Structural Analysis (I)

INTERNAL FORCES

2014

BEAMS

PART ( 6 )

Compound Beams Problems

@M@

Part (6)

2014

2014

Definitions/Reactions → Review Part (2) pages (4→7)

Required → Draw Internal Forces Diagrams (N.F.D , S.F.D & B.M.D).

Steps Of solution:

Calculate Reactions @ (supports/I.H) FOR (Main/Secondary) Beams Draw Internal Forces Diagrams

Step (1)Step (2)

→ Calculations (According to Section Position)

→ Drawing (On the same Datum)

6/1

Importatnt

Moment superposition Using Standard Cases

MP

8t3t/m`

M1= 0 M2 = 0

M3 = ??

=

P=8t

MwMw=wL²8=3*8²8

B.M.DMP=PL

4 =8*84 =16t.m

=24t.m

W= 3t/m`

WL²8 + PL 4 =40m.t

M3 = MP + MW

EX(1)

EX(2)3t/m`

M1= ??

M2 = 0

M3 = ??

B.M.DWL²8 = 24m.t

6t

M3 = MW - 0.5MP

12t.m

12t.m

WL²8 = 24m.t

6t

3t/m`

6*2t.m

+

+

=

2nd Par.

2nd Par.

2nd Par.

2nd Par.

Load/Shape Symmetry

2- REACTIONS

SEC. BEAM (1-2)

→ Y1 = Y2 = ∑ F2 = 4

2 = 2t

Main. BEAM (a-1)

∑ M a = 0.0→ Ma= (2*6+12*3)= 48t.m

∑ FY ↑ = 0.0→ Ya = 12+2 = 14t

Sec. BEAM (b-2)

Whole structure (Load/Shape) Symmetry

→ Ya = Yb = ∑ F2 = 4+12*2

2 = 14t

Note

CHECKWhole structure ∑ FY =(14+14)- (12+4+12)=0

...... O.K.

Solved Examples

EX(1)2t/m`

4t2t/m`

Xa Xb

a1 2

12t0t 0t

b Symmetric load

Stable → MAIN BEAM

Stable → MAIN BEAM

Unstable → SEC BEAM

1- StaticaL SYSTEM

No loads in X-direction → X a = Xb = 0t

2t/m`

4t

2t/m` b

12t

Ma48t.m

Ya 14t Yb 14t

Mb48t.m

Y1=2t2t

Y2=2t2ta

12t

N.F.D

14t

2t

2t

14t

-

+ Anti-Symmetric S.F.DS.F.D

WL²8=9m.t48t.m

PL 4=8m.t

WL²8=9m.t 48t.m

B.M.D

Symmetric B.M.D

4t12t

Internal Forces Diagrams

Similar to SEC. BEAM (1-2)

@M@

Part (6)

2014

2014

Structural Analysis (I) REVISION

6/2

ZERO

2nd Par.

@M@

Part (6)

2014

2014

aXa b c d1 21t/m`4t 12√2t12t

∑ FX → = 0.0 → Xa = 12t

11

12t

12t 18t

12t

a b c d21

Stable → MAIN BEAM

Unstable → SEC BEAM

Unstable → SEC BEAMI.H

→ stable → Main BEAM

2

a b

c

d

1

2

1t/m`4t12t

12t

Y2Yd

Y1Yc

Ya Yb

Y1

Y2

6/3

Structural Analysis (I) REVISION

18t

1- StaticaL SYSTEM

SEC. BEAM (1-2)

→ Y1 = P.aL = 18*4

6 = 12t

SEC. BEAM (c-1)

∑ M c = 0.0→ Y1= 12*3-6*2

6 =4t ∑ FY ↑ = 0.0

→ Yc = (12+6)- 4 = 14tMain. BEAM (a-b)

∑ M b = 0.0

∑ FY ↑ = 0.0→ Ya= 20*4-10*2

8 = 8t

→ Yb = (20+10)-8 = 20t

2

Y2=6tYd

c1

12t 6t

YcY1=6t

14t

a b1t/m`

4t12t

8tYa8t Yb 20t

Y1=6t

18t

→ Y1 = P.bL = 18*2

6 = 6t

2- REACTIONS

EX(2)

1 2

@M@

Part (6)

2014

2014

...... O.K.

a b

c

d

1

2

1t/m`8t12t

12t

12t

Y2=6t6t 6t

Y1=4t14t

8t 20t

12t

12t

12t-

4t 4t

8t4t

8t 12t

8t4t

8t

6t

8t- - -

Internal Forces DIAGRAMS

+ + +

16t.m

27t.m12t.m

12t.m

WL²8 + PL 4 =32m.t

PL 4=27m.t

PL 4=18m.t

N.F.D

S.F.D

B.M.D

14t20t

12t12t

6/4

Structural Analysis (I) REVISION

CHECKWhole structure

∑ FY =(12+8+4+12+12)- (8+20+14+6)=0

ZERO

2nd Par.

