Mechanics of solids by crandall,dahl,lardner, 3rd chapter

Chapter 3

Forces and Moments Transmitted by Slender

Members

Mechanics of Solids Vikas Chaudhari 2

Contents:

Slender members

Determination of Forces and moments under point loads

Sign conventions for shear force and Bending moment

Shear force and Bending moment Diagram

Distributed Loading

Differential Equilibrium relationships

Singularity functions


Slender members:

These are load carrying elements having much greater length (at least five times) than its lateral dimensions

Examples:

beam, columns, shafts, rods, stringers, struts, and links.

Slender members can be pulled, bent and twisted




1. A general method for determining the internal forces and moments acting across any section of a slender member which is in equilibrium is to cut and that part is isolated from the system.

GENERAL METHOD TO FIND INTERNAL FORCES AND MOMENTS

2. The isolated part is in equilibrium. Apply the equations of equilibrium to find internal forces and moments


x- Section : The section normal to x- axis

y- section : The section normal to y- axis

z- section : The section normal to z- axis


XZ

Y

MXZ

MXY

MXXFXZ

FXY

FXX

Forces and moments acting on a cross section of a member.

FXX: Force acting on x-

section and is along x- axis

Similarly FXY, FXZ, MXX, MXY and MXZcan be explained


Positive face of given sectionIf the outward normal points in a positive coordinate direction then that face is called as positive face

If the outward normal points in a negative coordinate direction then that face is called as Negative face

Negative face of given section

NOMENCLATURE

FXX :- Axial force :This component tends to elongate the member and is often given the symbol F or Fx.


FXY, FXZ :- Shear force:These components tend to shear one part of the member relative to the adjacent part and are often given the symbols V, or VY and VZ

MXX:- Twisting moment:This component is responsible for the twisting of the member about its axis and is often given the symbol MT or MTZ.

MXY, MXZ :- Bending moments:These components cause the member to

bend and are often given the symbols Mb, or Mby and Mbz.


sign convention for the axial force, shear force, and bending moment.

If force or moment component acts on a positive face in a positive coordinate direction then these components are treated as positiveIf force or moment component acts on a negative face in a negative coordinate direction then these components are treated as positiveIf force or moment component acts on a positive face in a negative coordinate direction then these components are treated as negativeIf force or moment component acts on a negative face in a positive coordinate direction then these components are treated as negative


Positive axial force Fx is a tensile force


P

P

P

P

Positive Force (Tensile)

Negative Force (Compressive)


If the plane of loading is the x-y plane then only three components occur:The axial force Fxx (F), the shear force Fxy, (V), and the bending moment Mxz (Mb),

Force and moment components in two dimensions


The shear force, V. The bending moment, M.

Sagging

Hogging

Sign conventions for SF and BM


The steps involved in solving the problems

1. Idealize the actual problem, i.e., create a model of the system, and isolate the main structure, showing all forces acting on the structure.

& 0 =ΣF 0 =ΣM

2.Using the equations of equilibrium calculate unknown external forces or support reactions

3. Cut the member at a point of interest, isolate one of the segments, and repeat step 2 on that segment.


Example 3.1As an example, let us consider a beam supporting a weight near the center and resting on two supports, as shown in Fig. 3a. It is desired to find the forces and moments acting at section C.

If the beam is not completely rigid, it will tend to bend slightly, as in Fig. b.


FY = RA+ RB – W = 0 (a)ΣF = 0

Step 1

Step 2

ΣM = 0

ΣMA = RBL – Wa = 0 (b)

Σ MB = Wb – RAL = 0 (c)


which gives RA = Wb/L and RB = Wa/L.Step 3To find the internal forces and moments consider the beam is cut at point C, which is point of interest

Calculation of shear force and bending moment at a considered section of a beam.


