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6.3 Trusses: Method of Sections
6.3 Trusses: Method of Sections Example 1, page 1 of 2
5 m
C D
EF
G H
5 m
5 m
B
3 m
A
1. Determine the force in members AC, CD, and DF, and
state whether the force is tension or compression.
1 Pass a section through the three
members whose forces are to be
determined.
A B
C D
EF
G H
4 kN
2 kN
6 kN 6 kN
2 kN
4 kN
6.3 Trusses: Method of Sections Example 1, page 2 of 2
F x = 0: 2 kN + 4 kN + F
CD = 0
F y = 0: F
AC F DF = 0
M D = 0: F
AC(3 m) (2 kN)(5 m + 5 m)
(4 kN)(5 m) = 0
Solving simultaneously gives
F AC = 13.33 kN (T) Ans.
F CD = 6.0 kN = 6.0 kN (C) Ans.
F DF = 13.33 kN = 13.33 kN (C) Ans.
+
++
We had assumed member CD to be in tension.
Calculations showed that F CD is negative, so our
assumption was wrong: CD must be in compression.
Similarly DF must be in compression.
5
2 kN
4 kN
C
Equations of equilibrium for the portion of the
truss (Note that moments are summed about
point D, even though point D is not part of the
free body):
4
F AC
F CD
F DF
At each cut through a
member, a force is shown
to represent the effect of
the portion of the member
on one side of the section
pulling on the portion on
the other side. It is
convenient to always
assume the force to be
tension.
3
Free-body diagram of portion of truss above the section2
3 m
5 m
HG
FE
D
5 m
6.3 Trusses: Method of Sections Example 2, page 1 of 3
2. Determine the force in members CD, CH, and GH, and state whether the force is tension or compression.
12 ft 12 ft 12 ft 12 ft
800 lb 800 lb 800 lb
A
F G H
B C D
E
800 lb
A
B
800 lb
C
F G
D
800 lb
H
E
1 Pass a section through the three
members whose forces are to be
determined.
8 ft
6.3 Trusses: Method of Sections Example 2, page 2 of 3
D
800 lb
H
E
F GH
F CH
F CD
E y
2 Free-body diagram of portion of truss to right of section
At each cut through a member, a force is shown.3
4 Equations of equilibrium for the portion of the
truss:
F x = 0: F
GH F CH sin F
CD = 0 (1)
F y = 0: F
CH cos 800 lb + E y = 0 (2)
M H = 0: F
CD (8 ft) + E y (12 ft) = 0 (3)
12 ft
+
++
5
Three equations but four unknowns, so another
equation is needed.
8 ft
12 ft
6
Geometry
= tan-1 = 56.31°
H
DC
8 ft
12 ft8 ft
6.3 Trusses: Method of Sections Example 2, page 3 of 3
800 lb
12 ft12 ft
7
A y
B
800 lb
C
Free-body diagram of entire truss.
F G
12 ft
D
800 lb
H
E y
EA x A
12 ft
8
+
9
Equilibrium equation for entire truss. This will give the needed fourth equation.
M A = 0: (800 lb)(12 ft) (800 lb)(2 12 ft) (800 lb)(3 12 ft) + E
y(4 12 ft) = 0
Solving gives E y = 1,200 lb.
Substituting E y = 1,200 lb into Eqs. 1, 2, and 3 and solving simultaneously gives
F CD = 1,800 lb (T) Ans.
F CH = 721 lb (T) Ans.
F GH = 2,400 lb = 2,400 lb (C) Ans.
8 ft
6.3 Trusses: Method of Sections Example 3, page 1 of 3
4 m 4 m 4 m
3. The diagonal members are not connected to each other where they cross. Determine the force
in members BG, CF, and FG, and state whether the force is tension or compression.
E
3 kN
F G H
D
CBA
3 kN 3 kN
BA
3 kN 3 kN
E F
C
3 kN
G
D
H
1 Pass a section through the three members
whose forces are to be determined.