@M@

Part (6)

2014

2014

2t/m`

2

Xa

a1

0t c2

b22t/m`

4t6t 4t.m

a

cbStable → MAIN BEAM Stable → MAIN BEAMUnstable → SEC BEAM

No loads in X-direction → X a = 0t

2t/m`

2

a1

c2

b

2

2t/m`4t6t

4t.m

1

2

Y2Y1

Y1 Y2

Load/Shape Symmetry

SEC. BEAM (1-2)

→ Y1 = Y2 = ∑ F2 = 8

2 = 4t

2t/m`21

Y2Y18t 4t4t

Main. BEAM (b-c)

∑ M c = 0.0→ Yb= 4*10-4+16*4-4*2

8 =11.5t ∑ FY ↑ = 0.0

→ Yc = (16+4+4)- 11.5 =12.5t Yc

cb2t/m`

4t

4t.m2

Y2

Yb 11.5t

4t

16t

Main. BEAM (a-1)

∑ M a = 0.0

∑ FY ↑ = 0.0→ Ma= (6*2+4*4)= 28t.m

→ Ya = 6+4 = 10t 2

1

2

6t Y1Ma28t.m

Ya 10t

a 4t

YcYb

Ma

Ya

∑ FY =(6+8+16+4)- (10+11.5+12.5)=0...... O.K.

12.5t

6/5

Structural Analysis (I) REVISION

EX(3)

1- StaticaL SYSTEM

2- REACTIONS

CHECKWhole structure

1 1 2 2

@M@

Part (6)

2014

2014

2t/m`

2

a1

c2

b

2

2t/m`4t6t

4t.m

1

2

8t

11.5t 12.5t

28t.m

10t

4t

4t

4t

4t

16t

N.F.D

10t

4t

4t8.5t

4t

28t.m

8t.m

4t.m4t.m

4t.mWL²8=16m.t

WL²8=4m.t8t.m

11.5t 7.5t 12.5t6t

4t.m

S.F.D

B.M.D

+ +--

6/6

Structural Analysis (I) REVISION

Internal Forces DIAGRAMS

ZERO

2nd Par.

@M@

Part (6)

2014

2014

Structural Analysis (I) REVISION

6/7

2t20t2t/m` 3t/m`

2t/m` 4t/m`

Y1=6tYa 6t

12t

6tY2=10t

2+10t

18t

Yd 8t

Yb 13t Yc 25t

N.F.D

6t

12t

S.F.D+

-

WL²8=9m.t WL²8=13.5m.tB.M.D

6t.m

6t.m

6t.m

6t.m 20t

6t

7t

13t

12t

++

--

6t.m 6t.m

6t.m

6t.m 6t.m

PL 4=30m.t

EX(4)

StaticaL SYSTEM/ Reactions

Internal Forces DIAGRAMS

MAINSEC SEC

1 2

1 1 2 2

1

1 2

2

ZERO

2nd Par.

6t 6t6m.t

11 t

12 t

36m.t

24m.t

S.F.D

B.M.D

+

-

+

-

11 t

1 t7 t

13 t

6 t 6 t

6 t12 t 12 t

36 m.t

9 m.t

6 m.t

24 m.t24 m.t

6 m.t

WL²8=9m.t

WL²8=9m.t

WL²8=2.25m.t

2nd Par.

@M@

Part (6)

2014

2014

6/8

EX(5)2t/m`

StaticaL SYSTEM/ Reactions

MAINSEC → MainMAIN SEC

Y1=5t5t

Y2=6t6t

Y3=6t6+6t

1 2 3

6t 6m.t

1 2

3

3

3 3221 1

1

2

Ma

Ya

Yb Yc

N.F.D

S.F.D

B.M.D

ZERO

6t 6t

12t

# END PART (6)