Positive ;

Negative ;

xL

WbxRM

LWbRV

Ab

A

==

==

} --- (d)

Shear Force and Bending Moment Diagrams:Graphs that show shear force and bending moment plotted against distance along beam are called as shear force and bending moment diagrams respectively


D

Part AD

V = RA = Wb/L (Negative shear)So for the part AD shear force is constant

Mb = (Wb/L) x (Positive BM)At x = 0 i.e. At point A; Mb = 0

At x = a i.e. At point D; Mb = Wab/L

x c

ci.e. 0 < x < a


D

Part DB

V = W - RA = Wa/L (Positive shear)So for the part DB shear force is constant

Mb = Wa – (Wa/L) x = (Wa/L) (L- x) (Positive BM)At x = a i.e. At point D; Mb =Wab/L

At x = L i.e. At point B; Mb = 0

x c

ci.e. a < x < L


LWbRa = L

WaRb =

W

0

0

LWa

LWb

LWab


Continuously Distributed Loads

In engineering most common distributions are Uniform Distribution and Linearly Varying Distributions

Δx

ΔF

x

q

xFq

x ΔΔ

=→Δ 0

limΔF is total force on the length Δx

Then Intensity of load


Resultants of Distributed loads1. Two system of forces are said to be statically

equivalent if it takes the same set of additional forces to reduce each system to equilibrium

2. A single force which is statically equivalent to a distribution of force is called the resultant of distributed force system

3. In solving problems with distributed loading, it is often more convenient to work with resultant of the distributed load. This is permitted while evaluating external reactions on the member and NOT when calculating internal forces and moments


Resultant of loadsFig. 10 A given loading (a), when replaced by itsresultant (b), produces the same support reactions butnot the same internal forces and moments nor the samedeflections.


Magnitude of resultant and its location for a distributed loading

Fig. 11 The resultant R of the distributed loading q(x)


• Consider a one-dimensional loading of parallel forces of intensity q(x) in Fig.11.

• To determine the magnitude of its resultant R and its location x, we write the equations of equilibrium twice, once using the actual load q(x) and again using the resultant R at x.

• The two sets of equations must give identical reaction forces if R is to be the resultant of the distributed load.


and

and

Thus, the conditions on R and x are

(3.2)


The first part of equation (3.2) states that the resultant is equal to the total area of the loading diagram

while the second part of equation (3.2) states that the line of action of the resultant passes through thecentroid of the loading diagram


The centroid of an area in the xy plane has the coordinates

(3.3)

The centroid of a volume has the coordinates

(3.4)

where the integrals extend over the volume

where the integrals extend over the area.


direction iseAnticlockw 2

wLM

UpwardVertically wLR2

A

A

=

=

w Force/ unit length

A

LAR

AM

x

x

x

Cantilever Beam supporting uniformly distributed load as shown in fig. Draw its shear force and bending moment diagrams


0wxRV Ax =−+

Shear force

At x = 0 ; V = -RA = - wL

x = L ; V = 0

Bending Moment

02

wxxRMM2

AAx =+−+

At x = 0 ; M = -MA = - wL2/2

x = L ; M = 0

w

AR xV

AR

AM xMw


w Force/ unit length

A

LAR

AM

0

0

wL

2wL2

Negative shear

Hogging/ Negative BM

Linear distribution

Parabolic Curve


DA BC

w kN/ lengthW kNW = 1.5 kN w= 1.5 kN/m

l (AC)= 1.5 m l (CB)= 2.25m l (BD)= 1.5m

RA = 0.45 kN RB = 3.3 kN

ΣFY = 0 = RA + RB – W – w x 1.5 = 0

ΣMA = 0 = RB x 3.75 – W x 1.5 – w x 1.5 x 4.5 = 0


1. Consider section x-x is taken in between A and C (0 < x < 1.5)

Shear Force

Vx + RA = 0

VA= - 0.45 kN

VC= - 0.45 kN

Bending Moment

Mx – RAx = 0

At x= 0 ; MA= 0

At x= 1.5m ; MC= 0.675 kNm


2. Consider section x-x is taken in between C and B (1.5 < x < 3.75)

Shear Force

Vx + RA – W = 0

VC= 1.05 kN

VB= 1.05 kN

Bending Moment

Mx + W (x – 1.5) – RAx = 0

At x= 1.5m ; MC= 0.675 kNm

At x= 3.75m ; MB= - 1.6875 kNm


Bending Moment

Mx – RAX – RB (x – 3.75) + W (x- 1.5) + w (x – 3.75)2/ 2

At x= 3.75m ; MB= -1.6875 kNm

At x= 5.25m ; MD= 0 kNm

3. Consider section x-x is taken in between B and D (3.75 < x < 5.25)Shear Force

Vx + RA + RB – W – w (x – 3.75) = 0

At x= 3.75m ; VB= - 2.25 kN

At x= 5.25m ; VD= 0


DA BC

w kN/ lengthW kN

Shear Force Diagram

Bending Moment Diagram

0- 0.45 kN

- 2.25 kN

1.05 kN

0

0.675 kNm

- 1.6875 kNm


Example 3.4A beam is subjected to Varying distributed load. Calculate internal forces and moments and draw shear force and bending moment diagrams.

