Sunday, May 3, 2020

SLOPE AND DEFLECTION FOR A CANTILEVER BEAM SUBJECTED TO EXTERNAL MOMENT


SLOPE AND DEFLECTION FOR A CANTILEVER BEAM SUBJECTED TO EXTERNAL MOMENT



Fig 1


Fig 2

Consider a cantilever beam PQ of length L subjected to external moment M on the free end side of the beam.
Fig 2 show the M/EI diagram for the beam subjected to external moment. The effect of moment remains same throughout the beam; hence there is invariability in M/EI diagram.       
Slope at the free end = Area of M/EI diagram
ϴQ = (L) (-M/ EI)
ϴQ = (-ML/ EI)
Deflection at Q = (Area of M/EI diagram) (Centroidal distance from Q to O)
YQ = (L) (-M/ EI) (1/2L)
YQ= -ML2/2EI
YQ= ML2/2EI (downward direction)


Saturday, May 2, 2020

SLOPE AND DEFLECTION FOR A CANTILEVER BEAM SUBJECTED TO UNIFORMLY DISTRIBUTED LOAD THROUGHOUT THE ENTIRE SPAN


SLOPE AND DEFLECTION FOR A CANTILEVER BEAM SUBJECTED TO UNIFORMLY DISTRIBUTED LOAD THROUGHOUT THE ENTIRE SPAN


Fig 1

Fig 2: M/EI diagram

Consider a cantilever beam subjected PQ (shown in fig 1) of span L, subjected to uniformly distributed load of w/m throughout the entire span. Fig 2 shows bending moment diagram of the cantilever beam with uniformly distributed load throughout the span.
Slope at the free end = Area of M/EI diagram (As per 1st moment area theorem)
Area of parabolic diagram = (1/3) (base) (height)
ϴQ = (1/3) (L) (-wL2/ 2EI)
Therefore,
ϴQ = -Wl3/ 6EI
ϴQ = Wl3/ 6EI rad (clockwise with tangent from P)

Consider the M/EI diagram in which O is the centroid point and X is the distance from free end to centroid (O) of the diagram.
Deflection at a point = Product of Area of M/EI diagram and its centroidal       
                                     distance from the reference point.
Here reference point is a point on which deflection has to be determined.
Therefore,
Deflection at Q = (Area of M/EI diagram)(Centroidal distance from Q to O)
YQ = (1/3) (L) (wL2/ 2EI)(X)
YQ = (1/3) (L) (wL2/ 2EI)(3/4(L))
YQ= -wL4/8EI
YQ= wL4/8EI (downward direction)

SLOPE AND DEFLECTION FOR A CANTILEVER BEAM SUBJECTED TO POINT LOAD AT FREE END


SLOPE AND DEFLECTION FOR A CANTILEVER BEAM SUBJECTED TO POINT LOAD AT FREE END

Fig 1



Fig 2: Deflected shape of beam



Fig 3: M/EI diagram of a beam




Consider a cantilever beam PQ (fig 1) of span L subjected to point load of magnitude W KN at free end. Fig 2 shows the deflected shape of the beam.Fig 3 shows bending moment diagram of the cantilever beam with concentrated load. Let ϴ be the slope and y is the deflection for the deflected beam.

Slope at the free end = Area of M/EI diagram (As per 1st moment area theorem)
ϴq= ½ (L) (-WL/EI)
Therefore,
ϴq = (-WL2/2EI)
ϴ= (WL2/2EI)(Clockwise with tangent from P)
Consider the M/EI diagram in which O is the centroid point and X is the distance from free end to centroid (O) of the diagram.
Deflection at a point = Product of Area of M/EI diagram and its centroidal        
                                     distance from the reference point.
Here reference point is a point on which deflection has to be determined.
Therefore,
Deflection at Q = (Area of M/EI diagram)(Centroidal distance from Q to O)
YQ = ½ (L) (-WL/EI)(X)
YQ= ½ (L) (-WL/EI)( 2/3(L))
YQ =  - WL3/3EI
YQ =  WL3/3EI(downward direction)
Note: Always for a cantilever beam slope and deflection is maximum in free end.

Wednesday, April 29, 2020

SIGN CONVENTION FOLLOWED IN MOMENT AREA METHOD FOR DETERMINING SLOPE AND DEFLECTION.


