Unit-2

Analysis of Beams and Frames

2.1 Analysis of Beams and Frames:

Analysis of beam and rectangular portal frames with indeterminacy up to second degrees

Case I):

1. Fixed beam carrying udl throughout:

Area & (A) @B

For fig (i) ……………..

(-)      (L) …………..(i)

(A) @B = 

= (-)     +  

=    (-)  ……………. (II)

Fig (2)

A =       (L)  =

AB =  =

As per principle (1)

Total area =0

 +  =0

  = ……….(I)

As per principle (2)

@B =0

  + =0

( ……………. (2) 

From (1),

MB =   = MA put in (2)

 2MA + MA +

MA =

 MA = 

 MB = 

2. Fixed beam carrying udl end moments are 

Case (II):

Fixed beam with point load.

Area &@B   for free BMD

A =    (L) 

A =

@B   =  

                  =  

As per principle (1)

Total Area =0

 (L) + =0

MA + MB = ……….. (1)

As per principle (2)

@B   = 0

(-) (2MA+2MB) +  =0

2MA+2MB =  (L + b) ………… (2)

From (1)

MB =  MA   (Put in eqnc2)

2MA +    (L +b)

MA =

=

MA =

Special cases

Case (III)

Fixed beam carrying partial udl

      = 

MA = 

MB =

      =

       =

       =

       =

MB=

Special case:

MA =

           =

    MB = 

=  

    MB = 

Case (IV)

 Fixed beam with partial uvl

 Diagram:

MB =

      = w - 

Case (V)

Fixed beam with sinking of support (A)

As per principle   (1)

Area of     M     dia. = 0

                  EI

(-)      MA + MB   (L)   = 0

              2 EI

MA + MB = 0

MA = - MB     (1)

As per principle     (2)

(AX)  @ B =0

tBA = (AX) @ B =   

(-) L2 (2MA + MB) = (-)  

6EI

3M – LM2 = ………..From (1)

Final Fig.

Case (VI)

Fixed beam with rotational fielding

Diagram

i) Area M dia. = B

                      EI

(-1)       1     MA + MB (L)         = B

             EI           2

MA + MB= (-) 2EI (B)       (1)

                                      L

Area     M     xXB = tBA = 0

              EI

 (-)    L2      2MA + MB   = 0

                6EI

2MA + MB   = 0     (2)

From (1)

…………..

2MA = 2EI (B) - MA = 0

                L

  MA = 2EI    (B)

                   L

MB = 4 EI    (B)

              L

For rotation B anticlockwise

Diagram

(Both end moment & rotation clockwise)

Fixed beam with couple

Diagram

Area & (AX) (a) B

A1 =    1        mA + mB (L)

           EI              2

A2 =    1        1 x ax m   (a)

           EI       2                  L

=   ma2

               2LEI

A3 =     x b x (b)

      =    -     mb2

                  2 LEI

(A x) 1 (a) B =   (-)     L2 2mA + mB

                                   6 EI

(A x) 2 (a) B =   (-)    Ma2 b + 1 a

                                   2 LEI               3

(A x) 3 (a) B =   (-)    Mb2 2 b

                                   2 LEI      3

  = (-)   Mb3

                                  3 LEI

As per principle (1)

Total Area = 0

∑ A = 0

(-)   mA + mB (L)         +   ma2 - mb2   - 0

               2                              2L        2L

mA + mB (L) =   m (a2 – b2)

                  2    2L

 mA+ mB = m (a-b)    (1)

                          L

(3) (Ax) (a)    B = 0

(-)   L2   (2 mA + mB) + ma2     b +     a - mb3          = 0

                   6                                 2L                   3     3L

     2MA + MB=      6 (m)       a2b+a  -   b3

                                   L2         2L    3       3L

2MA + MB = M     3a2b + a3 – 2b3

                        L                                                                            (2)

From (1),

MB= M (a-b) - MA  put in (2)

          L

2MA + M (a-b) – MA = M       3a2b + a3 – 2b3

             L                          L3

  MA = M      3a2b + a3 – 2b3       -    M (a-b)

                L3                                           L

Multiply &divide by ………….

MA = M         3a2b + a3 – 2b3 -(a-b)    ( ………..)

          L3

MA = M      3a2b + a3 – 2b3 -(a-b)    ( ………..)