2.5 m
6.3 Trusses: Method of Sections Example 3, page 2 of 3
F FG
F BG
F CF
2
3 At each cut through a member, a force is shown
4 m
4 Equations of equilibrium for the portion of the truss:
F x = 0: F
FG F BG sin F
CF sin = 0 (1)
F y = 0: F
BG cos + F CF cos + D
y 3 kN = 0 (2)
M G = 0: F
CF sin (2.5 m) + D y(4 m) = 0 (3)
+
5
Three equations but four unknowns, so
another equation is needed.
4 m
C
GF
6
= tan-1 = 58.0°
Geometry
D y
++
2.5 m
2.5 m
2.5 m4 m
G
3 kN
C
Free-body diagram of portion of truss to right of section
D
H
6.3 Trusses: Method of Sections Example 3, page 3 of 3
Free-body diagram of entire truss (This will give the needed fourth equation).
3 kN3 kN
E x
7
FE
A B
E y
D y
3 kN
HG
C
D
8
+
Equilibrium equation for entire truss.
M E = 0: (3 kN)(4 m) (3 kN)(2 4 m) + D
y(3 4 m) = 0
Solving gives D y = 3.0 kN. Then substituting = 58.0° and D
y = 3.0 kN into Eqs. 1, 2, and 3 and solving
simultaneously gives
F BG = 5.66 kN (T) Ans.
F CF = 5.66 kN (T) Ans.
F FG = 9.6 kN = 9.6 kN (C) Ans.
2.5 m
4 m4 m4 m
6.3 Trusses: Method of Sections Example 4, page 1 of 4
3 m1.5 m
3 m
6 m
3 m
3 m
3 m
4. Determine the force in members CE, EF, HF, and CF, and state whether the force is tension or compression.
I
G
E
C
A B
D
F
H
J
6 kN
4 kN
6 m
4 kN
6 kN
Pass a section through at
least some of the members
whose forces are to be
determined. The general
idea is to choose as few
members as possible --three
in this instance-- because
each time a member is cut
by a section, an additional
unknown is introduced into
the equilibrium equations.
1
I J
HG
E F
CD
BA
6.3 Trusses: Method of Sections Example 4, page 2 of 4
3 m
3 m
I
G
E F
H
J 6 kN
4 kN
L GH
L EF
2 Free-body diagram of portion of truss above section (Using
the upper portion of the truss rather than the lower
eliminates the need to calculate the reactions at the bottom
of the truss).
3 Equations of equilibrium for the portion of the truss:
MG = 0: (6 kN)(3 m) + FEF(3 m) F FH cos (L
GH) = 0 (1)
M F = 0: (6 kN)(2 3 m) 4 kN)(3 m) + F
CE cos (L EF) = 0 (2)
F x = 0: F
CE sin + F EF + F
FH sin + 4 kN + 6 kN = 0 (3)+
++
F EF
F CE
F FH
6.3 Trusses: Method of Sections Example 4, page 3 of 4
E
CD
G
I
F
H
J
BA
4 Geometry5
3 m
3 m
3 m
3 m
3 m
3 m
3 m
3 m
3 m 3 m
3 m
3 m
3 m
(6 m) tan
(3 m) tan
= tan-1 = tan-1 = 7.125°
L GH = 3 m + (3 m) tan + (3 m) tan = 3.75 m
L EF = 3 m + (6 m) tan + (6 m) tan = 4.50 m
6 Substituting these values for , LGH, and LEF into Eqs. 1, 2, and
3 and solving simultaneously gives:
F CE = 10.75 kN Ans.
F EF = kN = 7.33 kN (C) Ans.
F FH = 10.75 kN = 10.75 kN (C) Ans.
B'1.5 m 1.5 m
1.5 m
4 3 mJB'
BB'
6.3 Trusses: Method of Sections Example 4, page 4 of 4
8 Equilibrium equations for joint F
F x = 0: F
CF cos + F DF sin 7.125° + (10.75 kN)(sin 7.125°) 7.33 kN = 0 (4)
F y = 0: F
CF sin F DF cos 7.125° (10.75 kN)(cos 7.125°) = 0 (5)
+
+
L EF = 4.50 m
(3 m) tan
E F
DC
9 Geometry
= tan-1 = 31.61°
10 Substituting = 31.608° into Eqs. 4 and 5 and solving simultaneously gives:
F DF = 14.97 kN = 14.97 kN (C)
F CF = 7.99 kN (T) Ans.
