4.3) is

(4.4)

which is valid if no point couples act between z1 and z. Equations (4.1) and (4.3) are the differential equations of equilibrium for the beam. Equation (4.1) is valid at z if no point force acts there, and eq. (4.3) is valid at z if no point couple acts there. Considering separately the equilibrium of a point force F0 and a point couple with moment magnitude C0 acting at z = z0, as shown in Fig. 4.7, we obtain

(4.5)

Distributed loads may be replaced by a resultant force acting at the center of pressure. This procedure is con-venient in many situations. For example, the air load distribution on a wing may be replaced by lift and drag forces acting at the center of pressure. A segment of a beam from z = z1 to a typical value of z, z > z1, which has a

distributed load with intensity acting on it is shown in Fig. 4.8. Using ζ as a dummy variable to measure

the axial position, the resultant force is

(4.6)

and the center of pressure is given by

V y z( ) V y z1( ) py ζ( ) ζd

z1

z

∫–=

∆z 0→

zd

dMx V y=

Mx z( ) Mx z1( ) V y ζ( ) ζd

z1

z

∫+=

V y z0+( )

Mx z0+( )

V y z0 -( )

Mx z0 -( )

εz

y

z0

F0

C0

Fig. 4.7 Point force and couple acting at z0 on an inifintesimal beam element

V y z0+( ) V y z0

-( )– F0–= Mx z0+( ) Mx z0

-( )– C0–=

py z( )

Fy z( )

Fy z( ) py ζ( ) ζd

z1

z

∫=

zp z( )



(4.7)

Both the resultant force and center of pressure are functions of z. Equations (4.6) and (4.7) are con-

ditions of statical equivalence for the distributed load .

EXAMPLE 4.1 Cantilever wing with tip tank

Consider the cantilever wing with tip tank as shown in Fig. 4.9. Given the weight of the tip tank and its contents

W, the distance e of the weight W from the wing tip, the wing span L, and the value of the distributed load inten-sity at the wing root, determine the shear force and bending moment along the span. The solution to this

problem is given by a Mathematica 4.0 program listed below.

V y z1( )

Mx z1( )

V y z( )

Mx z( )

z1

zp z( ) z

Fy z( )V y z1( )

Mx z1( )

V y z( )

Mx z( )

z1

ζz

py z( )

Fig. 4.8 Resultant force at the center of pressure statically equivalent to a distributed load.

zp z( )Fy ζpy ζ( ) ζd

z1

z

∫=

Fy z( ) zp z( )

py z( )

py z( ) p0zL---=

z

L

e

W

Fig. 4.9 Cantilever wing with tip tank.

p0



1 2 3 4 5 6

ü Example 4.1Shear force and bending moment diagrams for a cantilever wing with tip tank.

Input the distributed load function.

py = p0zÅÅÅÅL;

The shear force and bending moment distributions are determined from eqs. (4.2) and (4.4) with z1 = 0. The shear force and bending moment at the wing tip (z = 0) are denoted by Vy0 and Mx0 , respectively.

Vy = Vy0 - ‡ py „z

-z2 p0ÅÅÅÅÅÅÅÅÅÅÅÅÅ2 L

+ Vy0

Mx = Mx0 + ‡ Vy „z

Mx0 -z3 p0ÅÅÅÅÅÅÅÅÅÅÅÅÅ6 L

+ z Vy0

Boundary conditions at the wing tip, obtained from equilibrium of the tip tank, deter-mined the shear force Vy0 and moment Mx0 .

bc1 = HVy ê. z Ø 0L - Wbc2 = HMx ê. z Ø 0L - e W

-W + Vy0

-e W + Mx0

slv1 = Solve@bc1 == 0, Vy0D

88Vy0 Ø W<<

Vy0 = Vy0 ê. slv1@@1DD

W

slv2 = Solve@bc2 == 0, Mx0D

88Mx0 Ø e W<<

Mx0 = Mx0 ê. slv2@@1DD

e W



Print@ "Shear force VyHzL = ", VyDPrint@"Bending moment MxHzL = " MxD

Shear force VyHzL = W -z2 p0ÅÅÅÅÅÅÅÅÅÅÅÅÅ2 L

Bending moment MxHzL = ikjje W + W z -

z3 p0ÅÅÅÅÅÅÅÅÅÅÅÅÅ6 L

y{zz

Plot the shear force and bending moment diagrams for the following parameter values: L = 144 in, p0 = 70 lb/in, W = 500 lbs, and e = 6 in. (Plots labled p1 and p2 have been suppressed using the DisplayFunction option.)

