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PART (7)

2014

2014

7/1

Structural Analysis (I) Internal Forces/BEAMS

Inclined Members Problems

الرأسي الطول

Beams

a

b c

Inclined Member (a-b)

a

b

ΘΘ

Steps of Analysis

المائل الطول

األفقي مع الميل زاويه

األفقي الطول

1- Calculate Reactions and Forces in Link members(IF found).

2- Sections سبق كما

3- Calcuate the Internal Forces a- Horizontal members سبق كما

b- Inclined members

الضلع إتجاه في الضلع علي المؤثره القوي جميع تحليل يتمShear / Normal Forces

(N.F.).(S.F.)

Bending Momentالضلع. إتجاه علي ليس و الفعل) ردود / األحمال ) القوي إتجاه علي يعتمد

4- Drawing the Internal Forces

عليه العمودي اإلتجاه وفي

مراعاة مع المنشأ خط علي عمودي السابقه(Datum)الرسم اإلشارات قواعد

+ve Signs

القوي تحليل1- Vertical Forces

Θ

cosΘ = l l` sinΘ = h l`

Θ

P PcosΘ

PsinΘ

N = - PsinΘ & Q = + PcosΘ

ΘP(4 5)

P(3 5)

N = + 3P5 & Q = - 4P5

34

5P

2- Horizontal Forces

ΘH

HcosΘ

HsinΘ

N = - HcosΘ & Q = -HsinΘ

Θ

H(2 √5)

H(1 √5)

N = + (H√5) & Q = +(2H√5)

21

√5

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PART (7)

2014

2014

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Structural Analysis (I) Internal Forces/BEAMS

Exercise (1)

3t/m`

12t6t.m

6t4t

Draw Internal Forces Diagrams

B.M.DS.F.DN.F.D

Reactions

Get Xa ∑ FX = 0.0 Xa = 4.0 t

∑ M @ a= 0.0 Yb(8)-6(10)+ 4(3)+6 - 12(4)- 6(1)= 0.0

∑ FY = 0.0 Ya = (6+6+12) - 12 = 12tYb = 12t

Get Ya & Yb

Check ∑ M @ b = 12(8) -4(3) - 6(7) - 12(4) -6 + 6(2)= 0.0 ...... OK.

Ya

Xa 4t

12t

12t Yb

b

a

6t

Bending Moment Calculations

Ma = 0.0 t.mM1 = 12(2) - 6(1) = +18 t.mM2 = 12(4)- 6(3) - 4(1) = +26 t.mM3 = 12(6)- 6(5) - 12(2) - 4(2)= +10 t.mM4 = M3 ± MConcentrated = 10 - 6 = + 4t.m

From Right

M4 = - 6(4) + 4(1) + 12(2) = +4 t.m (Check)

Mb = - 6(2) = -12 t.m

3t/m`

12t6t.m

6t4t

4t

12t

12t

b

a

6t

1

2

34

From Left

Free Body Diagram

12t6t.m

6t4t

4t

12t

6t

a

6t 6t

6t18t.m

18t.m

12t.m12t.m

6t

4t4t

4t4t

Check ∑ FY = 0.0 ∑ FX = 0.0 2

1√5

( sinΘ = 0.45 & cosΘ = 0.9 )

Check ∑ FY' = 0.0 ∑ FX' = 0.0

X

Y

X'Y'

6.3t

5.4t

0.9t

3.6t

10.8t 7.2t

Shear Forces

Normal Forces

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PART (7)

2014

2014

7/3

Structural Analysis (I) Internal Forces/BEAMS

Inclined Member

12t

6t

6t

4t

4t

Θ

1.8t

3.6t2.7t

5.4t

10.8t 5.4t1.8t

3.6t

2.7t5.4t

Normal / Shear Calculations

N1 = - 6.3 t (Comp.)

N2 = - 6.3 + 5.4 = - 0.9 t (Comp.)

N3 = N2 = - 0.9 t (Check)

1

2

3

1

2

3

Q1 = +3.6tQ2 = +3.6 - 10.8 = -7.2 tQ3 = Q2 = - 7.2 t (Check)

القوي تحليل

Internal Forces Diagrams

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PART (7)

2014

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Structural Analysis (I) Internal Forces/BEAMS

3t/m`

12t6t.m

6t4t

4t

12t

12t

b

a

6t

18t.m 18t.m26t.m

10t.m

12t.m 12t.m

B.M.D

4t 4t

4t 4t

6.3t6.3t

0.9t

N.F.D

6t 6t

6t

12t

3.6t

7.2t

7.2t

S.F.D

+

+

-

-

-

-

--

-

18t.m 18t.m

12t.m

12t.m

3*2²8 = 1.5t.m

4t.m

Standard Cases

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PART (7)

