COURSE:
CE 201 (STATICS)
LECTURE NO.:
24 to 26
FACULTY:
DR. SHAMSHAD AHMAD
DEPARTMENT:
CIVIL ENGINEERING
UNIVERSITY:
KING FAHD UNIVERSITY OF PETROLEUM
& MINERALS, DHAHRAN, SAUDI ARABIA
TEXT BOOK:
ENGINEERING MECHANICS-STATICS
by R.C. HIBBELER, PRENTICE HALL
LECTURE NO. 24 to 26
THE METHOD OF JOINTS AND
ZERO FORCE MEMBERS
Objectives:
►
To show how to determine the forces in the
members of a truss using the method of joints
►
To show how to
members of a truss
identify
the
zero-force
THE METHOD OF JOINTS
The method of joints for the analysis of a truss consists of
the following steps:
►
Determine the support reactions considering the equilibrium of
truss as a whole
►
Consider only one joint at a time
►
Draw the free body diagram for the joint into consideration,
indicating the magnitudes, directions, and the senses of the
known external forces, reactions, and member forces
►
Apply the conditions of equilibrium for the joint into consideration,
as:
Eq.1
ΣFx = 0
Eq.2
ΣFy = 0
►
Solve the equations of equilibrium (Eq.1 & 2) to determine the
forces in the members meeting at the joint into consideration
►
Go on considering the other joints of the truss one by one till the
forces in each of the members of the truss are determined
THE METHOD OF JOINTS
Important Points
► Always assume the sense of the unknown member forces
acting on the joint’s free-body diagram as “tension”
►
If after analysis
to be negative,
compression.
the magnitude of a member force is
reverse its sense, i.e. the sense is
found
to be
►
In all cases, the analysis should start at a joint having at
least one known force or reaction and at most two unknown
forces, for example joint B in the above figure.
ZERO FORCE MEMBERS
►
The zero-force members of a truss support “no
loading” and are used to increase the stability of
the truss during construction and to provide
support if the applied load is changed.
►
Identification of the zero-force members of a
truss greatly simplifies the analysis of the truss,
using the method of joints.
►
Following two general rules may be helpful in
identifying the zero-force members:
ZERO FORCE MEMBERS
Rule-I
“If only two members form a
truss joint and no external load
or support reaction is applied to
the joint, the members must be
zero-force members”.
For example:
+P∑Fy = 0 ⇒ FDC sin θ = 0 ⇒ FDC = 0
+O∑Fx = 0 ⇒ FDE + 0 = 0 ⇒ FDE = 0
Initially, 9 members but reduced to
only 5 members after removing the
zero force members.
+
→ ΣFx = 0; FAB = 0
+ ↑ ΣFy = 0; FAF = 0
ZERO FORCE MEMBERS
Rule-II
“If three members form a truss joint
for which two of the members are
collinear, the third member is a zeroforce member provided no external
force or support reaction is applied to
the joint”.
For example:
+O ∑Fx=0 ⇒ FDA = 0
+P ∑Fy = 0 ⇒ FDC = FDE
+O ∑Fx=0 ⇒ FCA sinθ = 0
+P ∑Fy = 0 ⇒ FCB = FCD
⇒ FCA =
0
Initially, 7 members but reduced to
only 3 members after removing the
zero force members.
PROBLEM SOLVING: Example # 1
Determine the force in each member of the truss and state if
the members are in tension or compression. Set P1 = 500-lb
and P2 = 100-lb.
PROBLEM SOLVING: Example # 1
The support reactions can be calculated
by applying equilibrium conditions to the
above free-body diagram, as follows:
∑Fx = 0
The free-body diagram of the
entire truss is shown below:
A
Ax
C
Cy
Ay
⇒ Ax – 100 = 0
⇒ Ax = 100 lb
(1)
∑Mabout A = 0
⇒ −14 Cy + 500 × 6 + 100 × 8 = 0
⇒ Cy = 271.428 lb
(2)
∑Fy = 0
P2
= 100lb
B
P1 = 500lb
⇒ Ay + Cy – 500 = 0
⇒ Ay + Cy = 500
⇒ Ay = 500 – 271.428 = 228.572 lb.
