undefined | unit 2 analysis of beams and frames

Unit-2

Analysis of Beams and Frames

Q1) Explain the condition when fixed beam carrying udl throughout.

A1) Fixed beam carrying udl throughout:

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Area & (A) @B

For fig (i) ……………..

(-) (L) …………..(i)

$C:\Users\TARUNA\Desktop\internship\january\20 jan 2020\UNIT 2 PART B\9.JPG$

(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 =

 Fixed beam carrying udl end moments are

$C:\Users\TARUNA\Desktop\internship\january\20 jan 2020\UNIT 2 PART B\10.JPG$

$C:\Users\TARUNA\Desktop\internship\january\20 jan 2020\UNIT 2 PART B\11.JPG$

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Q2) Explain the condition fixed beam with point load.

A2) Fixed beam with point load.

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Area &@B for free BMD

A = (L)

A =

$C:\Users\TARUNA\Desktop\internship\january\20 jan 2020\UNIT 2 PART B\16.JPG$

@B = =

(2) 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 =

$C:\Users\TARUNA\Desktop\internship\january\20 jan 2020\UNIT 2 PART B\17.JPG$

Special cases

$C:\Users\TARUNA\Desktop\internship\january\20 jan 2020\UNIT 2 PART B\18.JPG$

$C:\Users\TARUNA\Desktop\internship\january\20 jan 2020\UNIT 2 PART B\19.JPG$

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Q3) Explain the condition when fixed beam carrying partial udl.

A3) Fixed beam carrying partial udl

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MA =

MB =

MB=

Special case:

$C:\Users\TARUNA\Desktop\internship\january\20 jan 2020\UNIT 2 PART B\24.JPG$

MA =

MB =

= 

MB =

Q4) Explain the condition of fixed beam with partial UVL.

A4) Fixed beam with partial uvl

Diagram:

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MB =

= w -

$C:\Users\TARUNA\Desktop\internship\january\20 jan 2020\UNIT 2 PART B\28.JPG$

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Q5) Explain the condition of fixed beam of sinking of support.

A5) Fixed beam with sinking of support (A)

$C:\Users\TARUNA\Desktop\internship\january\20 jan 2020\UNIT 2 PART B\31.JPG$

As per principle (1)

Area of M dia. = 0

(-) 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.

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$C:\Users\TARUNA\Desktop\internship\january\20 jan 2020\UNIT 2 PART B\32.JPG$

$C:\Users\TARUNA\Desktop\internship\january\20 jan 2020\UNIT 2 PART B\34.JPG$

Q6) Explain the condition of fixed beam with rotational fielding.

A6) Fixed beam with rotational fielding

Diagram

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$C:\Users\TARUNA\Desktop\internship\january\20 jan 2020\UNIT 2 PART B\31.JPG$

i) Area M dia. = B

(-1) 1 MA + MB (L) = B

EI 2

MA + MB= (-) 2EI (B) (1)

Area M xXB = tBA = 0

(-) L2 2MA + MB = 0

6EI

2MA + MB = 0 (2)

From (1)

…………..

2MA = 2EI (B) - MA = 0

 MA = 2EI (B)

MB = 4 EI (B)

For rotation B anticlockwise

$C:\Users\TARUNA\Desktop\internship\january\20 jan 2020\UNIT 2 PART B\36.JPG$

(Both end moment & rotation clockwise)

Q7) Explain the condition of fixed beam with couple.

A7) Fixed beam with couple

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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)

(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)

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) ( ………..)

= M 2a2b – b3 – ab2

MA = MB (a – o) (………….)

MA = MB (2 a-b)

MB = M (a-b) - MB (2a-b)

L L2

MB = (-) MQ (2b – a)

Q8) Explain the condition of couple at centre.

A8) Special case:

Couple at centre

 B = 10mm (1)

E = 200 G pa

I = 8 x 106 mm4

EI = 1600 KN/M2

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(1) FEM

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

12 12

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MA = MB

Final MA =?

MB =?