p1 =Plot@HVy ê. 8 L Ø 144, p0 Ø 70, W Ø 500, e Ø 6<L,

8z, 0, 144<,PlotRange Ø 81000, -5000<,GridLines Ø Automatic,AxesLabel Ø 8"z,inches", "Vy, lbs"<,PlotLabel ØStyleForm@"Shear force diagram",

"Section"D,DisplayFunction -> IdentityD

Ü Graphics Ü

p2 =Plot@HMx ê. 8 L Ø 144, p0 Ø 70, W Ø 500, e Ø 6<L,

8z, 0, 144<,GridLines Ø Automatic,AxesLabel Ø 8"z,inches", " Mx, lb-in"<,PlotLabel ØStyleForm@"Bending moment diagram",

"Section"D,DisplayFunction -> IdentityD

Ü Graphics Ü

Show@GraphicsArray@88p1<, 8p2<<DD

1 2 3 4 5 6

The magnitude of the bending moment is largest at the wing root, and this vlaue is important for wing structural design. Its value in lb-in is

Mx ê. 8 L Ø 144, p0 Ø 70, W Ø 500, e Ø 6, z Ø 144<

-166920

(The plots are are shown on the next page.)



1 2 3 4 5 6

20 40 60 80 100 120 140z,inches

-50000

-40000

-30000

-20000

-10000

10000

Mx, lb-inBending moment diagram

20 40 60 80 100 120 140z,inches

-5000

-4000

-3000

-2000

-1000

1000Vy, lbs Shear force diagram

Shear force and bending moment diagrams for the cantilever wing with tip tank



The shear force and bending moment at the wing tip are determined from equilibrium of the tip tank, which gives Vy(0) = W and Mx(0) = eW. Note that eq. (4.1) shows that the slope on the shear diagram is equal to the

distributed load intensity. For example at z = 0, , so that the slope of the shear diagram is zero at z = 0.

Similarly, the slope on the moment diagram is equal to the shear force as given by eq. (4.3). In particu-

lar, at z = 45.36 inches the shear force is zero. Thus, in the bending moment diagram the moment is stationary at z = 45.36 in (i.e., it has a horizontal slope), and Mx may be either a local maximum, minimum, or a value corre-sponding to horizontal inflection point. It is important to compute the largest magnitude of the bending moment, and this is accomplished by checking the bending moments where Vy = 0, at the end points of the beam, and the locations where the bending moment is discontinuous. Note that the maximum bending moment magnitude occurs at the wing root, where . The bending moment changes sign, and hence vanishes, at

z = 81.4 inches. In a plot of the beam deflection versus z, which is not shown above, the location z = 81.4 inches is called an inflection point because the curvature is changing from concave down (positive Mx) to concave up

(negative Mx) as z increases through z = 81.4 inches.

EXAMPLE 4.2 The air load acting on a wing given as discrete data.

The problem statement and data for this example is taken from the aircraft structures text by Peery (1950). How-ever, the notation is changed to that of the this text, and the solution is given in terms of a Mathematica 3.0 pro-gram. Since the air load on the wing is given at discrete spanwise locations and not as a mathematical function, it is useful to use Mathematica’s list manipulation capabilities to effect the solution. Before giving the problem statement, we will discuss some aspects of list manipulations.

Lists provide a mechanism for representing arrays, vectors, matrices, and for grouping together objects such as data, variables, or expressions. A list is a collection of objects whose symbols are enclosed in braces, {}, and separated by commas, as in . It is usually more efficient to do operations on

lists rather than to do operations on individual items in the list. A function is applied separately to each element in the list if it has the attribute “Listable”. For example, addition, multiplication, and the logarithm have the attribute Listable, so that

That is, Listable functions in Mathematica are automatically distributed or “threaded” over lists that appear as its arguments. The number of elements in a list is given by the built-in function Length [ ]; e.g.,

. A summary of Mathematica’s built-in functions used for list manipulation is given in the table below.

