structural analysis ii _ t3

SUST

Collage of Engineering

School of Civil Engineering

4th Class – 7th semester – 2014

Structural Analysis II Moment Distribution Method

(Hardy Cross Method)

Tutorial No. (3)

Continuous Beams

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Structural Analysis II Moment Distribution Method (Beams)

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Example 1

Calculate the final end moments and then sketch the shear force diagram for the following beam using the Moment Distribution Method taking the Elastic Modulus (E) as constant.

Solution:

1) Fixed End Moments:

→ 𝑀𝐴𝐵𝐹 = −

𝑃𝑎𝑏2

𝐿2 = −60×4×22

62 = −26.67 𝑘𝑁. 𝑚

→ 𝑀𝐵𝐴𝐹 =

𝑃𝑎2𝑏

𝐿2 =60×42×2

62 = 53.33 𝑘𝑁. 𝑚

→ 𝑀𝐵𝐶𝐹 = −𝑀𝐶𝐵

𝐹 = −𝜔𝐿2

12= −

20×32

12= −15 𝑘𝑁. 𝑚

→ 𝑀𝐶𝐷𝐹 = 𝑀𝐷𝐶

𝐹 = −𝑃𝐿

8= −

30×8

8= −30 𝑘𝑁. 𝑚

2) Calculate the Stiffness:

→ 𝐾𝐴𝐵 =1.5𝐸𝐼

6=

𝐸𝐼

4 ⇛ 𝐿𝑒𝑡 𝑘 = 𝐸𝐼/4 ⇛ 𝐾𝐴𝐵 = 𝑘

→ 𝐾𝐵𝐶 =𝐸𝐼

3 ⇛ 𝐾𝐵𝐶 = 1.33𝑘

→ 𝐾𝐶𝐷 =2𝐸𝐼

8=

𝐸𝐼

4 ⇛ 𝐾𝐶𝐷 = 𝑘

3) Distribution factors:

⟹ 𝐽𝑜𝑖𝑛𝑡 𝐴 → 𝐷𝐴𝐵 = 0

⟹ 𝐽𝑜𝑖𝑛𝑡 𝐵 → 𝐷𝐵𝐴 =𝐾𝐴𝐵

𝐾𝐴𝐵+𝐾𝐵𝐶=

𝑘

𝑘+1.33𝑘=

1

1+1.33= 0.43

→ 𝐷𝐵𝐶 =𝐾𝐵𝐶


1.33𝑘

𝑘+1.33𝑘=

1.33

1+1.33= 0.57 𝑂𝑅 (𝐷𝐵𝐶 = 1 − 0.43 = 0.57)

⟹ 𝐽𝑜𝑖𝑛𝑡 𝐶 → 𝐷𝐶𝐵 =𝐾𝐵𝐶

𝐾𝐵𝐶+𝐾𝐶𝐷=

1.33𝑘

1.33𝑘+𝑘=

1.33

1.33+1= 0.57

→ 𝐷𝐶𝐷 =𝐾𝐶𝐷


𝑘

1.33𝑘+𝑘=

1

1.33+1= 0.43 𝑂𝑅 (𝐷𝐶𝐷 = 1 − 0.57 = 0.43)

⟹ 𝐽𝑜𝑖𝑛𝑡 𝐷 → 𝐷𝐷𝐶 = 0


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4) Moment distribution Table:

𝑱𝒐𝒊𝒏𝒕 A B C D

𝑴𝒆𝒎𝒃𝒆𝒓 AB BA BC CB CD DC

𝑫𝒊𝒕𝒓𝒊𝒃𝒖𝒕𝒊𝒐𝒏 𝑭𝒂𝒄𝒕𝒐𝒓

0 0.43 0.57 0.57 0.43 0

𝐹. 𝐸. 𝑀 -26.67 53.33 -15 15 -30 30

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 -16.48 -21.85 8.55 6.45

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 -8.24 4.28 -10.93 3.23 1

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 -1.84 -2.44 6.23 4.70

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 -0.92 3.12 -1.22 2.35 2

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 -1.34 -1.78 0.70 0.52

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 -0.67 0.35 -0.89 0.26 3

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 -0.15 -0.20 0.51 0.38

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 -0.08 0.26 -0.10 0.19 4

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 -0.11 -0.15 0.06 0.04

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 -0.06 0.03 -0.08 0.02 5

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 -0.01 -0.02 0.05 0.03

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 -0.01 0.03 -0.01 0.02 6

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 -0.01 -0.02 0.01 0.00

𝑭𝒊𝒏𝒂𝒍 𝑬𝒏𝒅 𝑴𝒐𝒎𝒆𝒏𝒕𝒔

-36.65 33.39 -33.39 17.88 -17.88 36.07

5) The shear force diagram:


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Example 2

Calculate the final end moments at joints for the following beam using the Moment Distribution Method considering the beam has a constant value of the Elastic Modulus (E).

Solution:


→ 𝑀𝐴𝐵𝐹 = −𝑀𝐵𝐴

𝐹 = −𝑃𝐿

8= −

20×4

8= −10 𝑘𝑁. 𝑚


𝐹 = −𝜔𝐿2

12= −

4×52

12= −8.33 𝑘𝑁. 𝑚


𝐹 = 0

2) Stiffness:

→ 𝐾𝐴𝐵 =𝐸𝐼

𝐿=

𝐸×(6000×10−8)

4= (6

4)(10−5𝐸) ⇛ 𝐿𝑒𝑡 𝑘 = (10−5𝐸) ⇛ 𝐾𝐴𝐵 = (6

4)𝑘 = 1.5𝑘

→ 𝐾𝐵𝐶 =𝐸×(5000×10−8)

5= (5

5)(10−5𝐸) = 𝑘

→ 𝐾𝐶𝐷 =𝐸×(4000×10−8)

7= (4

7)(10−5𝐸) = 0.57𝑘





1.5𝑘

1.5𝑘+𝑘=

1.5

1.5+1= 0.60



𝑘

1.5𝑘+𝑘=

1

1.5+1= 0.40



𝑘

𝑘+0.57𝑘=

1

1+0.57= 0.64



0.57𝑘

𝑘+0.57𝑘=

0.57

1+0.57= 0.36



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0 0.60 0.40 0.64 0.36 0

𝐹. 𝐸. 𝑀 -10 10 -8.33 8.33 0 0

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 -1.00 -0.67 -5.33 -3.00

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 -0.50 -2.67 -0.34 -1.50

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 1.60 1.07 0.22 0.12

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 0.80 0.11 0.54 0.06

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 -0.07 -0.04 -0.35 -0.19

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 -0.04 -0.18 -0.02 -0.10

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 0.11 0.07 0.01 0.01

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 0.06 0.01 0.04 0.01

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 -0.01 0.00 -0.03 -0.01

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 -0.01 -0.02 0.00 -0.01

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 0.01 0.01 0.00 0.00

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 0.01 0.00 0.01 0.00

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 0.00 0.00 -0.01 0.00


-9.68 10.64 -10.64 3.07 -3.07 -1.54

Example 3

Using the Moment Distribution Method calculate the bending moment at supports for the following beam. [𝐸𝐼 ≡ 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡]

Solution:



𝐹 = 𝑀𝐷𝐸𝐹 = −𝑀𝐸𝐷

𝐹 = −10×42

12= −13.33 𝑘𝑁. 𝑚


𝐹 = 𝑀𝐶𝐷𝐹 = −𝑀𝐷𝐶

𝐹 = −10×62

12= −30 𝑘𝑁. 𝑚

2) Stiffness:

→ 𝐾𝐴𝐵 = 𝐾𝐷𝐸 =𝐸𝐼

4 ⇛ 𝐾𝐴𝐵 = 𝐾𝐷𝐸 = 1.5𝑘

→ 𝐾𝐵𝐶 = 𝐾𝐶𝐷 =𝐸𝐼

6 ⇛ 𝐿𝑒𝑡 𝑘 = (𝐸𝐼/6) ⇛ 𝐾𝐵𝐶 = 𝐾𝐶𝐷 = 𝑘


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1.5𝑘

1.5𝑘+𝑘= 0.60



𝑘

1.5𝑘+𝑘= 0.40



𝑘

𝑘+𝑘= 0.50



𝑘

𝑘+𝑘= 0.50

⟹ 𝐽𝑜𝑖𝑛𝑡 𝐷 → 𝐷𝐷𝐶 = 𝐷𝐵𝐶 = 0.40 (𝑓𝑟𝑜𝑚 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑦)