          L3

      = M       2a2b – b3 – ab2

                     L3

            MA = MB (a – o)   (………….)

                       L3

MA = MB    (2 a-b)

           L3

MB =   M (a-b) - MB (2a-b)

             L                L2

MB = (-) MQ (2b – a)

                 L2

Note:

(Free end moment)

Show FEM in the direction of couple applied and after substituting a & b if answer is respective arrow shall be corrected sign changes.

Special case:

Couple at centre

 B = 10mm (1)

E = 200 G pa

I = 8 x 106 mm4

EI = 1600 KN/M2

(1) FEM

MA1 = (-) MB1 = WL2 = 15 X 16 = 20 KNM

                              12             12

MA = MB 

Final  MA =?

  MB =?

MA = VB (4) – 1

  VA = 27

∑Fy = 0

VA /VA =

BMP

BMA= (-) 26 KNM

BMB = (-)19 KNM

BMC = 33 (2.2) – 26 – 15      (2.2)2

                                                     Z

  BMC = 10.3 KNM

BMP

(1)              FEM’S

MA1= (-) Mb (2a-b)

                   L2

= (-) 24 x 5 (1)                       

                     (8)2

 = (-) (1.875) KNM

MA1 =

=

MA2=

MA2=

MA2 = 44.92 KNM

MB2 = (-) 56.320 KNM

 MA = MA1 + MA2

            = 1.875 + 44.92

 MA = 43.045 KNM

  MB = - 7.875 – 56.32 = - 64.195 KNM

X1 =           y1                            (a)

              y1 + y2

∑F y =VA + VB – 18  3 =0

VA =17.98 x 18KN

 VA = 18KN

BMD

BMA = - 43.05 KNM

BMB = - 64.19 KNM

BMC(R) = 18 (3) – 43.05 = 10.95 KNM

BMC(R) = 10.95 + 24 = 34.95 KNM

BMF = 18 (4) – 43.05 + 24 -   18 = 43.95 KNM

                                                    2

2.2 Moment Area Method (Mohr’s Theorem):

1. Principle (I):

Angle between tangents drawn to elastic curve at any two points A & B is equal to area of M/EI diagram between A & B

2. Principle II:

Position of B on elastic curve with respect to tangent drawn at A is equal to moment of area of M/EI diagram between A & B about B.

As per principle 1:ϴ = Area (A)

 Principle 2: t BA = (A) (xB)

3.  Area and CG

For rectangle

For parabola (udl)

For cubic parabola (UVL Load)

t AB = Area(AB) x xa

 t BA = Area(BA) x xb

 t CA = Area(CA) x xC

Example:1]

Derive slope & Deflection at free end of cantilever

Solution:

ϴB = Area of (M/EL) AB

     = ½ x L (PL/EI)

ϴB = PL2/ZEI (    )

t BA = ΔB = (Area)AB x xB

              = (PL2/2EI) (2/3L)

ΔB = PL3/3EI (   )

Example: 2]

Find:  ϴC, ΔC?

Solution:

BMD:

BM at c = 0

BMB = (-) 10 x 2 x 1

          = (-) 20 KNM

BMA = (-) 30 x 3 – 10 x 2(4)

          = (-) 170 KNM

A1 = 30/EI

A2 = ½ x 75/EI x 3 = 112.5/EI

A3 = 1/3 x 2 x 20 x 20/EI = 13.33/EI

Elastic curve

 Slope & Deflection:

ϴC = Area of (M/EI) dia. |CA

ϴC = A1 + A2 + A3 = 1/EI {30 + 112.5 +13.33}

ϴC = 155.83/EI (      )

ΔC = t CA = (A1x1 + A2x2 + A3x3) C

     = 1/EI (30 x (3.5) + 112.5(2 + 2/3 x 3) + 13.33 (3/4 x 2))

ΔC = 575/EI (     )

Example: 3]

Find:  ϴC&ΔC

Solution:

E = 2 x 105 M Pa

I = 250 x 3503/12

EI = 1.78 x 105 KNm2

 BMA = -50 x 4 x (2 + 2)

            = - 800 KNM

BMB = -50 x 4 x 2

         = -400 KNM

BMC = 0

BMD

 M/EI Dia.:

A1 = 100 x 2/EI = 200/EI

A2 = 1/ 2x 100/EI x 2 = 100/EI

A3 = 1/3 x 200/EI x 4 = 266.67/EI

 Elastic curve

 ϴC = Area of (M/EI) AC

      = A1 + A2 + A3

      = 200 + 100 + 266.67/EI

      = 566.67/EI

ϴC = 3.17 x 10-3 rad. (      )

ΔC = t CA = {Area}A-C x xC

      = (A1x1 + A2x2 + A3x3)@C

      = 200/EI x (4+1) + 100/EI x (4 + 2/3 x 2) + 266.67/EI(3/4 x 4)

      = 1000/EI + 533.33/EI + 800.01/EI

      = 2333.34/EI

      = 13.1 x 10-3 m

ΔC = 13.1 mm (    )

Example: 4]

Find ϴC&ΔC

Solution:

  BMD

A1 = M/2EI x 3 = 1.5M/EI

A2 = M/EI x 3 = 3M/EI

ϴC = Area (M/EI) A-C

      =A1 + A2

ϴC = 4.5M/EI

ΔC = t CA= (A1x1 + A2x2)@c 

      = 1.5M/EI x (3+1.5) + 3M/EI (1.5)

ΔC = 11.25M/EI (   )

Example: 5]

Find: ΔC  

Solution:

 BMA = M

BMB = 0

 BMD/‘M/EI Dia.

 Elastic Curve

 Slope & Deflection

t AB = {Area}A-B x(xA)

       = 1 / 2 x M/EI x L {1/3L}

t AB = ML2/6EI

t BA = {Area}A-B x(xB)

       = 1 / 2 x M/EI x L {2/3 L}

t BA = ML2/3EI

ϴA ≈ tan (ϴA) = t BA/L = ML/3EI  (    )

ϴB ≈ tan (ϴB) = t AB/L = ML/6EI (     )

t AB/L = C1C3/L/2

C1C3 = t AB/2 = ML2/12EI

t CB = Area C-B x (xc)

      = 1 / 2 x M/2EI x L/2 x {1/3 x L/2}

t CB  = ML2/48EI

ΔC     = C1– C2

        = C1C3 – C2C3

        = ML2/12EI – ML2/48EI

ΔC   = ML2/16EI (      )

Example:6]

Find: ϴC, ΔC=?

Solution:

BMA = BMB = 0

BMC = PL/4

          = 40 x 4/4

          = 40 KNM

A1 = 1 / 2 x 2 x 40/EI = 40/EI

A2 = 1 / 2 x 2 x 20/EI = 20/EI

 Elastic curve

Slope & Deflection

t AB = (Area)A-B x xA

       (A1x1 + A2x2)@A

        = 40/EI x (2/3 x 2) + 20/EI (2 + 1/3(2))

        = 53.33/EI + 53.33/EI

t AB  = 106.67/EI

t BA =  {Area}AB x xB

      = (A1x1 + A2x2)@B

      = 1/EI {40 x (2 + 1/32) + 20(2/3 x 2)}

    = 133.33/EI

ϴA ≈ tan (ϴA) = t BA/L = 133.33/4EI = 33.33/EI (      )

ϴB ≈ tan (ϴB) = tan AB/L = 106.67/4EI = 26.67/EI (      )

For ΔC (C1, C2)

t AB/4 = C1C3/2

C1C3 = t AB/2 = 106.67/2EI = 53.33/EI

t CB = {Area}CB x  (xC)

       = 20/EI x (1/3 x 2) = 13.33/EI

ΔC = C1C3 – C2C3

      = 53.33 – 13.33

ΔC = 39.99/EI

ΔC = 40/EI (      )

Example: 7]

 Find: ϴA& ΔC?