Summary of list manipulation functions in Mathematica (taken from Blachman, 1992)

Function Description

Range [min, max, step] Generates the list {min, .., max} using step (arithmetic progression)

Table [expr, {imax}] Generates a list of imax copies of expr (more general)

Array [s, dim] Generates a list of length dim with elements s[i]

Sort [list] Sorts elements of list into canonical order

py 0( ) 0=

dMx dz⁄

Mx 166920 lb-in–=

item1 item2 item3 … itemn, , , ,{ }

5 8 11, ,{ } 2 3– 6–, ,{ }+ 7 5 5, ,{ }=

5 8 11, ,{ }* 2 3– 6–, ,{ } 10 24– 66–, ,{ }=

Log 5 8 11, ,{ }[ ] Log 5[ ] Log 8 ][ ] Log 11 ][ ], ,{ }=

Length 5 8 11, ,{ }[ ] 3=



Problem statement: The aerodynamic loads on an airplane wing cannot be represented by a simple equa-tion. The load per inch of span, , of the airplane wing shown in Fig. 4.10is tabulated in column

two of the Table printed at line “Out[14]” in the Mathematica program below. Find the shear force and bending moment diagrams for the wing.

Solution: The values of the shear force and bending moment at various points along the wing are calculated in the Table (see Out[14] in the code). The points are called stations and are designated by their distances from

Reverse [list] Reverses elements in list

RotateLeft [ list, n] Cycles the elements n positions to the left

RotateRight [ list,n] Cycles the elements n positions to the right

Permutations [list] Generates a list of all possible permutations of the elements of list

Drop [list, n] Drops the first n elements from list

Take [list, n] Takes the first n elements from list

First [list] Give the first element of list

Last [list] Gives the last element of list

list [[n]] or Part [list, n] Gives the nth element

Rest [list] Returns all but the first element of list

Select [list, crit] Picks out elements in list which meet the criterion of crit

Append [list, elem] Returns a list with elem appended to the end of list

AppendTo [list, elem] Changes list by appending elem to the end

Prepend [list, elem] Returns a list with elem added to the from of list

PrependTo [list, elem] Changes list by adding elem to the front

Insert [list, elem, n] Inserts elem at position n in list

Length [list] Gives the number of elements in list

Dimensions [list] Gives the dimensions of a list or expression

Complement [ list1, list2, ... ] Gives the complement, i.e., those elements in list1 but not in list 2, ...

Intersection [ list1, list2, ... ] Gives a sorted list of all the elements common to all list1, list2, ...

Union [list1, list2, ... ] Gives a sorted list of the distinct elements

Join [list1, list2, ... ] Joins or concatenates lists together

Partition [list, n] Partition list into sublists of length n

Flatten [list] Flattens out nested lists, i.e., eliminates nested lists

Transpose [list] Transpose

Apply [f, list] Replaces the head of list with f

Map [f,list] Applies f to each element in list

Listable An attribute, if set, automatically maps a functions onto a list

ColumnForm[list] Prints list as a column

MatrixForm[ list] Prints elements in list in a regular array

Summary of list manipulation functions in Mathematica (taken from Blachman, 1992)

Function Description

py z( ) p z( )=



the centerline of the airplane, as shown in Fig. 4.10. These distances are measured along the wing rather than horizontally, since the air loads are perpendicular to the wing. The distances between stations, , are computed as a list in the code. The value of the shear at any point is obtained as the area under the load curve from that point out to the wing tip. The load curve is assumed to be a series of straight lines between the known points, and the area is obtained as the sum of the areas of the trapezoids. The area of the trapezoids are obtained as the prod-uct of the average height and the base . The change in the shear between two stations is equal to

the area of the load curve between the stations. The shear is then obtained by summation of the -values.

The change in the bending moment between two stations is equal to the area under the shear curve. This

area is also assumed trapezoidal and is obtained by multiplying the sum of the shears at the adjacent stations by one-half the distance between the stations. The bending moments are obtained by a summation of the -val-

ues. Plots of the air load, shear force, and bending moment distributions are shown at the end of the Mathematica program

z

Fig. 4.10

∆z

pave ∆z ∆V y

V y ∆V y

∆Mx

∆Mx

Example 4.2: Numerical quadrature for the shear force and bending moment in a wing (Peery, 1950, pp. 107-109)

In[1]:= Off@General::spell1D

Input the airload intensity at each z-station and each z-station coordinate as two separate lists. Dimensional units: z-list, inches; p-list, lb/in.