SIGN CONVENTION FOLLOWED IN MOMENT AREA METHOD FOR DETERMINING SLOPE AND DEFLECTION.

1.     For Right hand side of the support ,anticlockwise moments are taken as positive ,clockwise moments are taken as negative and vise versa in case of left hand side support.

2.   Slope from Right hand side is taken as positive when it makes anticlockwise angle with left hand side tangent and vise versa with slope of left hand side tangent.

3.     Deflection is taken as positive if the right hand side tangent is above the left hand side tangent and vise versa with deflection of left hand side tangent.

2nd MOMENT AREA THEOREM


2nd MOMENT AREA THEOREM

Consider the simply supported beam AB subjected to the load W. Let C and D be the two points between the supports A and B in order to determine the deflection for elemental length. Let ∆ be the deflection between the two points C and D. Let X be the distance from D to the meeting point of tangent. Let ϴCD be the angle between the tangents drawn from points C and D.
From property of circles,
Referring to the figure
∆ = x (ϴCD)
From 1st moment area theorem
W k t
ϴCD = CD   (M/ E I) (dx)
∆ = CD   (M/ E I) (x) (dx)
Therefore 2nd  theorem of moment area states that 
Deflection at a point in a beam in the direction perpendicular to its original straight line position measured from tangent to elastic curve at another point is given by moment of M/EI diagram about the point where deflection is required.



1ST THEOREM OF MOMENT AREA METHOD


1ST THEOREM OF MOMENT AREA METHOD



Consider a simply supported beam of span L with supports at A and B, subjected to point load of magnitude W.



Fig 1




    Fig 2


   Fig 3

Consider figure 2, which indicates the deflected shape of the simply supported beam subjected to point load. Let C and D be the two points between the supports A and B in order to determine the slope for elemental length. Let dx be the elemental length between CD to determine the slope value which resembles shape of an arc and projected to meet at point O making an angle dϴ. Let R be the radius of arc. Let ϴCD be the angle between the tangents drawn from points C and D.Fig 3 represents M/EI diagram of the over all beam and shaded portion represent for points CD.

We know that from the bending equation,

M/I = E/R............. (1)
Referring Fig 2, we know that from property of Circle
dx =  R dϴ
Therefore, R = dx / dϴ............. (2)
Substituting (2) in (1)
M/I = E/ (dx / dϴ)
dϴ = (M/ E I) ( dx)
This is for elementary length dx
For CD portion
ϴCD = CD  (M/ E I) ( dx)
Therefore 1st theorem of moment area states that 
change in the slope of a beam between two points is equal to the area under the curvature diagram between those two points.

Tuesday, April 28, 2020

MOMENT AREA METHOD - introduction


MOMENT AREA METHOD

Introduction
1.     Whenever a structure subjected to external load, due to action of external load on the structure, beyond elastic limit the structure will deform with an eccentric distance with reference to its initial position.
2.     The deformation values are most important to be known in order to design a structure.
3.     The deformation in a structure should be within the range and if its values are large then it causes crack and damage to the structure.
4.     The most important factor on which deflection of a structure depends on bending moment and flexural stiffness.
5.     Analysis of deflection is required to solve the statically indeterminate structure.

Moment area method
1.     Moment area method is one of the important and easy ways to determine slope and deflection in various structural elements.
2.     In this method area of M/EI diagram is used compute slope and defection in a structure.
3.     There are two important theorems used to determine the slope and deflection f a structure.
4.     1st theorem of moment area is used to determine slope of the deformed structure.
5.     2nd theorem of moment area is used to determine value of deflection for a deformed structure with respect to its initial position.
6.     Moment area method is basically depend on classical beam theory to analyse the deflected shape of the beam

Classical beam theory
·  This theory is used to determine the deformation of the structure subjected to transverse load.
·        It is also called as Euler –Bernoulli theory.
·        Assumptions in this theory for analyzing a structure.

1.     Plane sections remains plane even after loading: Consider the large span of beam subjected to external loading .If the small portion of the beam is sectioned; then the flat portion of the sectioned beam remains flat even after action of loading (deformation).This assumption is applied for bending of beams only for transverse loads which is symmetric in nature but not for the torsional force. This assumption is also valid for the sections perpendicular to the neutral axis remains perpendicular to the neutral axis even after loading.