In given problem varying distributed load is given. For calculating reactions at support the distributed load has to be replaced by a single resultant force at the location x.


Fig.12 Example 3.4. A distributed loading is replaced by its resultant.


The external supports RB and MB are now easily obtained by applying the conditions of equilibrium.


It is not permissible to use the above resultant R to calculate shear force and bending moments within the beams.

We can, however, use general method to find the internal forces and bending moments


(a)

(b)

(c)


At x= 0 ; V = 0

At x= L ; V = woL / 2

At x= 0 ; M = 0

At x= L ; M = - woL2 / 6


Parabolic distribution

Cubic distribution

+ ve Shear

Hogging

2Lw o


DIFFERENTIAL EQUILIBRIUM RELATIONSHIPS

2. Instead of cutting a beam in two and applying the

equilibrium conditions to one of the segments, very

small differential element of the beam will be considered

1. It is an alternative procedure for obtaining internal

forces and moments for the slender members

3. The conditions of equilibrium combined with a limiting conditions will lead us to differential equations connecting the load, the shear force, and the bending moment


4. Integration of these relationships for particular cases furnishes us with an alternative method for evaluating shear forces and bending moments.



If the variation of q(x) is smooth and if Δx is very

small then R is very nearly given by q Δx and the

line of action of R will very nearly pass through the

midpoint ‘o’ of the element.

The conditions of equilibrium applied to Fig. 14c

are then


Integrating above equations with appropriate conditions will give values of shear forces and bending moments


Example 3.5

In Fig.15a a beam carrying a uniformly distributed load

of intensity q = - wo is supported by a pinned joint at A

and a roller support at B. We shall obtain shear-force

and bending-moment diagrams by integration of the

differential relationships (3.11) and (3.12).


RA = RB = wo L / 2


External moments are absent at either end of the beam, hence

We have two boundary conditions available to find C1 and C2.

Mb= 0 at x = 0

Mb= 0 at x = LInserting these boundary conditions values of C1 and

C2 yield


The shear force and bending moment are then

given by

The bending moment is maximum when the shear force is zero.



Example 3.6

Consider the beam shown in Fig. 3.16a with simple

transverse supports at A and B and loaded with a

uniformly distributed load q = - w0 over a portion of the

length. It is desired to obtain the shear-force and

bending-moment diagrams.