(3)
PROBLEM SOLVING: Example # 1
∑Fy = 0
⇒ 271.43 – FCB sin 45 = 0
⇒FCB =
271.43
sin 45
= 383.86 lb (T)
Ans.
∑Fx = 0
⇒ − FCA – FCB cos 45 = 0
The magnitude and sense of the forces ⇒ FCA = – FCB cos 45 = –383.86 cos 45o
in each member of the truss may be = –271.43 lb = 271.43 lb (C)
Ans.
determined by considering free-body Joint A:
diagram of the joints, as follows:
The FBD of joint A is as follows:
Joint C:
8
271.43
100 lb
The free-body diagram of joint C is as
θ = tan −1 = 59.03ο
6
θ
follows:
228.53 lb
FCA
∑Fx = 0
45o
FCB
271.43 lb
F.B.D. of joint C
∴sinθ = 0.8 and cosθ = 0.6
FAB
⇒ 100 – 271.43 + FAB cosθ = 0
Ans.
⇒FAB = 285.71 (T)
PROBLEM SOLVING: Example # 2
Determine the force in each member of
the truss and state if the members are
in tension or compression. Set P = 4
KN.
PROBLEM SOLVING: Example # 2
Applying equilibrium conditions at
joint A,
∑Fy = 0
⇒ −4 – FAE sinθ = 0
4
−
⇒ FAE = 0.447 = − 8.948 kN
= 8.948 kN (C)
Ans.
∑Fx = 0
Analysis of the given truss may be
started from joint A without ⇒ FAB + FAE cosθ = 0
calculating the support reactions ⇒ FAB = − 0.894 FAE = −0.894 (− 8.948)
= 8 kN (T)
Ans.
because at joint A there are only two
Joint B:
unknowns.
The FBD of joint B is as follows:
Joint A:
∑Fx = 0 ⇒ − 8 + FBC = 0
The FBD of joint A is as follows:
⇒ FBC = 8 kN (T) Ans.
∑Fy = 0 ⇒ −8 – FBE = 0
⇒ FBE = −8 kN
⇒ FBE = 8 kN (C) Ans.
PROBLEM SOLVING: Example # 2
∑Fy = 0
⇒ − 8–8.948sinθ+FEC sinθ−FED sinθ = 0
8 + 8.948sinθ
= 26.836 kN (2)
⇒FEC–FED= sinθ
Eq. (1) + Eq. (2)
⇒ 2FEC = 17.88
⇒ FEC = 8.9458 kN (T) Ans.
Joint E:
The FBD of joint E is as follows:
From Eq. (1)
∴ FED = −8.948 – 8.948 = −17.896 kN
⇒ FED = 17.896 kN ( C ). Ans.
Joint D:
The FBD of joint D is as follows:
∑Fy = 0
⇒ FDC – 17.896 sinθ = 0
⇒ FDC = 8 kN (T) Ans.
∑Fx = 0
∑Fx = 0
⇒ 8.948 cosθ +FEC cosθ + FED cosθ = 0
⇒ 17.896 cosθ - Dx = 0
⇒ FEC + FED = −8.948
(1)
⇒ Dx = 17.896 × 0.894 = 16 kN
PROBLEM SOLVING: Example # 2
The magnitude and sense of the force
in each member of the truss is shown
below:
Joint C:
The FBD of joint C is as follows:
8 kN
4 kN
A
8 kN
4 kN
B
8 kN
8.9
48
kN
8 kN
C
Cy = 16 kN
kN
48
8.9
8 kN
E
∑Fx = 0
– 8 – 8.948 cosθ – Cx = 0
⇒ Cx = – 16 kN = 16 kN (→)
∑Fy = 0
⇒ – 4 – 8 – 8.948 sinθ + Cy = 0
⇒ Cy = 16 kN ( ↑)
Cx = 16 kN
17.
896
kN
D
Dx = 16 kN
PROBLEM SOLVING: Example # 3
Determine the force in each member of the truss
and state if the members are in tension or
compression considering only the self-weight of
the members @ 4 kg/m. Set P = 0.