Solution:

∑MA = 0

-50 x 1 + VB x 4 + 20 = 4VB = 30

VB = 7.5 KN

VA + VB = 50

VA = 42.5KN

BMA = 0

BMB = 0

BMC = 42.5 x 1 = 42.5 KNM

BMDL = 42.5 x 3 – 50 x 2 = 27.5 KNM

BMDR = 7.5 x (1) = 7.5 KNM

A1 = 1 / 2 x 1 x 42.5/EI = 21.25/EI

A2 = 1 / 2 x 2 x 15/EI = 15/EI

A3 = 27.5/EI x 2 = 55/EI

A4 = 7.5/2EI x 1 = 3.75/EI

Elastic curve:

Slope &Deflection:

t AB = {Area}A-B x xA

       = (A1x1 + A2x2 + A3x3 + A4x4)@A

      = 21.25/EI x [2/3 x (1)] + 15/EI(1 + 1/3 x 2) + 55/EI(1 + 1)+ 3.75/EI(3 + 1/3 x 1)

      = 14.16/EI + 25/EI + 110/EI + 12.5/EI

     = 161.66/EI

t BA = Area(B-A) x xB

       = {A1x1 + A2x2 + A3x3 + A4x4}@B

       = 1/EI {21.25(3 + 1/3 x 1) + 15(1 + 2/3 x 2) + 55(2) + 3.75 x 2/3}

       = 218.33/EI

ϴA ≈ tan (ϴA) = t BA/4

ϴA = 218.33/4EI = 54.58/EI (      )

ϴB ≈ tan (ϴB) = t AB/4 = 161.66/4EI = 40.41/EI (     )

To find ΔC

t BA/4 = C1C3/1

C1C3 = 54.58/EI

C1C2 = t CA = Area CA x xC

          = A1xC

         = 21.25/EI (1/3 x 1)

         = 7.08/EI

ΔC = C1C3– C1C2

      = 54.58/EI – 7.08/EI

      = 47.5/EI (     )

Key takeaways

t AB = Area(AB) x xa

 t BA = Area(BA) x xb

 t CA = Area(CA) x xC

2.3 Slope Deflection Method

Slope deflection methods: means it is a method or tool by which we find out how a structure or a member of a structure behaves when subjected to certain excitation.
In other words finding out internal forces (axial force, shear force, moment), stress, strain, deflection, etc in a structure under applied load conditions.

1. Fixed End Moments: Standard Cases Fixed End Moments

Standard Cases

1.

2.

3.

4.

5.

2. Derivation of slope Deflection Method

In beam, a continuous beam ABCD, consider AB part and take fixed end moments and rotation at support as given below 

As per the diagram

Slop deflection equations are

 3. Slope Deflection Method

Slope=

Deflection=

At support form

At the simply supported end

2.     is always zero.

3.     is always zero.

4.     M Cantilever

=0 at the fixed end at force end

2.      At fixed end

4. Types of numerical for slope deflection method

1. The beam is without sink ( is zero.)

2. Beam with sink (is given)

3. Analyze of the frame without sink

4. Analyze the frame with the sink.

Analyze of the continuous beam without sink

which is given.

      Steps for analysis of the slope deflection method.

1. Find unknown slope.

2. Find fixed end moment

3. Apply the slope deflection equation.

    4. Use the joint equilibrium equation at the joint.

    5. Find final moments.

    6. Find relation by equation &Draw SFI

    7. Draw BMD

1)     Analyze the continuous beam shown in figure Draw also B.M. and S.F.

Step-1 Find fixed end moments

Step-II Slope deflection method

Step-III Apply equilibrium condition at joint

Step-IV   Find final moments.

Step-V Find reaction

S.F. at A Take a moment at A is

SFD & BMD

2)     Determine support moments & draw BMD for the beam shown by using S.D.

2.     Find fixed end moments

For BC Beam

3.     Apply the SD equation

AB=>

BC=>

Apply equilibrium condition joint B

-------   due to Hinge support

Put in equation (1) (2) (3) & (4)

Draw BMD Diagram

4.     Analyze the continuous beam ABCD by S.D. method draw BMD

Modify Diagram

Apply S.D. equation

Joint equilibrium equation

Joint at B

5.     Analyze the continuous beam as shown in the figure by S.D. method Draw BMD

Find fixed end moment

KNM

S.D. equation

Span AB

Joint equilibrium equation

Joint B.

