Post on 14-Dec-2015
description
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)