chapter 3 trusses - موقع المهندس حماده شعبان 3 final2 (2).pdfb + r > 2 j...
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Chapter 3
Trusses
.
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Trusses
(members) *
(joints) (forces) (joints)
(member) (tension) (compression).
(compression) (tension) *
(members) (joint method) (member)
(section method)
*
.
(joint) (tension) (member) *
(compression) (joint).
.
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Analysis of Trusses
considering:
b = number of members.
r = number of reactions.
j = number of joints.
:tabilityS
:tructure will be unstable in the following casesS
1) b + r < 2 j
2) b + r 2 j (but all reactions are parallel)
3) b + r 2 j (but all reactions are concurrent)
: of stability) providedeterminacy (D
b + r = 2 j statically determinate
b + r > 2 j statically indeterminate
degree of indeterminacy = b + r 2 j
( )
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members (zero) truss *
(. )
(only 2 members) (joint) *
(.A, D) ((zero member
(only 3 members) (joint) *
(. C, D) ((zero member ((member
zero members
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.((members
.((support reactions 1
(. )
(only 2 members) (joint) 2
. 3
(joints) 3 ,2 4
2 - members)) .
.((members ( 4) 5
(tension) ((member 6
(compression) .
. 7
joint method
.
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((members
. ( truss)
.((support reactions 1
( )
2
members) .)
( ) 3
4
.
((members - 2 5
.
((members 6
.
( section method) ( joint method) 7
. ((members
ethodmection s
, ,
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a)
statically indeterminate to the 4th degree
b)
statically determinate.
c)
statically indeterminate to the first degree
d)
the left panel is a collapsible component
(6)
parallel are reactions all satisfied
d
be not can F
< r b
j
r
b
r b
j
r
b
r b
j
r
b
r b
j
r
b
x 0
11
6
2
9
9 2 19
9
4
15
8 2 16
8
3
13
10 2 24
10
4
20
unstable
3-2 classify each of the following trusses as statically determinate, statically indeterminate, or unstable.
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a)
stable, statically determinate.
b)
stable, statically indeterminate to 1.
c)
stable, statically indeterminate to 1.
d)
(r + b) 2 j = 1
stable, statically indeterminate to 1.
6
10
3
12)(
6
9
4
12)(
6
9
4
2)(
8
13
3
j
b
r
jbr
j
b
r
jbr
j
b
r
jbr
j
b
r
3-3 classify each of the following trusses as statically determinate, statically
indeterminate, or unstable. If indeterminate, state its degree.
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Gy
Gx FGF = 20 kN
FAG = 15 kN Ax
FAB = 10 kN
20 kN
18 kN
7.5kN
FEF
FBC = FCD
15 kN
FCD
FED = 8.33 kN
5 kN
FCD = 6.67 kN
5 kN
10 kN 6.67 kN
FBF
G
A C
F
B
D
FBE = 4.17 kN
FAf = 18 kN 15 kN FAF = 18 kN
;
members *
( C) ( T)
.
. ( , ) *
joint F:
3-8 Determine the force in each member of the roof truss. State if the members
are in tension or compression. Assume all members are pin connected.
56.3o
56.3o
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FAH
FAB
15
A
(From symmetry)
joint A:
joint B:
joint H:
3-9 Determine the force in each member of the roof truss. State if the members are in tension or compression. Assume all members are pin connected.
FBH
B FAB
10 kN
.
FBC
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from and:
joint G:
3.9
16.67 kN 16.67 kN
FGC
G
!!
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(From symmetry)
(FCG) is a zero-force member
(
+
E) FGF (
3-10 The Howe truss is subjected to the loading shown. Determine the force in members GF, CD, and GC, and state if the members are in tension or
compression.
5 kN
2 kN
9.5 kN
D
F
FGF
FDG
FCD
3 m
2 m 2 m
G
:
M )
.(
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(From symmetry)
.
3-11 The Howe truss is subjected to the loading shown. Determine the force in
members GH, BC, and BG, of the truss and state if the members are in tension or
compression. (Fig. 3-11)
B
9.5 kN
H
G
FBC
FGH
FBG
5 kN
2 m 2 m
3 m 2 kN
36.87o
56.31o
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.
Solution:
(From symmetry)
3-14 Determine the force in members IH, BC, and BH of the bridge truss. Solve
for each unknown using a single equation of equilibrium. State if the membrs are in tension or compression. Assume all members are pin connected.