In[2]:= z = 80, 20, 40, 60, 80, 100, 120, 140, 160, 180,200, 220, 225<;

p = 8125, 123, 120, 116, 111, 105, 98, 89, 80, 71,58, 35, 0<;

Compute distances between stations.

In[3]:= Dz = Take@HRotateLeft@zD - zL, Length@zD - 1D

Out[3]= 820, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 5<



. 1 2 3 4 5 6 7

Compute average airload intensity in each interval between stations.

In[4]:= pave = Take@N@HRotateLeft@pD + pL ê 2D, Length@pD - 1D

Out[4]= 8124., 121.5, 118., 113.5, 108.,101.5, 93.5, 84.5, 75.5, 64.5, 46.5, 17.5<

The trapezoidal rule of numerical integration is used to compute the change in the shear force over each interval from eq. (4.2). A list of DVy - values is computed by a direct multiplication of lists Dz and pave .

In[5]:= DVy = - Dz * pave

Out[5]= 8-2480., -2430., -2360., -2270., -2160.,-2030., -1870., -1690., -1510., -1290., -930., -87.5<

Since Vy HtipL - Vy HrootL = Ÿroottip HdVy ê dzL „ z, and Vy HtipL = 0, the shear force at the root is the negative of

the sum the elements in the DVy -list. A simple way to sum elements in a list is to change the Head of the list from "List" to "Plus" by using the Apply function.

In[6]:= Head@DVyD

Out[6]= List

In[7]:= Vy0 = - Apply@Plus, DVyD

Out[7]= 21107.5

The shear force at station i is the the partial sum of the of the DVy - values over the intevals from 1 to i; i.e.,

Vy HiL = ⁄ j=1i DVy H jL + Vy0 . We use the Table function to generate a list of the shear force values at each

station.

In[8]:= Vy = Table@HSum@DVy@@jDD, 8j, 1, i<D + Vy0L, 8i, 1, Length@DVyD<D

Out[8]= 818627.5, 16197.5, 13837.5, 11567.5, 9407.5,7377.5, 5507.5, 3817.5, 2307.5, 1017.5, 87.5, 0.<

Add the value of the shear force at the root to the beginning of this list in order to have the shear force at each zi - station, including the root.

In[9]:= Vy = PrependTo@Vy, Vy0D

Out[9]= 821107.5, 18627.5, 16197.5, 13837.5, 11567.5, 9407.5,7377.5, 5507.5, 3817.5, 2307.5, 1017.5, 87.5, 0.<



1 2 3 4 5 6

Compute the average force in each interval from the list of shear force values.

In[10]:= Vave = Take@N@HRotateLeft@VyD + VyL ê 2D, Length@VyD - 1D

Out[10]= 819867.5, 17412.5, 15017.5, 12702.5, 10487.5,8392.5, 6442.5, 4662.5, 3062.5, 1662.5, 552.5, 43.75<

The change in the bending moment is computed from eq. (4.4) using the trapezoidal rule of numerical integration.

In[11]:= DMx = Dz * Vave

Out[11]= 8397350., 348250., 300350., 254050., 209750.,167850., 128850., 93250., 61250., 33250., 11050., 218.75<

Since Mx HtipL - Mx HrootL = Ÿroottip HdMx ê dzL „ z, and Mx HtipL = 0, the bending moment at the root is the

negative of the sum the elements in the DMx -list.

In[12]:= Mx0 = -Apply@Plus, DMxD

Out[12]= -2.00547 µ 106

Compute the bending moment at each station.