MA = VB (4) – 1

 VA = 27

∑Fy = 0

VA /VA =

$C:\Users\TARUNA\Desktop\internship\january\20 jan 2020\UNIT 2 PART B\45.JPG$

BMP

BMA= (-) 26 KNM

BMB = (-)19 KNM

BMC = 33 (2.2) – 26 – 15 (2.2)2

 BMC = 10.3 KNM

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(1) FEM’S

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

= (-) 24 x 5 (1)

(8)2

= (-) (1.875) KNM

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MA1 =

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

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X1 = y1 (a)

y1 + y2

∑F y =VA + VB – 18  3 =0

VA =17.98 x 18KN

 VA = 18KN

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BMD

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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

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Q9) Explain moment area method in detail.

A9) Moment area method (Mohr’s Theorem):

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

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.

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As per principle 1:ϴ = Area (A)

Principle 2: t BA = (A) (xB)

Area and CG

For rectangle

$D:\MY DOCUMENTS\internship\february\11 feb2020\91.JPG$ $D:\MY DOCUMENTS\internship\february\11 feb2020\92.JPG$

For parabola (udl)

$D:\MY DOCUMENTS\internship\february\11 feb2020\93.JPG$

For cubic parabola (UVL Load)

$D:\MY DOCUMENTS\internship\february\11 feb2020\94.JPG$

$D:\MY DOCUMENTS\internship\february\11 feb2020\95.JPG$

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$D:\MY DOCUMENTS\internship\february\11 feb2020\98.JPG$

Q10) Find out slope and deflection at free end of cantilever.

A10) Derive slope & Deflection at free end of cantilever

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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 ( )

Q11) Find out slope and deflection of given beam.

A11)

$D:\MY DOCUMENTS\internship\february\11 feb2020\101.JPG$

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

$D:\MY DOCUMENTS\internship\february\11 feb2020\102.JPG$

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

$D:\MY DOCUMENTS\internship\february\11 feb2020\103.JPG$

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 ( )

Q12) Find out ϴC& ΔC .

$D:\MY DOCUMENTS\internship\february\11 feb2020\104.JPG$

Find: ϴC&ΔC

A12)

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 $D:\MY DOCUMENTS\internship\february\11 feb2020\105.JPG$

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

$D:\MY DOCUMENTS\internship\february\11 feb2020\103.JPG$

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 ( )

Q13) Find ϴC&ΔC

$D:\MY DOCUMENTS\internship\february\11 feb2020\106.JPG$

A13)

BMD $D:\MY DOCUMENTS\internship\february\11 feb2020\107.JPG$

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 ( )

Q14) Find: ΔC

Find: ΔC

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A14)

BMA = M

BMB = 0

BMD/‘M/EI Dia.

$D:\MY DOCUMENTS\internship\february\11 feb2020\109.JPG$

Elastic Curve

$D:\MY DOCUMENTS\internship\february\11 feb2020\110.JPG$

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 ( )

Q15) Find: ϴC, ΔC=?

Find: ϴC, ΔC=?

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A15)

BMA = BMB = 0

BMC = PL/4

= 40 x 4/4

= 40 KNM

$D:\MY DOCUMENTS\internship\february\11 feb2020\112.JPG$

$D:\MY DOCUMENTS\internship\february\11 feb2020\113.JPG$

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

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

Elastic curve

$D:\MY DOCUMENTS\internship\february\11 feb2020\114.JPG$

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 ( )

Q16) Find: ϴA& ΔC?

$D:\MY DOCUMENTS\internship\february\11 feb2020\115.JPG$

A16)

∑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

$D:\MY DOCUMENTS\internship\february\11 feb2020\116.JPG$ $D:\MY DOCUMENTS\internship\february\11 feb2020\117.JPG$

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:

$D:\MY DOCUMENTS\internship\february\11 feb2020\118.JPG$

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 ( )

Q17) Explain slope and deflection method in detail.

A17)

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.

Fixed End Moments: Standard Cases Fixed End Moments

Standard Cases