PROBLEM SOLVING: Example # 3
PA = ½ (mAB + mAE)
= ½ (157 + 175) = 166 N
PB = ½ (mAB + mBE + mBC)
= ½ (157 + 78 + 157) = 196 N
PC = ½ (mBC + mCE + mCD)
= ½ (157 + 175 +157) = 245 N
PE = ½ (mAE + mBC + mEC + mED)
= ½ (175 + 78 + 175 + 175) = 302 N
Masses of members are calculated PD = ½ (mED + mCD)
= ½ (175 + 157) = 166 N
below using the unit weight and the
length of the members:
FBD of the truss showing loads at
joints is given below:
mAB = 4 × 4 = 16kg = 157N
mAE = 4 ( 4 + 2 ) = 17.88 kg = 175 N
2
2
mBE = 4 × 2 = 8 kg = 78 N
mBC = 4 × 4 = 16 kg = 157 N
mCE = 4 ( 4 + 2 ) = 17.88 kg = 175 N
2
2
Loads at joints of the truss are
calculated as follows:
166 N
196 N
245 N
B
C
A
Cx
Cy
E
302 N
D
166 N
Dx
PROBLEM SOLVING: Example # 3
166 N
196 N
245 N
B
C
A
∑Fy = 0 ⇒ −166 – FAE sinθ = 0
Cx
Cy
−166
⇒ FAE = 0.449 = −371.36 N
⇒ FAE = 371.36 N (C)
E
302 N
D
166 N
Dx
Ans.
∑Fx = 0 ⇒ FAB + FAE cosθ = 0
⇒FAB = −FAE cosθ
Analysis of the given truss may be = −(−371.36)×0.894 = 332 N (T)
started from joint A without
Joint B:
calculating the support reactions
The FBD of joint B is as follows:
because at joint A there are only two
unknowns.
Joint A:
The free-body diagram of joint A is as
follows:
∑Fx = 0⇒ FBC – 332 = 0
⇒ FBC = 332 N (T) Ans.
∑Fy = 0 ⇒ −196 −FBE = 0
Ans.
⇒ FBE = −196 ⇒FBE = 196 N (C) Ans.
PROBLEM SOLVING: Example # 3
166 N
196 N
245 N
B
C
A
Eq. (1) + Eq. (2)
⇒ 2FEC = 1114.09
⇒ FEC = 557 N (T)
Cx
Cy
E
From Eq. (1),
FED =−371−557=−928N=928N (C) Ans.
302 N
D
166N
Dx
Joint D:
The FBD of joint D is as follows:
Joint E:
The FBD of joint E is as follows:
∑Fx = 0
⇒ FEC cosθ+ FED cosθ + 371 cosθ = 0
⇒ FEC + FED = −371
(1)
∑Fy = 0
⇒−196–371sinθ –302+FECsinθ−FED sinθ = 0
⇒FEC–FED= 302 + 196 + 371× 0.447 =1485.09N
0.447
Ans.
(2)
∑Fy = 0 ⇒ FDC – 928 sinθ - 166 = 0
⇒ FDC = 581 N ( T )
∑Fx = 0 ⇒ 928 cosθ - Dx = 0
⇒ Dx = 928 × 0.894 = 830 N
PROBLEM SOLVING: Example # 3
166 N
196 N
245 N
B
C
A
Cx
Cy
E
302 N
D
166N
Dx
Joint C:
The FBD of joint C is as follows:
The magnitude and sense of the force
in each member of the truss is shown
below:
196N
166N
A
332N
B
196N
371
.36
N
245N
332N
Cy = 1075N
581N
928
N
D
166N
∑Fx = 0 ⇒ Cx – 332 – 557 cosθ = 0
⇒Cx = 830 N
∑Fy = 0 ⇒−245 – 581–557 sinθ +Cy = 0
⇒ Cy = 1075 N
Cx = 830N
N
5 57
E
302N
C
Dx = 830 N
PROBLEM SOLVING: Example # 4
Determine the force in each member of the truss and
state if the members are in tension or compression.