2014

2014

7/5

Structural Analysis (I) Internal Forces/BEAMS

Load Resultant / Bending Moment

Concentrated LoadsΘ

P

P L┴

Mmax = PL4

P L┴

Mmax = PL`4

P h┴

Mmax = Ph4

P

Distributed Loads

w L┴

Mmax= WL²8

wt/m`

WL

w h┴

Mmax= Wh²8

wt/

m`

Wh

w L┴

Mmax= WL`²8

WL

wt/m`

w L┴

Mmax= WLL`8

WL

wt/m`

Moment Superposition

Mmax= WL²8 + Ph²

8

wt/m`

WL

Pt/m

`

Ph

Mmax= WL²8 - Ph²

8

wt/m`

WL

Pt/m

`

Ph

assume Ph < Wl

P

WL`

WL

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PART (7)

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7/6

Structural Analysis (I) Internal Forces/BEAMS

Exercise (2)3t/m`

2t/m

`

4.5t

4.5t3t

9t

3t

3t/m`

2t/m

`

6.75t.m

Ya

Xa

Ma

∑ FY = 0.0 Ya = 9 t

∑ FX = 0.0 Xa = 3 t

∑ M @ a = 0.0Ma + 4.5(0.75)+4.5(2.25)-3(2.25) = 0.0

Ma = 6.75 t.m

Reactions

Bending Moment Calculations

Ma = 6.75 t.m from left

M1 = 4.5(0.75) - 3(0.75) = 1.125 t.m from right

1

2

M2 = 0.0 t.m

Normal / Shear Calculations

4.5t

4.5t

3t

9t

3t

( sinΘ = 0.7 & cosΘ = 0.7 )

Θ

2.1t

6.3t

2.1t

6.3t

3.15t 3.15

t

3.15t 3.15

t

2.1t

2.1t

8.4t

4.2t

3.15t

3.15t

1.05t

5.25t

a

6.75t

.m1.13

t.m3*1.5²

8 =0.84 t.m

3*1.5²8 - 2*1.5²

8 =0.28 t.m

B.M.D

8.4t

5.25t

4.2t

1.05t

N.F.D

S.F.D

+

-

1

2

3

القوي تحليل

Solved Examples

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PART (7)

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7/7

Structural Analysis (I) Internal Forces/BEAMS

-

+

++

+

- -

2nd Par.

2nd Par.

2t/m` 1t/m`

2t/m`

6.125t

Exercise (1)

2t/m` 1t/m`

2t/m`

Ya

Yb

Xa

Draw Internal Forces Diagrams

0.0t

Ya = Yb = ∑P2 = 6.125t

∑ FX = 0.0 Xa = 0.0 tReactions

Load Symmetry

4t

4.25t

4t

a

b

Inclined Member

3*3*4.258 = 1.6t.m

8.25 t.m

8.25 t.m

8.25 t.m8.25

2*2²8 =1 t.m

B.M.D

2.125t1.5t 1.5t

2.125t

1.5t1.5t4.25t

3t3t

( sinΘ = 0.7 & cosΘ = 0.7 )

1t/m`

6.125t

0.0t4t

4.25t

2.125t

2.125t

2.125t

1.5t

1.5t

3t

0.0t

1.5t

1.5t

0.0t N.F.D

1

2

8.25 t.m

6.13t2.13t

1.5 t

6.13t

1.5 t

2.13t

6.125t

S.F.D

القوي تحليلΘ

1.5t

1.5t

3t

1

2

Shear ForcesNormal Forces

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PART (7)

2014

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7/8

Structural Analysis (I) Internal Forces/BEAMS

Exercise (2)2t/m` 1t/m`

1t/m`

∑ FX = 0.0 Xa = 0.0 tReactions

4.5t

Xa =0.0t

Ya

a

b

11.5tYb

∑ M @ a= 0.0Yb(7)-5(8)-1(6.5)-8(2)-2(1)= 0.0

∑ FY = 0.0 Ya = 4.5tCheck

∑ M @ b = 4.5*7-2*6-8*3+5*1-1*0.5 = 0.0 ...... OK.

2t

8t

1t 5t

Yb = 11.5t

Bending Moment Calculations

1

2

Ma = 0.0 t.m from left

M1 = 4.5(2)- 2(1) = 7 t.m

Mb = 5(2) =10 t.m from Right

Inclined Member

8t

34

5

( sinΘ = 0.6 & cosΘ = 0.8 )

Θ

2.5t

5.5t

2.0t1.5t

4.8t6.4t 4.4t

3.3t

1.5t

4.8t3.3t

2.0t

6.4t 4.4t

القوي تحليل

N. Forces

S.Forces

M2 = 11.5(1) -5(2)-1(.5)=1 t.m

5 t

6.5 t6.5 t4.4

t

2.5t

5 t

4.5 t

2 t

7 t.m

5 t.m

7t.m

1t.m

1 t.m

3.3t

1.5t

0.0t

0.0t

B.M.D

N.F.D

S.F.D

1*2²8 =0.5 t.m

2*4²8 =4 t.m

4t/m`

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PART (7)