F EF = 7.33 kN (C)
F CF
F DF
F FH = 10.75 kN (C)
3 m
(3 m) tan 7.125° + 4.50 m
3 m = 7.125°
7 Free-body diagram of joint F.
This free body will enable us to
calculate the remaining unknown
force the force in member CF.
F
6.3 Trusses: Method of Sections Example 5, page 1 of 4
G
M
1 Pass a section through the four
members whose forces are to be
determined. It does not appear
possible to find a section that cuts
only three of these members.
5. Determine the force in members RS, LS, FL, and EF, and state whether the force is tension or compression.
N O P Q R S T
H I J K L M
AB C D E F
G
2 m
3 m
2 m
3 m3 m3 m3 m3 m
4 kN 4 kN 4 kN
C
N O
A
H
B
I L
P Q R S
4 kN 4 kN
D
J
4 kN
E F
K
T
6.3 Trusses: Method of Sections Example 5, page 2 of 4
3 m
F
S
2 mG
T
M2 m
2 Free body diagram of truss portion to right of section line
F RS
F LS
F FL
F EF
G y
3 Equations of equilibrium for the portion of the truss:
M S = 0: F
EF(2 2 m) + G y(3 m) = 0 (1)
M F = 0: F
RS(2 2 m) + G y(3 m) = 0 (2)
F y = 0: F
FL F LS + G
y = 0 (3)++
+
Three equations with five unknowns so two more
equations are needed.4
6.3 Trusses: Method of Sections Example 5, page 3 of 4
4 kN
3 m
P
3 m
A
3 m
B C
N
H
O
I J
3 m3 m
D
4 kN
3 m
4 kN
E F
Q SR
K L
2 m
G
T
M2 m
5 Free-body diagram of entire truss (This free body will enable us to calculate the reaction at G).
6 Equation of equilibrium for the entire truss.
M A = 0: (4 kN)(2 3 m) (4 kN)(3 3 m) (4 kN)(4 3 m) + G
y(18 m) = 0 (4)
Solving gives
G y = 6 kN
Substituting G y = 6 kN into Eqs. 1 and 2 and solving gives:
F EF = 4.5 kN (T) Ans.
FRS = 4.5 kN = 4.5 kN (C) Ans.
A x
A y G
y
+
6.3 Trusses: Method of Sections Example 5, page 4 of 4
S
7 Free-body diagram of joint S. This free body will
enable us to calculate the force in member LS.
F LS
F MS
F ST
F RS = 4.5 kN (C)
8 Equations of equilibrium for joint S. Note that there
are three unknowns but only two equations.
F x = 0: 4.5 kN + F
ST + F MS cos = 0 (5)
F y = 0: F
LS F MS sin = 0 (6)
+
+
9 Geometry
S T
M
2 m
3 m
= tan-1 = 33.69°
10 Free body diagram of joint T
F ST
F MT
T
11 Two members meet at joint T, they are not collinear and no
external force acts at joint T, so members ST and MT are zero-
force members.
Substituting F ST = 0 in Eq. 5 and solving Eqs. 5 and 6
simultaneously gives:
F MS = 5.41 kN = 5.41 kN (C)
F LS = 3.0 kN (T) Ans.
12 Substituting F LS = 3.0 kN and G
y = 6 kN into Eq. 3 and
solving gives:
F FL = 3.0 kN = 3.0 kN (C) Ans.
2 m3 m
6.3 Trusses: Method of Sections Example 6, page 1 of 4
5 m
5 m
T
N
A
B C D E F G H I J K L
M
O P Q R S
U V WX
10 kN12 panels @ 4 m each
6. Determine the force in members TU, EF, and EU. State whether the force is tension or compression.
SQ RPON
A
CB ED
10 kN
HGF JI
M
LK
T U V W X
1 Even though we were not asked to determine the force in member EP,
we have to pass the section through it because we must make the
section go completely through the truss.