Set P1 = P2 = 4 kN.
PROBLEM SOLVING: Example # 4
Applying equilibrium conditions at
joint C,
∑Fy = 0 ⇒ FCB sinθ −4 = 0
⇒ FCB = 8 kN (T)
Ans.
∑Fx = 0 ⇒−FCD – FCB cos 30 = 0
⇒FCD= −6.93 kN = 6.93 kN (C) Ans.
Joint D:
Analysis of the given truss may be The free-body diagram of joint D is as
started from joint C without follows:
calculating the support reactions
because at joint C there are only two
D
unknowns.
Joint C:
The FBD of joint C is as follows:
∑Fx = 0 ⇒ −FDE – 6.93 = 0
⇒ FDE= −6.93 kN= 6.93 kN ( C) Ans.
∑Fy = 0 ⇒ FDB – 4 = 0
⇒ FDB = 4 kN ( T) Ans.
PROBLEM SOLVING: Example # 4
∑ Fy = 0
⇒ +FBA sin 30 – FBE sin 30 – 4 – 8 sin 30 = 0
Joint B:
The FBD of joint B is as follows:
4+4
⇒ FBA – FBE = 0.5 = 16
(2)
Solving, Eqs. (1) and (2),
FBA = 12 kN (T) and FBC = 4 kN (C)
Ans.
Support A
The FBD of support A, as follows:
Ay
Ax
12
=
=
By Sine Rule, Sin120 Sin150 Sin90
∑ Fx = 0
⇒ −FBA cos30 – FBE cos30+ 8 cos30 = 0
(1)
⇒ FBA + FBE = 8
From the above equation, Ax and Ay
can be determined as:
Ax = 10.392 kN and Ay = 6 kN
PROBLEM SOLVING: Example # 4
The magnitude and sense of the force
in each member of the truss is shown
below:
Ay=6kN
Support E
FBD of support E, as follows:
A
Ax=10.392kN
12k
N
B
4kN
8kN
4kN
Ex=10.392kN
E
∑Fx = 0 ⇒ Ex – 6.93 – 4 cos30 = 0
⇒ Ex = 10.392 kN
∑Fy = 0 ⇒ Ey – 4 sin30 = 0
⇒ Ey = 2 kN
Ey=2kN
6.93kN
D
4kN
6.93kN
C
4kN
PROBLEM SOLVING: Example # 5
For the given loading, determine the zero-force
members in the Pratt roof truss. Explain your
answers using appropriate joint free-body diagrams.
PROBLEM SOLVING: Example # 5
FBC
B
FBK
FBA
F.B.D. of Joint B
Since, there is no external force or
reaction at joint B and FBA and FBC are
collinear, FBK = 0 (according to Rule
II)
FLB
FKC
FLA
L
FLK
FKL
K
FKS
F.B.D. of Joint L
F.B.D. of Joint K
Since, there is no reaction or external
force at joint L and FLA and FLK are Since, there is no external force or
collinear, FLB = 0 (according to Rule II) reaction at joint K, and FKL and FKS
are collinear, FKC = 0 (according to
Rule II)
PROBLEM SOLVING: Example # 5
Therefore, the zero-force members of
the given of truss are as follows:
LB, BK, KC, HF, FI, IE and EJ
Ans.
After
removing
the
zero-force
members, the truss diagram is shown
below:
Similarly, considering joints H, F, I,
and E,
FHF = 0
FFI = 0
FIE = 0
FEJ = 0
Multiple Choice Problems
1. The zero-force members of the truss as shown in
the following figure are
(a) EA and ED
(c) CA and CB
(b) DA and CA
(d) BA and BC
Ans: (b)
Feedback:
The members DA and CA are the zero-force
members by applying Rule II at joints D and C
Multiple Choice Problems
2. The magnitude and sense of the force in member
CE of the truss as shown in the following figure are
(a) 0
(c) 5 kN (C)
Ans: (c)
Feedback:
∑F
x
= 0 (at joint C)
⇒ 5 + FCE = 0
⇒ FCE = −5 kN = 5 kN (C)
(b) 5 kN (T)
(d) none of these