In[13]:= Mx = Table@HSum@DMx@@jDD, 8j, 1, i<D + Mx0L,8i, 1, Length@DMxD<D;

Mx = PrependTo@Mx, Mx0D

Out[13]= 9-2.00547 µ 106, -1.60812 µ 106, -1.25987 µ 106,

-959519., -705469., -495719., -327869., -199019.,

-105769., -44518.7, -11268.8, -218.75, 0.=

In[14]:= TableForm@Transpose@8z, p, Vy, Mx<D,TableHeadings Ø

8None, 8"z,in", "p,lbêin","Vy,lb", "Mx,lb-in"<<D



1 2 3 4 5 6 7

Out[14]//TableForm=

z,in p,lbêin Vy,lb Mx,lb-in

0 125 21107.5 -2.00547 µ 106

20 123 18627.5 -1.60812 µ 106

40 120 16197.5 -1.25987 µ 106

60 116 13837.5 -959519.

80 111 11567.5 -705469.

100 105 9407.5 -495719.

120 98 7377.5 -327869.

140 89 5507.5 -199019.

160 80 3817.5 -105769.

180 71 2307.5 -44518.7

200 58 1017.5 -11268.8

220 35 87.5 -218.75

225 0 0. 0.

Plots of the airload, shear force, and bending moment distributions. (Intermediate plots have been suppressed.)

In[15]:= p11 = ListPlot@Transpose@8z, p<D,PlotStyle Ø [email protected]<,

DisplayFunction -> IdentityDp12 = ListPlot@Transpose@8z, p<D,

PlotJoined -> True,DisplayFunction -> IdentityD

p1 = Show@p11, p12,AxesLabel Ø 8"z, in.", "lbêin"<,PlotLabel Ø "Airload distribution",

DisplayFunction -> IdentityDp21 = ListPlot@Transpose@8z, Vy<D,

PlotStyle -> [email protected]<,DisplayFunction -> IdentityD

p22 = ListPlot@Transpose@8z, Vy<D,PlotJoined -> True,

DisplayFunction -> IdentityDp2 = Show@p21, p22,

AxesLabel Ø 8"z, in.", "lb"<,PlotLabel Ø "Shear force",

DisplayFunction -> IdentityDp31 = ListPlot@Transpose@8z, Mx<D,

PlotStyle -> [email protected]<,DisplayFunction -> IdentityD



In[24]:= Show@GraphicsArray@88p1<, 8p2<, 8p3<<DD

50 100 150 200z,in

-2µ106

-1.5µ106

-1µ106

-500000

lb-in Bending Moment

50 100 150 200z, in.

5000

10000

15000

20000

lb Shear force

50 100 150 200z, in.

20

40

60

80

100

120

lbêin Airload distribution



4.3 Semi-graphical method

The semi-graphical method to draw the shear force and bending moment diagrams is best illustrated by doing an example.

EXAMPLE 4.3 Uniform barge with symmetric load

Consider a barge at rest in still water with a uniform immersed cross section, and subjected to the symmetrical loads shown in Fig. 4.11. This is an example of a structure with no boundary supports, and is typical of aero-

space and ocean vehicle structures. We wish to sketch the shear force and bending moment diagrams for the barge. In this example there is a distributed load acting on the barge due to buoyancy forces produced by displac-ing the water. Let represent the distributed load intensity due to buoyancy, and is a constant along the

barge because the immersed cross section is uniform and the water is still.

Solution: A semi-graphical method is used to sketch the shear force and bending moment diagrams. In this approach we first sketch the distributive load , then the shear force Vy(z), and finally the bending moment

Mx(z). Equations (4.1) and (4.3) are used to note that the slope of shear diagram at z is the negative of the distrib-uted load intensity at z, and the slope of the moment diagram at z is the shear force at z. In addition, eqs. (4.2) and (4.4) set at z = z2 give

(4.8)

(4.9)

Equation (4.8) is interpreted in a graphical sense to mean that the difference in the shear force between z2 and z1 is the area under the distributed loading diagram from z1 to z2. This is not geometrical area. The area

between the curve and the z-axis has units of force, and may be positive, zero, or negative. Similarly eq.

(4.9) is interpreted to mean the difference in the bending moment is the area under the shear force diagram.

Vertical equilibrium of the entire barge requires the buoyant upthrust equals 40kN, so that .

The total distributed load intensity is the difference between and the magnitude of the downward acting

15kN (total) 15kN (total)10kN

5m 5m 5m 5m

Fig. 4.11Uniform section barge in still water with symmetric load.

pb pb

py z( )

V y z2( ) V y z1( )– py z( ) zd

z1

z2

∫–=

Mx z2( ) Mx z1( )– V y z( ) zd

z1

z2

∫=

py z( )

pb 2kN( ) m⁄=

pb


Semi-graphical method

applied loading intensity. The distributed loading intensity diagram is constructed in this manner as shown in Fig. 4.13.