→ 𝐷𝐷𝐸 = 𝐷𝐵𝐴 = 0.60 (𝑓𝑟𝑜𝑚 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑦)

⟹ 𝐽𝑜𝑖𝑛𝑡 𝐸 → 𝐷𝐸𝐷 = 0


𝑱𝒐𝒊𝒏𝒕 A B C D E

𝑴𝒆𝒎𝒃𝒆𝒓 AB BA BC CB CD DC DE ED


0 0.60 0.40 0.50 0.50 0.40 0.60 0

𝐹. 𝐸. 𝑀 -13.33 13.33 -30 30 -30 30 -13.33 13.33

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 10.00 6.67 0.00 0.00 -6.67 -10.00

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 5.00 0.00 3.33 -3.33 0.00 -5.00

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 0.00 0.00 0.00 0.00 0.00 0.00


-8.33 23.33 -23.33 33.33 -33.33 23.33 -23.33 8.33

Notes:

For continuous beams those symmetrical about the vertical Centre-line of the beam with odd number of supports, we can analyze just one half of the beam treating the support that located at the Centre-line of the continuous beam as edge fixed support.

For continuous beams with symmetrical spans and edge fixed supports, the Final End Moments equal the Fixed End Moments. Because there is no un-balanced moment will develop at the interior supports to be distributed.


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Example 4 (Edge simple support)

Using the Hardy Cross’s Method calculate the distributed and final end moments at the joints for the following beam. [𝐸𝐼 ≡ 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡]

Solution:


→ 𝑀𝐴𝐵𝐹 = 𝑀𝐵𝐴

𝐹 = 0


𝐹 = −𝑃𝐿

8= −

16×5

8= −10 𝑘𝑁. 𝑚


𝐹 = −𝜔𝐿2

12= −

3×52

12= −6.25 𝑘𝑁. 𝑚

2) Stiffness:

→ 𝐾𝐴𝐵 = 𝐾𝐵𝐶 =𝐸𝐼

5⟹ 𝑘 & 𝐾𝐶𝐷 = 0.75 (

𝐸𝐼

5) = 0.75𝑘 (

𝑇ℎ𝑒 𝑠𝑡𝑖𝑓𝑓𝑛𝑒𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑑𝑔𝑒 𝑏𝑒𝑎𝑚𝑤𝑖𝑡ℎ 𝑒𝑑𝑔𝑒 𝑠𝑖𝑚𝑝𝑙𝑒 𝑠𝑢𝑝𝑝𝑜𝑟𝑡 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙𝑡𝑜 75% 𝑜𝑓 𝑖𝑡𝑠 𝑎𝑐𝑡𝑢𝑎𝑙 𝑠𝑡𝑖𝑓𝑓𝑛𝑒𝑠𝑠

)





𝑘

𝑘+𝑘= 0.50 & 𝐷𝐵𝐶 = 1 − 0.50 = 0.50



𝑘

𝑘+0.75𝑘= 0.57 & 𝐷𝐶𝐷 = 1 − 0.57 = 0.43





𝑫. 𝑭 0 0.50 0.50 0.57 0.43 1

𝐹. 𝐸. 𝑀 0 0 -10 10 -6.25 6.25

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 -6.25

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 -3.13

𝑁. 𝐹. 𝐸. 𝑀 0.00 0.00 -10.00 10.00 -9.38

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 5.00 5.00 -0.35 -0.27

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 2.50 -0.18 2.50

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 0.09 0.09 -1.43 -1.08

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 0.05 -0.72 0.05

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 0.36 0.36 -0.03 -0.02

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 0.18 -0.02 0.18

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 0.01 0.01 -0.10 -0.08

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 0.01 -0.05 0.01

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 0.03 0.03 -0.01 0.00

𝑭𝒊𝒏𝒂𝒍 𝑬. 𝑴 2.74 5.49 -5.48 10.82 -10.83 0.00

User

Sticky Note

.75 k لان البيم طرفي مع ساند بسيط


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Example 5 (Two edge simple supports)