6.3 Trusses: Method of Sections Example 6, page 2 of 4
T
F EF
F EP
F EU
F TU
A y
A x
5 m
5 m
4 m 4 m 4 m 4 m
2 Free-body diagram of portion of the truss to the left of the section
3 Equations of equilibrium for the portion of the truss:
F x = 0: A
x + F TU + F
EP cos + F EF = 0 (1)
F y = 0: A
y + F EU + F
EP sin = 0 (2)
M E = 0: A
y (4 4 m) F TU(2 5 m) = 0 (3)
+
++
4 Three equations with six unknowns so three more
equations are needed.
5 Geometry
P
F
E
4 m
5 m 4 m5 m = tan-1 = 51.34°
ON
A
B C D E
6.3 Trusses: Method of Sections Example 6, page 3 of 4
12 panels @ 4 m each
CB
A
N
T
RQPO
10 kN
GFED JIH
S
LK
M
XU V W
6 Free-body diagram of entire truss (This free body will enable us to calculate the reactions at support A).
A x
A y
M y
7 Equations of equilibrium for entire truss. Note that we
only write two equations because we only need to
calculate Ax and Ay, since only Ax and Ay appear in
Eqs. 1, 2, and 3.
F x = 0: A
x = 0 (4)
M M = 0: (10 kN)(6 4 m) A
y(12 4 m) = 0 (5)
Solving gives A x = 0 and A
y = 5 kN.
Consideration of joint F shows that member FP is a zero-force
member, so F FP = 0.
But if member FP is removed (because it is a zero-force
member), consideration of joint P shows that member EP is also
a zero-force member, so F EP = 0.
8
+
+
6.3 Trusses: Method of Sections Example 6, page 4 of 4
9 Substituting
= 51.34°,
A x = 0,
A y = 5 kN,
and
F EP = 0
into Eqs. 1, 2, and 3, and solving gives:
F TU = 8 kN = 8 kN (C) Ans.
F EF = 8 kN (T) Ans.
F EU = 5 kN = 5 kN (C) Ans.
6.3 Trusses: Method of Sections Example 7, page 1 of 6
7. Determine the force in
members KM, LM, and
DK. State whether the
force is tension or
compression.
1 We choose a section that cuts at least some of the
members whose forces are to be determined. But the
section should cut as few other members as possible,
since each time a member is cut, an additional
unknown appears in the equilibrium equations.
L
2 kN
2 kN
A
G
2 kN
2 kN
B C
I
D
KJ
M
E
H
2 kN
2 kN
F
2 kN
I
2 m 2 m 2 m 2 m 2 m 2 m 2 m 2 m 2 m 2 m
I
6 m2 kN
2 kN
2 kN
2 kN
2 kN
2 kN
2 kN
H
L
M
KJI
G
FEDCB
A
6.3 Trusses: Method of Sections Example 7, page 2 of 6
ED
K L
2 kN
F
H
2 kN
2 kN
F KM
F ML
F CD
F y
2 Free-body diagram of portion of truss to right of section.
It is not essential but we can save some work if we use
the principle of transmissibility as shown in Step 3.
F CD
D
K
F y
2 kN
E
H
L
2 kN
F
2 kN
F KM cos
F KM sin
F ML cos
F ML sin
3 Same free body as in Step 2, but now the force F KM has been
moved along its line of action to joint D (principle of
transmissibility) and then expressed in terms of vertical and
horizontal components. Similarly F ML is moved to joint F.
4 Equations of equilibrium for free body in Step 3. Note that
because we were not asked to determine F CD, we choose two
moment equations in which F CD does not appear.
M F = 0: (2 kN)(2 m) + (2 kN)(3 2 m) F
KM sin (4 2 m) = 0 (1)
M D = 0: (2 kN)(2 m) (2 kN)(3 2 m) (2kN)(4 2 m) + F
y(4 2 m) + F ML sin (4 2 m) = 0 (2)
5 Two equations but three unknown forces, so another
equilibrium equation is needed.