5 10 15 20z, m

Shear force diagram

-4

-2

2

4

kN

Fig. 4.12Shear force and bending moment diagrams for the barge in still water

0

2

-1

py

z, m5 10

1520

kN/m

10 kN

5 10 15 20z, m

Bending moment diagram

2.5

5

7.5

10

12.5

15

17.5

kN-m



The point force of 10kN acting at z = 10m is shown schematically in the -diagram as a downward

pointing arrow. Actually, as , because a point force is a finite load acting over zero length.

Point forces are idealizations to actual loads and introduce discontinuities in the mathematical descriptions of some of the dependent variables. The reader should verify the distributive loading intensity diagram of Fig. 4.13.The shear force diagram is drawn below the loading intensity diagram in Fig. 4.13. Equilibrium at z = 0 requires Vy(0) = 0, and the slope dVy/dz at z = 0 is equal to 1kN/m. The slope is constant between , thus Vy(z) is a straight line in this range of z. The difference in the shear force between z = 5m and z = 0 is equal

to the negative of the area under the curve which is 5kN. Thus Vy(5) = 5kN since Vy(0) = 0. At z = 5+m the

loading intensity jumps to +2kN/m. The slope of the shear force jumps from 1kN/m to -2kN/m at z = 5m, but the shear force is itself continuous. The difference Vy(10) - Vy(5) is equal to the negative of the area between the

-curve and the z-axis between z = 5m and z = 10m. Thus Vy(10) - Vy(5) = -10kN, so Vy(10) = -5kN. Note

the shear force is zero at z = 7.5m. At z = 10m the point force of 10kN acts. According to the first of eqs. (4.5)

Vy(10+) - Vy(10-) = 10kN, so that Vy(10+) = 5kN. The slope of the shear at z = 10m is +2kN/m, and remains con-

stant until z = 15m. The difference Vy(15) - Vy(10+) = -10kN, so that Vy(15) = -5kN. Finally, the slope changes to

-1kN/m at z = 15+m and remains constant in the range 15 < z < 20. The difference Vy(20) - Vy(15) = 5kN, so that Vy(20) = 0. Checking vertical equilibrium at z = 20m verifies that Vy(20) should be zero.

Moment equilibrium at z = 0 shows Mx(0) = 0. The slope of Mx at z = 0 is equal to the shear force at z= 0.

Hence at z = 0, as shown in Fig. 4.13. The slope on the moment diagram increases linearly

from zero at z = 0 to 5kN at z = 5m. Thus Mx(z) is parabolic from z= 0 to z = 5. The difference Mx(5) - Mx(0) is equal to the area under shear diagram from z = 0 to z = 5m. Hence, Mx(5) - Mx(0) = 12.5 kNm, and Mx(5) = 12.5 kNm since Mx(0) = 0. From z = 5 to z = 7.5 the slope of the moment decreases from 5kN to zero. At z = 7.5, Mx is a local maximum with a magnitude of 18.75 kNm. The slope of Mx(z) for 7.5 < z < 10 is negative, decreasing linearly from zero to -5kN. The difference Mx(10) - Mx(7.5) = -6.25 kNm, so that Mx(10) = 12.50 kNm. The slope of Mx(z) at z = 10m jumps from a -5kN to a +5kN as shown in Fig. 4.13, but the moment itself is continu-ous. The bending moment diagram in the range 10 < z < 20 is completed in a manner similar to the description of its construction in the range 0 < z < 10.

In this example the shear force diagram is antisymmetric about z = 10m and the bending moment is symmet-

ric about z = 10m. This follows from the symmetrical loading on the barge and equilibrium eqs. (4.1) and (4.3).

4.4 Buoyancy Force Distribution on Ships

The simple uniform buoyancy distribution acting on the barge in Example 4.3 is an exception to the buoyancy distributions found in practice. It is true that equilibrium requires the total buoyant upthrust to equal the weight of the ship and its contents. However, the distribution of the buoyancy and weight along the length of the ship is not necessarily the same. The difference in the magnitudes of the buoyancy and weight distribution intensities is the applied load intensity . In ship design three conditions are recognized to compute for the same ship.