KNM

SFD

BMD

6.     Analyze the continuous beam by S.D. equation

Apply S.D. equation

AB=>

2)

BC=>

Joint equilibrium condition

BMD

7.     Analysis of fixes beam by S.D method

Step 1) Dki = 2 

Step 2) Find Fixed End Moments

Solution:

For AB

For BC

For CD

Apply S.D. equation

Joint equilibrium condition

8.     Determine the final moments by slope deflection method take EI are constant.

Step 1) Dki = 2

Step 2) Find fixed end moments

Step 3) Apply S.D. equation

For beam BC

--------(3)

-------(4)

OR

For member, BC apply modified equation

Step4)

9.5 + 1.46 EI + 0.33EI-------(A)

0-------(B)

Equating equation A & B get

OR

Apply this condition to the modified equation

Step 6) BMD

2.     Problems based on sink

Sink means deflection which occurs due to load which is acting on the beam

Note: always take EI in terms of KNM2

1)     Analyze continuous beam ABCD having

            Point B is sunk by 2.5mm also draw BMD

Solution:

Step 1) Dki = 2  

Step 2) Fixed end moments

STEP 3) Apply S.D. equation

AB=>,l = 3m

BC=>

Apply equilibrium condition

Joint B

Joint C

KNM

KNM

KNM

KNM

KNM

BMD

2)     Determine the moment at the end of the fixed support A of the propped cantilever shown

Take   EI Draw BMD

CASE 1)… because support at the same level

STEP 1) Dki = 1

Step 2) Fixed end moments

AB =>

  simply supported end

Case 2)

AB =>

Equilibrium Equation

Frames

Non-sway-The frame is not deflected 

Sway-The frame is deflected 

  occurred only in column

  is zero in the beam

AB =>

Equilibrium condition

Joint at B

2.     Analyze the frame by SD method by BMD

     Find fixed end moments

   ----due to no loading

S.D. equation

AB =>

BD =>

3.     Equilibrium condition joint B

BMD

3.     Analyze the frame shown in Figure support A is fixed B& C are winged. Draw BMD

Find a fixed end

 S.D. equation

For AB Span l =6m, ,I =2I,

For BD Span l =4m, ,I =I,

For BC Span l =4m, ,I =1.5I,

Equilibrium condition joint B

+ + =0

-16.67+3.83++ ---------(A)

------(B)

--------(C)

Equating Equation A, B & C you will get

3.     Problem on sway Frame

Example: 1] Analyze the frame shown in fig slope deflection method

Step 1) Sway Frame: due to lateral loud frame in sway

Find Fixed Ends Moments:

M FAB =   =             = 7.5KKNM

MFBA =   =            = -22.5KNM

M FBC      = M FCB = = = 60 KNM..

Apply Slope Deflection equations

For AB: ƟA=0, ƟB  =?

MAB = MFAB      (2ƟA+ ƟB-)

MAB = 7.5 +    ------------1)

MBA = -22.5 +   ---------------2)

For BC: =?

MBC = +60+ ()   -------3)

MCB = -60+(ƟB)    -------------4)

MDC = MF DC(2ƟD+ ƟC-)

MDC   = (ƟC-) ---------5)

MCD   = (2ƟC-)   --------6)

4) Apply joint equilibrium equation

MBA+MBC = 0

-22.5 + = 0---------------(A)

MCB+MCD = 0

-60+(ƟB) + ( 2ƟC-)= 0   ------------(B)

Shear equation:           HA +HD -40= 0   --------------7)

For AB Column

- HA 4 +MAB +MBA+ 40 *1=0

4HA = MAB + MBA+ 40 *1

 Put Equation 1&2

  4HA =7.5 + -----(8)

This Fig     4HD = MDC+MCD

Put Equation 5 & 6

4HD = (ƟC-)  + ( 2ƟC-)    ---------9)

Put equation 8 & 9 in Equation (7) Equation u will get equation (C)

Then Equating Equation (A,B,&C)

QB = -39.25/EI

QC = 19.25/EI

      = 110/EI  

Final moments

MAB = 29.12 KNM

MBA = -20.5 KNM

MBC = 20.5 KNM

MCB = -60.51 KNM

MCD = 60.51KNM

MDC = +50.875 KNM

Key takeaways

Steps for analysis of the slope deflection method

      1. Find unknown slope.

      2. Find fixed end moment

      3. Apply the slope deflection equation.

         4. Use the joint equilibrium equation at the joint.

         5. Find final moments.

         6. Find relation by equation &Draw SFI

         7. Draw BMD

2.4 Three moments Equation

1. Concepts – i) Moment at simply supported end is always zero.

                           ii) Moment at fixed & intermediate is created

2. Formula:   

For span ABC

MAl1 + 2MB (l1+l2) + Mcl2 = 6a1x1  + 6a2x2

                                                                 l 1           l2

For different I -

MAl1 + 2MB (l1+l2) + Mcl2 = 6a1x1  + 6a2x2

                   I1              I1   I2            I2      l1I1          l2 I2

Area due to loading:

Case (1) Both ends are simply supported end

2)  6a1x1 = A = 2 x 6 x 90 = 360   A2 = 1 x 4 x 10 = 20m2

                           3                                               2

x1 =  6/2 = 3m                          x2 =  4/2 = 2m           

  6a1x1 = 6 x 360 x 3     = 1080

     l1                             6

  6a2x2 = 6 x 20 x 3     = 60

      l2                   4

3)     Apply 3 moments theorem

For span ABC

MAl1 + 2MB (l1+l2) + Mcl2 = 6a1x1 + 6a2x2

                                                    l 1           l2

 MA = MC = 0   l 1=6m   l 2 =4m

 0 + 2MB (6+4) + 0 = 1080 + 60

 2MB (10) = 1140

S.F.D. Due to external load

RA = Wl = 20x6 = 60 KN  RB = W = 10 = 5 KN

           2           2                                              2       2

 RB = Wl = 20x6 = 60KN  RB = W = 10 = 5 KN

           2           2                                              2       2

2)  Due to moment

EMA=0     EMB = 0

RB x 6 – 57 = 0    RC x 4 + 57 = 0

RB =    9.5 KN     RC = 14.25 KN

Efi = 0

RA + RB = 0     RC + RB = 0

 RA = - 9.5 KN      RB = 14.25 KN

Total reaction:

RA = 50.5 KN

RB= 69.5 KN

RC = 5-14.25 = - 9.25 KN

RB = 5 + 14.25 = 19.25 KN

-2  One end is fixed & other is Hinge

A1 = 2 x wl2 x 12  = 2  x 30 x 122 x 12   = 4320 mm2

         3                              3                8

x1 = l1 = 12  = 6  = 6m

         2      2

A2 = 1  x Wab x 12  =  1  x   240 x 4 x 8   x   12   = 3840 mm2

          2                              2                12

x1  = l2 = 12 = 6m    x1 =  L + a     x2 = l +b

         2      2                                                      3          3 

Apply 3 Moment Theorem

A’AB

MA1l1 + 2MA (l1+ l1 ) + MB l1     =    6a1x1  + 6a1x1

                                                                   l1               l1

0 + 2MA (0+12) + 12MB = 0 + 6 x 4320 x 6

                                                                12

24MA + 12MB = 12960    (1)

Span ABC

MAl1 + 2MB (l1+ l2) + MCl2     =    6a1x1  + 6a2x2

                                                              l1               l2

12MA + 2 MB (12 + 12) + 0 =    6 x 4320 x 6   + 6 x 3840 x 6

     12                12

12MA + 48 MB = 24480           (2)

Add eqn.  (1)  & (2)

  MA = 325.71 KNM

     MB = 428.57 KNM

Draw S.F.D.

1) Due to external load

RA = wl = 30 x 12 = 180 KNW   RB = wb  = 160 KN

           2             2                                                       l

RB = wl = 30 x 12 = 180 KN   RC = wa =120 KN =

          2             2                                                                     l      

2) Due to moment

EMA=0      EMB=0

325.71 – 428.57 + RB x 12 = 0   428.57 + RC x 12 = 0

RB = 8.57       RC = 35.71 KN

RA = 8.57 KN      RB = 35.71 KN

Total Reaction:

RA = 180 – 8.57 = 171.43 KN

RB = 180 + 8.57 = 188.57 KN

RB = 160 + 35.71 = 124.29 KN

RC = 120 + 35.71 = 155.71 KN

(3) Both ends are fixed

A1 = 0

xi = 0

A1 =  2  x  wl2  x 6 =  112.5    A2 =  2  x  wl2 x  5  = 125

           3        8                                                                     3        8  

x1 = 6/2 =  3     x2 = 5/2 = 2.5

A12 = 0       x12 = 0

Span A1AB

MAl1+ 2MB (l11 + l1) + MBl1    =  6a11x11  +  6a1x1

                                               l11                       l1

0 + 2MA (0+6) + 6MB = 0 + 6 x 112.5 x 3 

           6 

12MA + 6MB = 337.5  

Span ABC

MAl1+ 2MB (l11 +l1 ) + MC =  6a1x1  +  6a2x2

                                        l1                      l2

6MA + 2 MB (6+ 5) + MC x 5 = 6 x 112.5 x 3  +  6 x 175 x 2.5

                      6    5

 6MA + 22MB + 5MC = 712.5   (2)