Using the Moment Distribution Method calculate the moment at supports for the following beam. [𝐸 ≡ 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡]

Solution:



𝐹 = −20×32

12= −15 𝑘𝑁. 𝑚


𝐹 = −40×4

8= −20 𝑘𝑁. 𝑚

→ 𝑀𝐶𝐷𝐹 = −

40×3×22

52 = −19.2 𝑘𝑁. 𝑚 & 𝑀𝐷𝐶𝐹 =

40×2×32

52 = 28.8 𝑘𝑁. 𝑚

2) Stiffness:

→ 𝐾𝐴𝐵 = 0.75 (𝐸𝐼

3) =

𝐸𝐼

4 ⟹ 𝑘


4 ⟹ 𝑘

→ 𝐾𝐶𝐷 = 0.75 (2𝐸𝐼

5) =

3

4(

2𝐸𝐼

5) ⟹ 1.2𝑘





𝑘

𝑘+𝑘= 0.50 & 𝐷𝐵𝐶 = 1 − 0.50 = 0.50



𝑘

𝑘+1.2𝑘= 0.45 & 𝐷𝐶𝐷 = 1 − 0.45 = 0.55





𝑫. 𝑭 1 0.50 0.50 0.45 0.55 1

𝐹. 𝐸. 𝑀 -15 15 -20 20 -19.2 28.8

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 15.00 -28.80 𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 7.50 -14.40

𝑁. 𝐹. 𝐸. 𝑀

22.50 -20.00 20.00 -33.60

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 -1.25 -1.25 6.12 7.48

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 3.06 -0.63

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 -1.53 -1.53 0.28 0.35


𝐵𝑎𝑙𝑎𝑛𝑐𝑒 -0.07 -0.07 0.35 0.42


𝐵𝑎𝑙𝑎𝑛𝑐𝑒 -0.09 -0.09 0.02 0.02


𝐵𝑎𝑙𝑎𝑛𝑐𝑒 -0.01 -0.01 0.02 0.03

𝑭𝒊𝒏𝒂𝒍 𝑬. 𝑴 0.00 19.55 -19.56 25.30 -25.30 0.00


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Example 6 (Edge simple support + Cantilever)

Using the Moment Distribution Method sketch the shear force and bending moment diagrams for the beam shown below. [𝐸 ≡ 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡]

Solution:



𝐹 = −20×82

12= −106.67 𝑘𝑁. 𝑚


𝐹 = −60×4

8= −30 𝑘𝑁. 𝑚

→ 𝑀𝐶𝐷𝐹 = −20 × 2 = −40 𝑘𝑁. 𝑚

2) Stiffness:

→ 𝐾𝐴𝐵 =2𝐸𝐼

8=

𝐸𝐼

4 ⟹ 𝑘

→ 𝐾𝐵𝐶 = 0.75 (𝐸𝐼

4) ⟹ 0.75𝑘





𝑘

𝑘+0.75𝑘= 0.57 & 𝐷𝐵𝐶 = 1 − 0.57 = 0.43

⟹ 𝐽𝑜𝑖𝑛𝑡 𝐶 → 𝐷𝐶𝐵 = 1


𝑱𝒐𝒊𝒏𝒕 A B C

𝑴𝒆𝒎𝒃𝒆𝒓 AB BA BC CB CD

𝑫. 𝑭 0 0.57 0.43 1

𝐹. 𝐸. 𝑀 -106.67 106.67 -30 30 -40

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 10.00

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 5.00

𝑁. 𝐹. 𝐸. 𝑀 -106.67 106.67 -25.00

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 -46.55 -35.12


𝐵𝑎𝑙𝑎𝑛𝑐𝑒 0.00 0.00

𝑭𝒊𝒏𝒂𝒍 𝑬. 𝑴 -129.95 60.12 -60.12 40.00 -40.00


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Example 7

(Edge simple support + Cantilever + Actual joint moment)

Use the Hardy Cross’s Method to calculate the support reactions and draw the bending moment diagram for the beam shown below. [𝐸𝐼 ≡ 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡]

Solution:



40×3×22

52 = −19.2 𝑘𝑁. 𝑚 & 𝑀𝐵𝐴𝐹 =

40×2×32

52 = 28.8 𝑘𝑁. 𝑚

𝑀𝐵𝐹 = −50 𝑘𝑁. 𝑚


𝐹 = −20×4

8= −10 𝑘𝑁. 𝑚

→ 𝑀𝐶𝐷𝐹 = −𝑀𝐷𝐶

𝐹 = −10×32

12= −7.5 𝑘𝑁. 𝑚

→ 𝑀𝐷𝐸𝐹 = −10 × 2 = −20 𝑘𝑁. 𝑚

2) Stiffness:


5 ⟹ 0.8𝑘


4 ⟹ 𝑘

→ 𝐾𝐶𝐷 = 0.75 (𝐸𝐼

3) =

3

4(

𝐸𝐼

3) =

𝐸𝐼

4 ⟹ 𝑘





0.8𝑘

0.8𝑘+𝑘= 0.44 & 𝐷𝐵𝐶 = 1 − 0.44 = 0.56



𝑘

𝑘+𝑘= 0.50 & 𝐷𝐶𝐷 = 1 − 0.50 = 0.50



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𝑴𝒆𝒎𝒃𝒆𝒓 AB BA Joint Moment

BC CB CD DC DE

𝑫. 𝑭 0 0.44 0.56 0.50 0.50 1

𝐹. 𝐸. 𝑀 -19.2 28.8 -50 -10 10 -7.5 7.5 -20

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 12.50


𝑁. 𝐹. 𝐸. 𝑀 -19.20 28.80 -50.00 -10.00 10.00 -1.25

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 13.73 17.47 -4.38 -4.38

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 6.87 -2.19 8.74

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 0.96 1.23 -4.37 -4.37

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 0.48 -2.19 0.62

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 0.96 1.23 -0.31 -0.31

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 0.48 -0.16 0.62

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 0.07 0.09 -0.31 -0.31

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 0.04 -0.16 0.05

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 0.07 0.09 -0.03 -0.03

𝑭𝒊𝒏𝒂𝒍 𝑬. 𝑴 -11.33 44.59 5.41 10.63 -10.65 20.00 -20.00

5) The support reactions and the bending moment diagram:


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Example 8

(Two edge simple supports + Cantilever)

Calculate the final end moments at the supports using the Moment Distribution Method for the depicted beam. 𝐴𝑠𝑠𝑢𝑚𝑒 [𝐸𝐼 ≡ 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡]

Solution:


→ 𝑀𝐵𝐴𝐹 =

12×1.52

2= 13.5 𝑘𝑁. 𝑚


𝐹 = −12×52

12= −25 𝑘𝑁. 𝑚

→ 𝑀𝐶𝐷𝐹 = −𝑀𝐷𝐶

𝐹 = − [12×52

12+

60×5

8] = −62.5 𝑘𝑁. 𝑚

2) Stiffness:

→ 𝐾𝐵𝐶 = 𝐾𝐶𝐷 = 0.75 (𝐸𝐼

5) ⟹ 𝑘


⟹ 𝐽𝑜𝑖𝑛𝑡 𝐵 → 𝐷𝐵𝐶 = 1



𝑘

𝑘+𝑘= 0.50 & 𝐷𝐶𝐷 = 0.50



𝑱𝒐𝒊𝒏𝒕 B C D

𝑴𝒆𝒎𝒃𝒆𝒓 BA BC CB CD DC

𝑫. 𝑭 1 0.50 0.50 1

𝐹. 𝐸. 𝑀 13.5 -25 25 -62.5 62.5

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 11.50 -62.50


𝑁. 𝐹. 𝐸. 𝑀

30.75 -93.75

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 31.50 31.50

𝑭𝒊𝒏𝒂𝒍 𝑬. 𝑴 13.50 -13.50 62.25 -62.25 0.00


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Example 10 (Support Settlement)

Calculate the final end moments and draw the shear force diagram if the support (B) sinks by (15 𝑚𝑚) using the Moment Distribution Method. [𝑇𝑎𝑘𝑒 𝐸𝐼 = 3000 𝑘𝑁. 𝑚2 𝑎𝑛𝑑 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑎𝑙𝑜𝑛𝑔 𝑡ℎ𝑒 𝑏𝑒𝑎𝑚]

Solution:

1) The displacement sign:

→ ∆𝐴= ∆𝐶= ∆𝐷= 0 & ∆𝐵= 15 𝑚𝑚

→ 𝐹𝑜𝑟 𝑏𝑒𝑎𝑚 (𝐴𝐵); ∆𝐴𝐵= +15 𝑚𝑚 = +0.015 𝑚

→ 𝐹𝑜𝑟 𝑏𝑒𝑎𝑚 (𝐵𝐶); ∆𝐵𝐶= −15 𝑚𝑚 = −0.015 𝑚

→ 𝐹𝑜𝑟 𝑏𝑒𝑎𝑚 (𝐶𝐷); ∆𝐶𝐷= 0



𝑃𝑎𝑏2

𝐿2 −6𝐸𝐼∆

𝐿2 = −100×2×32

52 −6×3000×0.015

52 = −82.8 𝑘𝑁. 𝑚

𝑀𝐵𝐴𝐹 =

𝑃𝑎2𝑏

𝐿2 −6𝐸𝐼∆

𝐿2 =100×3×22

52 −6×3000×0.015

52 = 37.2 𝑘𝑁. 𝑚

→ 𝑀𝐵𝐶𝐹 = −

𝜔𝐿2

12−

6𝐸𝐼∆

𝐿2 = −50×52

12−

6×3000×−0.015

52 = −93.37 𝑘𝑁. 𝑚

𝑀𝐶𝐵𝐹 =

𝜔𝐿2

12−

6𝐸𝐼∆

𝐿2 =50×52

12−

6×3000×−0.015

52 = 114.97 𝑘𝑁. 𝑚


𝑃𝐿

8−

6𝐸𝐼∆

𝐿2 = −30×4

8− 0 = −15 𝑘𝑁. 𝑚

𝑀𝐷𝐶𝐹 =

𝑃𝐿

8−

6𝐸𝐼∆

𝐿2 =30×4

8− 0 = 15 𝑘𝑁. 𝑚

3) Stiffness:


5 ⟹ 𝑘


5 ⟹ 𝑘

→ 𝐾𝐶𝐷 =𝐸𝐼

4 ⟹ 1.2𝑘





𝑘

𝑘+𝑘= 0.50 & 𝐷𝐵𝐶 = 0.50


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𝑘

𝑘+1.2𝑘= 0.45 & 𝐷𝐶𝐷 = 1 − 0.45 = 0.55





𝑫. 𝑭 0 0.50 0.50 0.45 0.55 0

𝐹. 𝐸. 𝑀 -82.8 37.2 -93.37 114.97 -15 15

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 28.09 28.09 -44.99 -54.98

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 14.05 -22.50 14.05 -27.49

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 11.25 11.25 -6.32 -7.73

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 5.63 -3.16 5.63 -3.87

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 1.58 1.58 -2.53 -3.10

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 0.79 -1.27 0.79 -1.55

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 0.64 0.64 -0.36 -0.43

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 0.32 -0.18 0.32 -0.22

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 0.09 0.09 -0.14 -0.18

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 0.05 -0.07 0.05 -0.09

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 0.04 0.04 -0.02 -0.03

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 0.02 -0.01 0.02 -0.02

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 0.01 0.01 -0.01 -0.01

𝑭𝒊𝒏𝒂𝒍 𝑬. 𝑴 -61.94 78.90 -78.86 81.46 -81.46 -18.24

6) The support reactions and the shear force diagram:


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Example 10 (Support Settlement + Edge simple support)

Calculate the final end moments, if the support (B) sinks by (10 𝑚𝑚) using the Hardy Cross’s Method, taking the Bending Rigidity (𝐸𝐼 = 4000 𝑘𝑁. 𝑚2).