2.     The deformations are small compared to length of the beam.

3.     The material of the structure is elastic.

4.     The cross section of the beam remains constant throughout.

5.     The material of the beam should be homogeneous and isotropic.

6.     The length of the beam is should be greater than its cross sectional dimensions.
Under these Assumptions the relation between deflection and bending moment is given by the equation
dy(x)/dx2 = M(x)/EI

Where y = deflection in mm
 x = span to determine deflection
M = Bending moment
E = young’s modulus
I = moment of inertia of cross section of the beam.

Also watch explanation on Youtube channel,

Monday, April 27, 2020

LENGTH OF CABLE SUBJECTED TO UNIFORMLY DISTRIBUTED LOAD


LENGTH OF CABLE SUBJECTED TO UNIFORMLY DISTRIBUTED LOAD





Consider the figure in which cable of span l is subjected to uniformly distributed load of w/m throughout the entire span. Since cable is a flexible structure, it deflects in a parabolic way when subjected to udl. Let h is the central dip of the cable. The equation of cable is considered by taking point C as origin.
We know that the equation for dip for a parabola is given by
Y = 4hx (l -x)/ (l)2 ..........(1)      
Considering a section X- X of span x, for which the deflected length of the cable has to be calculated. Let s be the length of the arc for span x
Therefore EQ (1) becomes
Y = 4hx (x-0)/ (l)2
Y = 4hx 2/ (l) 2
We know that slope is given by tanϴ
tanϴ = dY / dX = 8hx/(l) 2
Deflected length of cable of X-X section is given by
 Sec ϴ = √ (1+ tan2ϴ)
ds/ dx = √ (1+ (dy/dx)2)
ds/ dx = √ (1+ (8hx/(l) 2)2)
ds/ dx = √ (1+ (64 h2x2/(l) 4)
Neglecting the smaller terms, length of deflected arc for section x- x is given by
ds = [1+ (32 h2x2/(l) 4]dx
Total deflected length of cable
L = 2 {  l /20 [1+ (32 h2x2/(l) 4]dx }        (limits 0 to (l/2))
Solving the above equation
L = l + 8/3(h2/l)
Where l = length of cable
L = deflected length of cable
h = central dip

CABLES SUBJECTED TO UNIFORMLY DISTRIBUTED LOAD


CABLES SUBJECTED TO UNIFORMLY DISTRIBUTED LOAD





Consider a cable of span L, subjected to uniformly distributed load of w/m throughout the entire span. Let A and B are the two pinned supports which have the vertical reactions Va and Vb, horizontal reaction H at both the ends. Let h is the central dip (vertical distance) of the cable.
Due to symmetry, the reactions Va and Vb are equal
Therefore, Va = Vb = wL/2
Taking moment about point C to determine the horizontal thrust
Va (L/2) – H(h) -w(L/2) (L/4) = 0
(Note: The value of the moment is taken as zero, since the cable structure will always be free from moments)
(wL/2) (L/2) – H(h) -w(L/2) (L/4) = 0
Solving the above Eq
H = wL2/8h
Determination of tension forces
At supports: Since, there are two forces in each support.i.e., one vertical reaction and one horizontal reaction. Tension in the cable can be determined by calculating the resultant of the above forces


Tension at supports T = Resultant of V and H

                                T = √ (V2 +H2)
                                T = √ ((wL/2)2 +(wL2/8h)2)
Simplifying the above equation,
Therefore T = (wL/2) √ (1 +(L2/16h2))
Tension at center span of the cable = The total shear force at that span from either         
                                                             of the support side
Shear force due to vertical load   V = Va – w(L/2) = w(L/2) – w(L/2) = 0
Shear force due to Horizontal thrust   = - H (Since it is in left direction hence taken
                                                                          as negative)
                                                         = -wL2/8h
Total tension force at the center = T = √ (V2 +H2)
                                                       T = √ (02 +(-wL2/8h) 2)
Therefore, Tension force at center = T = H = wL2/8h
Hence by determining the tension at supports and center of span we can say that the tension force is always maximum at supports and minimum at center.
Therefore, Tmax = (wL/2) √ (1 +(L2/16h2))
                 Tmin = H = wL2/8h