These conditions are called

• the still water condition,

• sagging condition, and

• the hogging condition.

py z( )

py ∞–→ z 10m→

0 z 5m< <

py z( )

py z( )

dMx( ) dz( )⁄ 0=

py z( ) py z( )


Buoyancy Force Distribution on Ships

A more detailed account of these conditions on the longitudinal bending of the ship is given by Muckle (1967) and Zubaly (1996), and here we only summarize the basic ideas.

A ship in still water is shown in Fig. 4.13, and a section between z and z + dz is shown in Fig. 4.14.

Archimedes' principle asserts that the buoyant upthrust is equal to the weight of the fluid displaced. Let A(z) denote the submerged cross section at z, and let γ denote the specific weight (force per volume) of the fluid. The differential buoyancy force dFb acting on the ship over a differential length dz is

(4.10)

Consequently, the buoyant upthrust per unit ship length, which we designate , is equal to γA(z); i.e.,

(4.11)

A curve of for a ship as well as the weight per unit length is shown in Fig. 4.15. Overall equilibrium requires

the area under these curves to have the same magnitude. If the submerged cross section is uniform in z, as is the

z

Fig. 4.13

A z( )dzdFb

Fig. 4.14

dFb γA z( )dz=

pb

pb

dFb

dz--------- γA z( )= =

pb

weight/length

buoyancy/length

Fig. 4.15



case for the barge in Example 4.3, the distribution of the buoyancy per unit length is a constant.

At sea a ship is subjected to waves, and this alters the buoyancy distribution. For longitudinal bending of the ship two extreme static conditions are assumed: sagging and hogging. In each condition, the length of the wave is assumed to be the length of the ship. This is an “accepted” assumption for the worst buoyancy distribution caus-ing the most severe bending of the ship.

The sagging condition is shown in Fig. 4.16. (Also see Fig. 1.3 on page 5.) The wave crests are at the bow and stern, and the wave trough is amidships. A schematic of the buoyancy per unit length is shown below the ship in Fig. 4.16. The immersed cross section is the largest at or near the wave crests, and is least near the trough. The intensity of the buoyancy distribution reflects this. In this condition the deck sags and is in compression while the bottom is in tension. The worst location to concentrate the cargo in the ship is amidships, as this will result in the largest bending moment.

The hogging condition is depicted in Fig. 4.17. Here the wave troughs are at bow and stern, and the crest is amidships. The immersed cross section is greatest near amidships and is least near bow and stern. The distribu-tion of the buoyancy per unit length , shown in Fig. 4.17, reflects this situation. In hogging the deck is in ten-

sion and the bottom is in compression. The worst possible locations to concentrate cargo is fore and aft, as this will produce the greatest bending moment in the ship.

pb

pb

Fig. 4.16Sagging

concentrated weight

pb

pb

Fig. 4.17Hogging

concentrated weights


References

4.5 References

Blachman, N.R., 1992, Mathematica: A Practical Approach, Prentice Hall, Englewood Cliffs, New Jersey, p. 133.

Muckle, W., 1967, Strength of Ship’s Structures, Edward Arnold Ltd., London, pp. 27-69.

Peery, D.J., 1950, Aircraft Structures, McGraw-Hill, New York, pp. 107-108.

Zubaly, R.M., 1996, Applied Naval Architecture, The Society of Naval Architects and Marine Engineers, Cornell Maritime Press, Inc., Centreville, Maryland, pp. 195-237.

4.6 Problems

1. The cantilever wing is subjected to a distributed air load , where the total lift (2

wings) at cruise, wing length ft., and . Also, the wing supports an

engine weighing 1000 lbs. Plot the loading diagram, shear force diagram , and bending moment diagram

as functions of z for ft. Partial answer: lb. and lb-ft.