Solution:


→ 𝐹𝑜𝑟 𝑏𝑒𝑎𝑚 (𝐴𝐵); ∆𝐴𝐵= +10 𝑚𝑚 = +0.01 𝑚

→ 𝐹𝑜𝑟 𝑏𝑒𝑎𝑚 (𝐵𝐶); ∆𝐵𝐶= −10 𝑚𝑚 = −0.01 𝑚



𝜔𝐿2

12−

6𝐸𝐼∆

𝐿2 = −20×82

12−

6×2(4000)×0.01

82 = −114.17 𝑘𝑁. 𝑚

𝑀𝐵𝐴𝐹 =

𝜔𝐿2

12−

6𝐸𝐼∆

𝐿2 =20×82

12−

6×2(4000)×0.01

82 = 99.17 𝑘𝑁. 𝑚


𝑃𝐿

8−

6𝐸𝐼∆

𝐿2 = −60×4

8−

6×4000×−0.01

42 = −15 𝑘𝑁. 𝑚

𝑀𝐶𝐵𝐹 =

𝑃𝐿

8−

6𝐸𝐼∆

𝐿2 =60×4

8−

6×4000×−0.01

42 = 45 𝑘𝑁. 𝑚

3) Stiffness:


8=

𝐸𝐼

4 ⟹ 𝑘

→ 𝐾𝐵𝐶 = 0.75 (𝐸𝐼

4) ⟹ 0.75𝑘





𝑘

𝑘+0.75𝑘= 0.57 & 𝐷𝐵𝐶 = 1 − 0.57 = 0.43




𝑴𝒆𝒎𝒃𝒆𝒓 AB BA BC CB

𝑫. 𝑭 0 0.57 0.43 1

𝐹. 𝐸. 𝑀 -114.17 99.17 -15 45



𝑁. 𝐹. 𝐸. 𝑀 -114.17 99.17 -37.50

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 -35.15 -26.52



𝑭𝒊𝒏𝒂𝒍 𝑬. 𝑴 -131.75 64.02 -64.02 0.00


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Example 11 (Support Settlement + Edge simple support)

Calculate the final end moments if the support (B) sinks by (20 𝑚𝑚) and support (C) by (12 𝑚𝑚) using the Moment Distribution Method. [𝑇𝑎𝑘𝑒 𝐸𝐼 = 5000 𝑘𝑁. 𝑚2]

Solution:


→ ∆𝐴= ∆𝐷= 0 & ∆𝐵= 20 𝑚𝑚 & ∆𝐶= 12 𝑚𝑚

→ 𝐹𝑜𝑟 𝑏𝑒𝑎𝑚 (𝐴𝐵); ∆𝐴𝐵= +20 𝑚𝑚 = +0.02 𝑚

→ 𝐹𝑜𝑟 𝑏𝑒𝑎𝑚 (𝐵𝐶); ∆𝐵𝐶= −(20 − 12) = −8 𝑚𝑚 = −0.008 𝑚

→ 𝐹𝑜𝑟 𝑏𝑒𝑎𝑚 (𝐶𝐷); ∆𝐶𝐷= −12 𝑚𝑚 = −0.012 𝑚



𝑃𝑎𝑏2

𝐿2 −6𝐸𝐼∆

𝐿2 = − [30×2×42

62 +30×4×22

62 ] −6×2(5000)×0.02

62 = −73.33 𝑘𝑁. 𝑚

𝑀𝐵𝐴𝐹 =

𝑃𝑎𝑏2

𝐿2 −6𝐸𝐼∆

𝐿2 = [30×2×42

62 +30×4×22

62 ] −6×2(5000)×0.02

62 = 6.67 𝑘𝑁. 𝑚


𝜔𝐿2

12−

6𝐸𝐼∆

𝐿2 = −15×52

12−

6×1.5(5000)×−0.008

52 = −16.85 𝑘𝑁. 𝑚

𝑀𝐶𝐵𝐹 =

𝜔𝐿2

12−

6𝐸𝐼∆

𝐿2 =15×52

12−

6×1.5(5000)×−0.008

52 = 45.65 𝑘𝑁. 𝑚


𝑃𝐿

8−

6𝐸𝐼∆

𝐿2 = −50×4

8−

6×5000×−0.012

42 = −2.5 𝑘𝑁. 𝑚

𝑀𝐷𝐶𝐹 =

𝑃𝐿

8−

6𝐸𝐼∆

𝐿2 =50×4

8−

6×5000×−0.012

42 = 47.5 𝑘𝑁. 𝑚

3) Stiffness:


6=

𝐸𝐼

3 ⟹ 1.78𝑘

→ 𝐾𝐵𝐶 =1.5𝐸𝐼

5=

3𝐸𝐼

10 ⟹ 1.6𝑘

→ 𝐾𝐶𝐷 = 0.75 (𝐸𝐼

4) =

3𝐸𝐼

16 ⟹ 𝑘





1.78𝑘

1.78𝑘+1.6𝑘= 0.53 & 𝐷𝐵𝐶 = 1 − 0.53 = 0.47


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1.6𝑘

1.6𝑘+𝑘= 0.62 & 𝐷𝐶𝐷 = 1 − 0.62 = 0.38





𝑫. 𝑭 0 0.53 0.47 0.62 0.38 1

𝐹. 𝐸. 𝑀 -73.33 6.67 -16.85 45.65 -2.5 47.5



𝑁. 𝐹. 𝐸. 𝑀 -73.33 6.67 -16.85 45.65 -26.25

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 5.40 4.78 -12.03 -7.37

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 2.70 -6.02 2.39

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 3.19 2.83 -1.48 -0.91

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 1.60 -0.74 1.42

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 0.39 0.35 -0.88 -0.54

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 0.20 -0.44 0.18

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 0.23 0.21 -0.11 -0.07

𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 0.12 -0.06 0.11

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 0.03 0.03 -0.07 -0.04

𝑭𝒊𝒏𝒂𝒍 𝑬. 𝑴 -68.71 15.91 -15.91 35.18 -35.18 0.00

Example 12

(Support Settlement + Edge simple support + Cantilever)

The settlement of (16 𝑚𝑚) of the support (C) is due to the applied load shown. Calculate the final end moments if the Flexural Rigidity (𝐸𝐼 = 200 𝑘𝑁. 𝑚2) along the whole length of the beam.

Solution:


→ ∆𝐴= ∆𝐵= 0 & ∆𝐶= 16 𝑚𝑚

→ 𝐹𝑜𝑟 𝑏𝑒𝑎𝑚 (𝐴𝐵); ∆𝐴𝐵= 0

→ 𝐹𝑜𝑟 𝑏𝑒𝑎𝑚 (𝐵𝐶); ∆𝐵𝐶= +16 𝑚𝑚 = +0.016 𝑚



𝐹 = −𝑃𝐿

8= −

100×8

8= −100 𝑘𝑁. 𝑚 (

𝑇ℎ𝑒𝑟𝑒 𝑖𝑠 𝑛𝑜 𝑎𝑑𝑑𝑖𝑡𝑖𝑜𝑛𝑎𝑙 𝑓𝑖𝑥𝑒𝑑𝑒𝑛𝑑 𝑚𝑜𝑚𝑒𝑛𝑡𝑠 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 ∆𝐴𝐵= 0

)


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𝜔𝐿2

12−

6𝐸𝐼∆

𝐿2 = −24×82

12−

6×200×0.016

82 = −128.3 𝑘𝑁. 𝑚

→ 𝑀𝐶𝐵𝐹 =

𝜔𝐿2

12−

6𝐸𝐼∆

𝐿2 =24×82

12−

6×200×0.016

82 = 127.7 𝑘𝑁. 𝑚

→ 𝑀𝐶𝐷𝐹 = −20 × 2 = −40 𝑘𝑁. 𝑚

3) Stiffness:


8 ⟹ 𝑘

→ 𝐾𝐵𝐶 = 0.75 (𝐸𝐼

8) ⟹ 0.75𝑘





𝑘

𝑘+0.75𝑘= 0.57 & 𝐷𝐵𝐶 = 1 − 0.57 = 0.43




𝑴𝒆𝒎𝒃𝒆𝒓 AB BA BC CB CD

𝑫. 𝑭 0 0.57 0.43 1

𝐹. 𝐸. 𝑀 -100 100 -128.3 127.7 -40

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 -87.70 𝐶𝑎𝑟𝑟𝑦 𝑂𝑣𝑒𝑟 -43.85

𝑁. 𝐹. 𝐸. 𝑀 -100.00 100.00 -172.15

𝐵𝑎𝑙𝑎𝑛𝑐𝑒 41.13 31.02



𝑭𝒊𝒏𝒂𝒍 𝑬. 𝑴 -79.43 141.13 -141.13 40.00 -40.00

structural analysis ii _ t3

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