Shear Force1 0O

dV wdx

− =

1 1OV w x C− =

Bending Moment1 1

10 0b bO

dM dMV w x Cdx dx

+ = ⇒ + + =

21 1 2

12b OM w x C x C+ + =

Part CBShear Force

2 0dVdx

=

2 3V C=

Bending Moment2 2

30 0b bdM dMV Cdx dx

+ = ⇒ + =

2 3 4bM C x C+ =

Part AC


For 4 Constants i.e. C1, C2, C3 and C4 we need to have 4 boundary conditions

At x = 0; MA = 0 At x = L; MB = 0

At x = a; V1 = V2 = VC and

Mb1 = Mb2 = MC

By applying these BC we get values of C1, C2, C3 and C4 as follows

2 0C =

11 ( 2)2 O

aC w aL

= −2

312

Ow aCL

=

24

12 OC w a=


Shear Force1

1 ( 2)2O O

aV w x w aL

= + −

Bending Moment2

11 1 ( 2) 02 2b O O

aM w x w a xL

+ + − =

Part CBShear Force

2

212

Ow aVL

=

Bending Moment2 2

21 1

2 2b O OM w a x w aL

+ =

Part AC

1 1 ( )( 2)2 2A O O

a L bV w a w aL L

+= − = −

2

2O

Cw aV

L=

0AM =

21 ( )2C O

bM w aL

=

Shear force will be constant in betn C to B

C BV V= =

21 ( )2C O

bM w aL

=

0BM =


11 ( 2) 02O O

aV w x w aL

= + − =

( )2

a L bxL+

=

2 2

max 2

( )8

Ob

w a L bML+

=


SINGULARITY FUNCTION METHOD

nn axxf >−=<)(

0(x)f thea, xIf n =<

nn ax (x)f thea, xIf >−<=>

0 1

1

≥+>−<

=>−<∫∞−

+

nn

axdxaxx n

n


is called unit concentrated moment2axfunction −>−<

1axfunction −>−<is called unit concentrated load


function stepunit called is axfunction 0>−<

function rampunit called is axfunction 1>−<



Solve the following example using singularity function

2

2

02

2

A B O

B O

aM R L w

aR wL

= = −

=

∑

2 2

0 ( )2

2

B A O

A O

aM R L w a b

L bR wL

= = − + +

−=

∑

C


4321 )()()()()( xqxqxqxqxq +++=

11)( −><= xRxq A0

02)( ><−= xwxq0

03)( >−<= axwxq

14)( −>−<= LxRxq B



2 2

2

2

2

2

2

A A O

C O

B O

L bV R wL

aV wL

aV wL

−= − = −

=

=

2

0

20

A

C O

B

M

a bM wL

M

=

=

=


11 ( 2) 02O O

aV w x w aL

= + − =

( )2

a L bxL+

=

2 2

max 2

( )8

Ob

w a L bML+

=


20kN

20kN

12kN

RA RB

A BC D

l (AC)= l (CD)= l (DB)= 2mkNRR BA 12=+

06.0204126 =×−×−×BR

kNRA 2=

kNRB 10=

Distance betn 20kN forces is 0.6m


4321 )()()()()( xqxqxqxqxq +++=

11 0)( −>−<= xRxq A

202 2)( −>−<= xMxq

13 412)( −>−<−= xxq

14 6)( −>−<= xRxq B

∫∞−

−=x

dxxqV )(

102 −>−<= x

2212 −>−<= x

1610 −>−<= x


[ - V = 002 >−< x 1212 −>−<+ x0412 >−<− x ]610 0>−<+ x

V for AC

[ - VAC = ]02 0>−< x kN212 −=×−=V for CD

[ - VCD = ]02 0>−< x kN212 −=×−=


V for DB

[ - VDB = 002 >−< x

kN1011212 =×+×−=]412 0>−<+ x

∫∞−

−=x

VdxM

[ - V = 002 >−< x 1212 −>−<+ x0412 >−<− x ]610 0>−<+ x


∫∞−

−=x

M 002[ >−<− x 1212 −>−<+ x0412 >−<− x ]610 0>−<+ x

=M 102[ >−< x 0212 >−<+ x1412 >−<− x ]610 1>−<+ x

M for AC

=ACM xx 202 1=>−<

0=AM kNmM C 4=


M for CD

=CDM 102 >−< x 0212 >−<+ x

122 += xM CD

kNmM C 161222 =+×=

kNmM D 201242 =+×=


M for DB

=DBM 102 >−< x 0212 >−<+ x1412 >−<− x

)4(12122 −−+= xxM DB

kNmM D 201242 =+×=

02121262 =×−+×=BM


20kN

20kN

12kN

RA RB

A BC D



BR P=

2BPLM =

A BV V P= = −

2

2

2

b

bA

bB

PLM Px

PLM

PLM

= −

= −

=


SFD

BMD

0

-P

2PL

−

2PL

0


Exercise ProblemsFind the values of shear force and bending moment for the give beams. Draw the shear force and bending moment diagrams. Use the general method for analysis.


Ex. 1

Ex. 2


Ex. 4

Ex. 3


Ex. 6

Ex. 5


Ex. 7


Solve the Exercise problems 1- 7 using singularity function method.Solve the Exercise problems 2, 3, 6, and 7 using differential equilibrium relationships. Ignore the values of point loads and concentrated moments given in those problems.

Note: The values of SF & BM for the problem solved by General method and Singularity function method will be same. But values of SF and BM of the problem solved by differential equilibrium method will not be same as we are considering only uniform distributed loads and neglecting the point loads and concentrated moments.

Mechanics of solids by crandall,dahl,lardner, 3rd chapter

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