Sunday, April 26, 2020

NUMERICAL ON THE DETERMINATION OF TENSILE FORCES WHEN CABLE SUBJECTED TO THE CONCENTRATED LOAD


NUMERICAL ON THE DETERMINATION OF TENSILE FORCES WHEN CABLE SUBJECTED TO THE CONCENTRATED LOAD



A cable supported at its ends with span 40m apart carries a load of 20KN,10KN and 12KN at the distances of 10m, 20m and 30m respectively from left support. If the vertical distance of the point where the 10KN load is carried is 13m below the level of end supports. Determine 
1. Support reactions 
2. Tension forces at different parts of the cable
3. Total length of cable



Step 1: Determination of the support reactions

V a + V b = 20 + 10 + 12= 42KN…. (1)

Taking moment about support A

V b (40) –12 (30) -10 (20) - 20 (10) = 0…. (2)

Solving Eq 2

V b= 19KN

Substitute the value of V b in Eq 1

V a= 23KN

Taking moment about D

V a (20) – H (13) – 20(10) = 0

23 (20) – H (13) – 20(10) = 0…. (3)

Solving Eq 3

Therefore, H = 20KN

Step 2: Determination of dip distance Yc and Ye

Bending Moment about C 

V a (10) – H (Yc) = 0 (Since the moment is zero in cables)

23(10) -20(Yc) = 0

Yc = 11.5m

Similarly Bending moment about E

Vb (10) – H (Ye) = 0

19(10) – 20(Ye) = 0

Ye = 9.5m

Step 3: Calculation of tension forces

At node C                                                


Tan Ꝋ1 = Yc /10

Tan Ꝋ1 = 11.5 /10

Ꝋ1 = Tan-1(1.15)
Ꝋ1= 490

Tan Ꝋ2 = (Yd – Yc)/10

Tan Ꝋ2 = (13-11.5)/10

Ꝋ2 = Tan-1(0.15)

Ꝋ2= 8.530

Applying Sin rule

T1/ Sin (90 - Ꝋ2) = T2 / Sin (90 + Ꝋ1) = 20 / Sin (180- Ꝋ1+ Ꝋ2)

Equating any two terms 

T2 / Sin (90 + 49) = 20 / Sin (180- 49+ 8.53)

T2 = 20.21KN

Similarly 

T1/ Sin (90 - Ꝋ2) = 20 / Sin (180- Ꝋ1+ Ꝋ2)

T1/ Sin (90 – 8.53) = 20 / Sin (180- 49+ 8.53)

T1 = 30.47KN

At Node D


Tan Ꝋ3= (Yd -Ye) /10

Tan Ꝋ3 = (13-9.5) /10

Ꝋ3 = Tan-1(0.35)

Ꝋ3= 19.30

Applying Sin rule

T2/ Sin (90 +Ꝋ3) = T3 / Sin (90 + Ꝋ2) = 10 / Sin (180- Ꝋ2- Ꝋ3)

Equating any two terms 

T3 / Sin (90 + 8.53) = 10 / Sin (180- 8.53- 19.3)

T3 = 21.18KN

At Node E 


Tan Ꝋ4= (Ye) /10

Tan Ꝋ4 = (9.5) /10

Ꝋ4 = Tan-1(0.95)
Ꝋ3= 43.530
Applying Sin rule

T3/ Sin (90 +Ꝋ4) = T4 / Sin (90 - Ꝋ3) = 12 / Sin (180- Ꝋ4+Ꝋ3)

Equating any two terms 

T4 / Sin (90 – 19.3) = 12 / Sin (180- 43.53+19.3)

T4 = 27.59KN

Step 4:

Determination of length of cable

Total deflected length of cable = AC + CD + DE + EB

AC = 10 Sec Ꝋ1

AC = 10 Sec 49 

AC = 15.24m

CD = 10 Sec Ꝋ2

CD = 10 Sec 8.53

CD = 10.11m

DE = 10Sec Ꝋ3

DE = 10Sec 19.3

DE = 10.6m

EB = 10Sec Ꝋ4

EB = 10Sec 43.53

EB = 13.53m

Total deflected length of cable = 15.24+10.11+10.6+13.53

                                        = 49.74m