2014

2014

Structural Analysis (I) REVISION

4t/m`8t

12t

Ya

a b

Yb 20t24t

16t

20t

12tXb

∑ M b= 0.0 Ya(8) - 16(6) - 20(2) + 8(2) -12(6) = 0.0

∑ FX = 0.0 → Xb = 12 t

→ Ya = 24 t ∑ FY = 0.0 → Yb = (16+20+8) - 24 = 20t CHECK ∑ M a= 8*10-12*6-20*8+20*6+16*2 = 0 .......O.K.

Force in LINK MEMBERS

Apply joint stability @ joints a and b

JOINT A

a

24t

24tF1

∑ FY = 0.0→ F1 = 24 t

JOINT B

20t

12t

F2F3

القــوي ــل تحليـ

θ3

4

5cosθ = 45

sinθ = 35

20t

12t

F2F3

F2cosθ

F2sinθ

stability equations

∑ FX = 0.0 → F2cosθ = 12 t45

→ F2 = 15 t

∑ FY = 0.0 → -F3 + F2sinθ = -20 t

→ F3 = 29 t35

Exercise (3)

7/9

Reactions

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PART (7)

2014

2014

4t/m`

4t/m`8t

12t

24t

16t20t

29t

15t9t

12t

Replace link members with theirforces in X and Y directions

Inclined member

20t

∑HL=12t

Sum of horizontalforces

∑VL =1t

Sum ofvertical forces

∑HR=12t

∑VR =21t

عليــه ــودي العم و ــلع الض ــاه إتج فــي القــوي ــل تحليـ

NQ

θ

θ

12cosθ

θ

1sinθ 1co

sθ

12sinθ

12sinθ

21cosθ21s

inθ12cosθ

20sinθ

20cosθ

عليــه ــودي العم و ــلع الض ــاه إتج فــي القــوي ــع تجميـ

θ9.6

t

0.6t 0.8

t

7.2t

16.8t12

.6t12

t16t

cosθ = 45

sinθ = 35

9.6t

7.2t

θ9.0

t

8t

8t12

t16t

3t

N=-9tQ=-8t

N=+3tQ=-24t

∑F x'=0

∑F y'=0

9t

3t

12t

-

-

+

NONE

8t

24t

24t

8t+

+

-

WLL'8=10m.t

16t.m

64t.m64t.m

WL²8=8m.t

Structural Analysis (I) REVISION

7/10

B.M.D

N.F.D

S.F.D

Internal Forces Diagrams

9t

8t6t

2 t/m`9 t.m

2 t/m`

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PART (7)

2014

2014

7/11

Structural Analysis (I) Internal Forces/BEAMS

Exercise (4)

9t

8t6t

2 t/m`9 t.m

2 t/m`

Reactions 12t

5t

5t

0.0t

5t3t

4t0.0t

3.5t12t

12.5t

21t

∑ FX = 0.0 X1 = 0.0 t

Secondary Beam

∑ M @ 1= 0.0Yc(9)-9(12)-12(6)-9 = 0.0

∑ FY = 0.0 Y1 = 0.0tCheck

∑ M @ c = +9+12*3+9*3 = 0.0 ...... OK.

Yc = 21t

∑ FX = 0.0 Xb = 12.0 t

Main Beam

∑ M @ b= 0.0Ya(6)-10(2.5)-10(2.5)= 0.0

∑ FY = 0.0 Yb = 3.5tCheck ∑ M @ a = 3.5*4+12*3-10*2.5-10*2.5

= 0.0 ...... OK.

Ya = 12.5t

Inclined Member 34

5

( sinΘ = 0.6 & cosΘ = 0.8 )

a

b

1a

a

c

1 c

5t

5t

3.5t

12.5t

12t8t

6t5t

5t

10t 7.2t

2.8t2.1t

9.6t

7.5t

10t

7.5t

7.5t

5t

10t 10t

10t5t

N. Forces S.Forces

8t6t

0.0t

3.5t12t

12.5t

a

b

6t

8t 10t

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PART (7)

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Structural Analysis (I) Internal Forces/BEAMS

-

7.5t

7.5t

10t

10t5t

5t

--

+

+

12t

9t 9t

27m.t

9m.t

2nd Par.

WL²8 =9m.t

Internal Forces Diagrams

9t

8t6t

2 t/m`9 t.m

2 t/m`

12t

5t

5t

0.0t0.0t

3.5t12t

12.5t

21t

a

b

1 c

End Part (7)

WL'²8 +PL'

4 =10m.t

B.M.D

N.F.D

S.F.D