2. A proposed solar airplane called Centurion is being designed to achieve semi-perpetual flight (Aviation Week & Space Technology, May 4, 1998, p.54). Centurion is a flying wing with a span of 206 ft., an 8-ft. chord, and no taper or sweep. The wing has five sections, one center, two mid-span, and two tips. It is supported by four landing pods. The tip sections have a dihedral to assist in turning and washout twist to prevent tip stall. The empty weight is predicted to be 1,105 lb., comprising 630 lb. for structure, 160 lb. for engines and propellers, 150 lb. for avionics, 75 lbs for batteries, 20 lb. of miscellaneous and 70 lb. for 7% growth. The aircraft should be able to take a 100 lb payload to 100,000 ft. It is powered by 14 electric motors producing a maximum of 2 hp. each. Assume the following: The span-wise airload distribution acting on the wing is as given in problem 1. Each engine is modeled as a concentrated weight acting its location on the wing. The payload, avionics, batteries, etc., lumped are together as a concentrated load at the center, and that the structural weight is uniformly distributed

py z( )2L

πzmax------------- 1 z( )2–=

L 20 000lbs,= zmax 32.5= z z zmax⁄=

V y z( )

Mx z( ) 0 z 32.5≤ ≤ V y 0( ) 9 000,= Mx 0( ) 131934–=

y

z

py z( )

MxMx

V ypy

+V y

6 ft.

32.5 ft.

1000 lb engine

fuselage



along the span. For steady level flight, determine the shear force and bending moment diagrams from the center-line of the wing to its tip, and show them in a sketch. Label significant points. The front view of half of the Cen-turion is shown below.

3. The barge shown below has a uniform cross section along its length and is subjected to a uniformly distrib-uted load of intensity , in which P has dimensional units of force/length. Also it is subjected to

buoyancy for the extreme hogging condition . Draw the shear force,

, and bending moment, , diagrams. Label significant points. Note that .

4. A barge has a plan view as shown. All waterplanes are identical. Cargo is loaded evenly in the four rectangu-lar holds as shown. Neglecting the weight of the barge itself, construct curves of weight, buoyancy, load, shear, and bending moment for the loaded barge in still sea water. Label the values of each curve at each bulkhead, and identify the maximum shear and bending moment. (Zubaly, 1996)

z1

z2

z3z4z5

z6

z7z8

z9

z10

z

z1 4.0 ft.

z2 12.0 ft.

z3 19.2 ft.

z4 32.7 ft.

z5 46.3 ft.

z6 60.0 ft.

z7 67.1 ft.

z8 80.6 ft.

z9 93.4 ft.

z10 103.0 ft. Centerline

py z( ) P–=

py z( )buoyancy

P 1 πzL---

cos–=

V y z( ) Mx z( ) Mx max2P L π⁄( )2=

zL L

P

waterline


Problems

5. A barge of uniform rectangular construction has a length of 30m, breadth of 10m, depth of 5m, floats at an even keel in fresh water at a draft of 2m when unloaded. The barge is transversely divided into three equal com-partments. These compartments are uniformly loaded as follows:

No. 1 hold, 200 tonne; No. 2 hold, 155 tonne; No. 3 hold, 245 tonne

(Note: one metric ton, or tonne, is equal to 1000 kg, and the mass density of fresh water is 1 tonne/m3)

You will plot the loading intensity diagram, shear force diagram, and the bending moment diagram for the loaded barge in a column format. Do not neglect the weight of the barge itself.

a) Since the moments of the weight about amidships are not equal for the loaded barge, the barge trims. Assume the trim angle is small. Show from overall equilibrium that the draft at z = 0 is , and

that the draft at z = L = 30m is

b) Plot the loading intensity diagram, , where the loading intensity is in N/m, Newton/meter.

(The specific weight in N/m3 is g times the mass density in kg/m3. If for simplicity g is taken as 10

(instead of 9.8) then specific weight of fresh water is .)

c) Determine the shear force, and plot it directly below the loading intensity diagram. Note; the dimen-sional unit of the shear force is N, or Newtons.

d) Determine the bending moment, and plot it directly below the shear diagram. Note; the dimensional units of the bending moment are Nm, or Newton-meters.

40 ft 40 ft 40 ft 40 ft 40 ft 40 ft

30 ft400tons

950tons

400tons

950tonsempty empty

d0 3.7 m=

dL 4.3m=

py z( )

10m/sec2 1000kg/m3× 10 000N/m3,=

d0 dL200t 155t 245t

y

ztrim angle

2m

unloaded loaded

5m

30m

axial force-shear force & bending moment diagrams

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