5 kN 3 kN
O
H FIH
FBC
FBH
2 m
4 m
4 m
B
I
63.43o
26.56o
A
2 m
Ey
Ay
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.
.
(reactions )
3-17 Determine the force in members GF, FC, and CD of the cantilever truss. State if the membrs are in tension or compression. Assume all members are pin
connected.
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Intire truss:
;
.
3-18 Determine the forces in members KJ, CD, and CJ of the truss. state if the
members are in tension or compression. (Fig. 3-18).
5 kN 15 kN 15 kN
3 m
FKJ
FCJ
FCD
Ay
A
C
33.69o
18.43o
J
3 m 3 m
3 m
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moment 2) members )
joint D:
(G FDE )
3-19 Determine the forces in members JI, JD, and DE of the truss. state if the
members are in tension or compression.
FDJ
97.5 kN 97.5 kN
D
J H G
FJI
5 10 20
49.17
43.18FDE
E
J
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from symmetry
joint A:
joint D:
3-20 Determine the forces in each member of the truss in terms of the load P and
state if the members are in tension or compression.
FDB
P
0.687 P 0.687 P
P/2
FAB
FAD
A
D
!!
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For maximum loads, choose the minimum, Pmax = 4.25 kN
From symmetry
Ay = Cy = 0.5 P
joint A:
joint B:
FAD cos FAB cos 45 = 0
FAD cos 14 = FAB cos 45
FAD = 0.729 FAB
FAD sin + .5 P FAB sin 45 = 0
(0.729 FAB * sin 14) + 0.5 P - FAB sin 45 = 0
FAB = 0.942 P ( C )
FAD = 0.687 P ( T )
- FBD + 2 FAB sin 45 = 0
FBD = 1.33 P ( T )
FAB = 0.942 P = 4 kN, Pmax = 4.25 kN
FBD = 1.33 P = 10 kN, Pmax = 7.52 kN
3-22 Members AB and BC can each support a maximum compressive force of 4 kN, and members AD, DC, and BD can support a maximum tensile force of 10
kN. If a = 1.8m determine the greatest load P the truss can support.
.
Ay = 0.5 P
FAD
FAB
FBC
FBD
FAB
B
= 14
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from symmetry
Due To Symmetry
.
3-24 Determine the forces in members GF, FB, and BC, of the fink truss and
state if the members are in tension or compression.
FBC
FGF
FFB
4 kN
30o 60
o
B
3 cos2 30
o
4.5 tan 30
o
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.
3-25 Determine the force in members GF, CF, and CD, of the roof truss and
indicate if the members are in tension or compression.
FGF
Ey
FCD
FCF
1 m 1 m
F
1.5 m
C
E
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This is a compound truss
Where ACB and FED are connected by
Three bars FA, EC, and DB.
Since
b + r = 2 j
9 + 3 = 2(6) = 12
Statically determinate
Truss is internally unstable.
.
3-26 Classify the truss and determine if it is stable.
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From symmetry:
joint A:
joint B:
joint F:
6 kN
A FAB
FAF
8 kN
FBE
FBF
8 kN
B
3-30 Specify the type of compound truss and determine the force in each member.
State if the members are in tension or compression. Assume all members are pin connected.
.
2 kN
2 kN 10 kN
F FEF
FCF
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3-31 Determine the forces in members IH , AH, and BC, of truss. State if the
members are in tension or compression.
5 m 5 m
FIH
FAH
FBC
10 kN 14 kN
5 m
B
J I
A
26.56o
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Supports Reaction:
Method of joints: By inspection,
members BN, NC, DO, OC, HJ, JE, and LG are zero Force members.
. . . .
3-39 Determine the force in members EF, EB, and LK, of the Baltimore bridge truss and state if the members are in tension or compression. Also, indicate all
zero-force members. (Fig.3-39).
FIK
FEP
FEF
Iy 2 kN 3 kN
K
P
F
4 m 2m
4 m
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(
BCF
BGF
BA
kN9 kN6
HGF
G
X X X m3 m3
X
X
m3
m5.1
m3X X
X
X
m5.1
(
6.26
3
5.1tan 1
5.4
33.56
3
5.4tan 1
3-41 Determine the force in members BG, HG, and BC, of the truss and state if the members are in tension or compression. Also, indicate all zero-force
members.
.
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