Span BCC

MBl2+ 2MC (l2 + l12) + MC1l12 = 6a2x2  +  6a12x12

l2                      l12

5MB + 2MC (5+0)  + 0 =   6 x 125 x 2.5

5MB + 10 MC = 375    (3)

MA = 17.33 KNM

MB = 21.59 KNM

MC = 26.70 KNM

Reaction due to external load

RA = wl = 150    RB = wl    = 100 KN

          2      2

RB = wl    = 150    RC = wl        = 100 KN

          2                                                                      2

Reaction due to moments

17.33 – 21.59 + RB x 6

          RB= 1.022

  RB = 0.71 RA = 0.71   RC = 1.022

Total Reaction:

RA = 150 – 0.71 = 149.29 KN

RB = 150 + 0.71 = 150.71 KN

RB = 100 – 1.02 = 98.98 KN

RC = 100 + 1.02 = 101.02 KN

Overhung Numerical:

A1 = 1 x 3 x wab     A2 = 1  x  wl  x 4

          2           l               2     4

1 x 3 x (20 x 1 x 2) = 20m2   A2 = 60 m2

2                     3

x1 =  4 /3    3 +1  = 4/3  = 1.33

                        3

Span ABC

MAl1+ 2MB (l1 + l2) + MCl2 = 6a1x1 + 6a2x2

                                          l1                      l2

0 + 2 MB (3+ 4) + 15 x 4 =    6 x 20 x 1.33 +    6 x 60 x 2

              3          4

 14MB + 60 = 53.2 + 180  

  MB = 173.2 KN

Key take ways

1. Find the reaction

2. Apply three moment theorem

3. Draw SFD

4. Draw BMD

2.5 Moment Distribution Method

It is used for the analysis of indeterminate structures. In this method solution of simultaneous equations of the slope, the election method is replaced by an iterative distribution procedure.

1. Carryover moment: A moment is allowed to a member permitting rotation at the point of application and keeping the other end fixed additional moment develops at the far end this is called.

2. Carryover factor:-the ratio of carryover moment to applied moment is called carry-over factor.

Carryover factor=

3. Stiffness:- Moment required to rotate an end by a unit angel when rotation is permitted at that end is called stiffness of the beam.

Stiffness of the beam AB=

4. Distribution factor:-When a moment is applied to a right joint where a number of members are meeting, the applied moment is shared by the members meeting at the joint.  The ratio of the moment shared by a member to the applied moment at the joint is called.

Examples:

Then consider at the B end

Carryover factor =

Put    

Distribution factor =>

2.     Continuous beam with simply supported ends lets CD lost span let

 M act at joint C.

5. Carryover factor:-

Types of toe and support.

Relative stiffness factor (I/l)

If for the end is fixed=

If for the end is simply roller internal hinged I overhung

5.     Distribution factor: A moment which tends to rotate without translation a joint to which several members are connected will be divided amongst connected member in proportion to their stiffness.

6. Carryover moment:-when a movement is applied to produce rotation without translation at the near supported and B of a beam whose father end A is fixed the carry over the moment at the further and is ½ are applied movement m I is of the same as the applied moment.

7. Application of MD method to continuous beam with fixed ends:-

8. Steps for analysis of moment distribution method.

Distribution factor is=

K=stiffness -it depends on the support condition.

1. Far end fixed &intermediate support =

2. The far end simply supports =

COF -> support

O -> simply support

½ -> fixed

½ -> intermediate

Examples:

AB=>

1)     Draw distribution table

2)     Moment distribution Table

SFD

BMD

2.     Analyze the given beam ABCD by Moment Distribution method

1) Find fixed moment

KNM

KNM

KNM

KNM

KNM

Step 2)

3) Moment Distribution Table

KNM

KNM

KNM

KNM

KNM

KNM

BMD

3.     Analyze the given beam by M.D. method

Step 1) Find fixed end moment

Step 2) Distribution Factor

Moment distribution table