++
2 m2 m 2 m 2 m
6.3 Trusses: Method of Sections Example 7, page 3 of 6
6 m
6 Geometry
= tan-1 ( 2 m ) = 71.56°
= tan-1 ( 2 m + 8 m
) = 30.96°
2 m
8 m
6 m
6 m
M
DF
6.3 Trusses: Method of Sections Example 7, page 4 of 6
C D
I
2 kN
LJ K
M
2 kN
FE
F y
2 kN
H
2 kN
8 Equilibrium equation for entire truss
M A = 0: (2 kN)(2 m) (2 kN)(3 2 m) (2 kN)(5 2 m)
(2 kN)(7 2 m) (2 kN)(9 2 m) (2 kN)(10 2 m) + F y(10 2 m) = 0 (3)
9 Solving gives F y = 7 kN. Substituting F
y = 7 kN, = 71.56°, and
= 30.96° in Eqs. 1 and 2, and solving simultaneously gives:
F KM = 2.11 kN (T) Ans.
F ML = 5.83 kN = 5.83 kN (C) Ans.
+
I
2 m 2 m 2 m 2 m 2 m 2 m 2 m 2 m 2 m 2 m
Free-body diagram of entire
truss. This free body will
enable us to calculate the
reaction at support F.
7
2 kN
2 kN
AA x
A y
2 kN
G
B
6.3 Trusses: Method of Sections Example 7, page 5 of 6
F DK
10 Free-body diagram of joint K (This free body will enable
us to calculate the force in member DK).
11 Since there are only two unknown forces, F KL and F
DK,
we could write force-equilibrium equations in the x
and y directions and then solve them simultaneously.
However, we can save work by noticing that a
zero-force member is present.
2 kN
L
2 kN
A
G
2 kN
B C
I
D
KJ
2 kN
2 kN M
E
H
2 kN
2 kN
FA
x
A y
F y
12 Free- body diagram of entire truss13 Consideration of joint K shows that KL must
be a zero-force member, so F KL = 0.
I
K
y
x
F KM = 2.11 kN (T)
F KL
6.3 Trusses: Method of Sections Example 7, page 6 of 6
14 Free-body diagram of joint K (repeated)
F y = 2.11 kN F
DK = 0 (4)
Solving gives
F DK = 2.11 kN (T) Ans.
+
F KM = 2.11 kN (T)
y
F KL
F DK
K
x
6.3 Trusses: Method of Sections Example 8, page 1 of 3
A
B
F
C D E
G H I
Cable
J30°
1 The section must pass through the
cable. Otherwise the portion of the
truss to the left of the section could
not be isolated as a free body.
5 ft 5 ft 5 ft 5 ft 5 ft
JIHGF
EDCB
A
Cable
8. Determine the force in members GH,
CD, and CH. State whether the force is
tension or compression. Also, find the
tension in the cable.
5 kip
3 ft
5 kip
30°
6.3 Trusses: Method of Sections Example 8, page 2 of 3
5 kip
A
B
F
C
G
2 Free-body diagram of portion of truss to left of section
T
F GH
F CH
F CD
3 The tension in the cable is
one of the unknowns.
4 Equations of equilibrium for the portion of the truss:
M C = 0: T cos 30°(3 ft) F
GH(3 ft) + (5 kip)(2 5 ft) = 0 (1)
M G = 0: (5 kip)(2 5 ft) + F
CD(3 ft) + F CH sin (3 ft) = 0 (2)
F y = 0: 5 kip + T sin 30° + F
CH cos 0 (3)++
+
C
G H
5 Geometry
3 ft
5 ft
= tan-1 ( 3 ft
) = 59.04°
6 Three equations but four
unknown forces, so another
equilibrium equation is needed.
3 ft
5 ft 5 ft
5 ft
30°
6.3 Trusses: Method of Sections Example 8, page 3 of 3
J x
8
5 kip
7
Equation of equilibrium for the entire truss. Only one equation is
used because we need to calculate T only; the reactions at J are not
needed.
M J = 0: (5 kip)(5 5 ft) T sin 30°(3 5 ft) = 0 (4)
T = 16.67 kip Ans.
Substituting = 59.04° and T = 16.67 kip into Eqs. 1, 2, and 3 and
solving simultaneously gives:
F GH = 2.23 kip (T) Ans.
F CD = 11.11 kip = 11.11 kip (C) Ans.
F CH = 6.48 kip = 6.48 kip (C) Ans.
+
9
5 ft5 ft5 ft5 ft5 ft
3 ft
30°
E
Free-body diagram of entire truss
A
B
F
C D
G
T
H
J y
I J