PART I
SOLID MECHANICS
1
Stability and Equilibrium
of Plane Frames
A pinjointed frame is a structure made up of slender (cross-sectional dimensions quite small
compared to length) members pin connected at ends and is capable of taking loads at joints.
Frames are used as roof trusses to support sloping roofs and as bridge trusses to support deck.
In many machines, steel frames are used. Transmission tower is another example of frame. In
the case of wooden frames, the ends are connected by making suitable wooden joints or by
nailing/bolting whereas steel frames are by riveting or by welding.
A frame in which all the members lie in a single plane, is called a plane frame. In such
a frame, loads act along the plane of the frame. Roof trusses and bridge trusses are the examples
of plane frames. If all the members of a frame do not lie in a single plane, then the frame is
called a space frame. Tripod and transmission towers are examples of space frame. In this
chapter, the analysis of only plane frame is considered.
1.1
PERFECT, DEFICIENT AND REDUNDANT FRAMES
A pinjointed frame which has got just sufficient number of members to
2
resist the loads without undergoing appreciable deformation in shape,
is called a prefect frame. Triangular frame is the simplest perfect
1
2
frame and it has three joints and three members (Fig. 1.1). Perfect
frames with four and five joints are shown in Fig. 1.2 and 1.3
1
3
3
respectively.
It may be observed that to increase joint in a perfect frame, two
Figure 1.1
more members are required. Hence, the following expression may be
written as the relationship between number of joints j, and the number of member m, in a
perfect frame.
4
4
2
m = 2j – 3
(1.1)
However, the above equation gives only a necessary, but
1
3
not a sufficient condition of a perfect frame. For example, the
5
two frames shown in Fig. 1.4 (a) and (b) have the same number
1
3
2
of members and joints. The frame shown in Fig. 1.4(a) is perfect
whereas the one shown in Fig. 1.4 (b) is not capable of retaining
Figure 1.2
its shape if loaded at the joint marked 6. Therefore, the only
3
4
Solid and Fluid Mechanics
necessary and sufficient condition of a perfect frame is that it should retain its shape when
load is applied at any joint in any direction.
4
2
P
4
2
4
6
2
4
6
1
6
2
1
3
5
3
5
7
1
3
(a)
Figure 1.3
5
1
3
(b)
5
Figure 1.4
A frame is said to be deficient if the number of members in
it are less than that required for a perfect frame. Such frames cannot
retain their shape when loaded. A deficient frame is shown in
Fig. 1.5.
A frame is said to be redundant if the number of members
Figure 1.5
in it, are more than that required in a perfect frame. Such frames
cannot be analysed by making use of the equations of equilibrium alone. Thus, a redundant
frame is statically indeterminate. Each extra member adds one degree
of indeterminancy. For the analysis of such members, the consistency
of determinations is to be considered. The truss shown in the Fig. 1.6
is a typical redundant truss. In this truss, one diagonal member in
each panel is extra. Hence, it is a two-degree redundant frame.
Figure 1.6
In this chapter, only the analysis of perfect frames is considered.
1.2
ASSUMPTIONS
In the theory that is going to be developed in this chapter, the following assumptions are
made:
(1) The ends of the members are pin-connected (hinged);
(2) The loads act only at the joints;
(3) Self-weights of the members are negligible;
(4) Cross-sections of the members are uniform;
If at all the cross-section varies, the centre of gravity of the section is assumed to be
located along the same longitudinal line.
In reality, the members are connected by bolting, riveting or by welding. No special care
is taken to ensure perfect pin-connections. However, experiments have shown that assuming
pin-connected ends is quite satisfactory since the members used are slender.
In most of the frames, the loads act at the joints. Even if a load is not acting at a joint, it
can be replaced by its reaction at the joint and a local bending effect on the member. The frame
may be analysed for the joint loads and the local bending effect on the member superposed in
the design of that member.
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In most of the trusses, the self-weight is really small compared to the loads they carry.
Hence, self-weight of the members may be neglected.
It is the duty of construction engineer to see that the centroid of all cross-sections lie
along a single axis so that the member is held in equilibrium by the two forces acting at its ends.
Because of the assumption of pin-connected ends, it is more appropriate to call the theory
that is going to be developed in this chapter as stability and equilibrium of pin-connected
plane trusses. Analysis of rigid frames is not covered in this book.
1.3
NATURE OF FORCES IN MEMBERS
The members of a truss are subjected to either tensile or compressive forces. A typical truss
ABCDE loaded at joint E is shown in Fig. 1.7(a). The member BC is subjected to compressive
force C as shown in Fig. 1.7(b). Effect of this force on the joint B (or C) is equal and opposite to
the force C as shown in Fig. 1.7(b).
C
B
C
C
C
T
A
C
T
D
E
T
Figure 1.7(a)
T
Figure 1.7(b)
The member AE is subjected to tensile force T. Its effect on the joints A and E are as
shown in Fig. 1.7(b). In the analysis of frame, we mark
Compression
the forces on the joints, instead of the forces in the
members as shown in Fig. 1.7(c). It may be noted that
compressive force in a member is represented in a
figure by two arrows going away from each other and
Tension
a tensile force by two arrows coming towards each
Figure 1.7(c)
other. In other words, if the arrows are towards joint
member force is compressive and if it is away from joints the member force is tensile. This is
quite logical considering the fact that the markings on the members represent the internal
reactive forces developed which are opposite in direction to the applied forces.
1.4
METHODS OF ANALYSIS
The following three methods are available for the analysis of pin-connected frames:
(1) Method of joints
(2) Method of section
(3) Graphical method.
The first two are analytical methods, only method of joints is dealt in this chapter.
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Chapter 1
5
Stability and Equilibrium of Plane Frames
6
Solid and Fluid Mechanics
1.5
METHOD OF JOINTS
At each joint, the forces in the members meeting and the loads acting, if any, constitute a
system of concurrent forces. Hence, two independent equations of equilibrium can be formed
at each joint. First, a joint is selected where there are only two unknown forces. Many times
such a joint can be identified only after finding the reaction at the support by considering the
equilibrium of the entire frame. Then making use of the two equations of equilibrium at that
joint, the two unknown forces are found. Then, the next joint is selected for analysis where
there are now only two unknown forces. Thus, the analysis proceeds from joint to joint to find
the forces in all the members.
It may be noted that if there are j number of joints, 2j number of equations can be
formed. There will be three reactions in a general determinate truss. The force in each member
is unknown. Hence, if there are m number of members, the total number of unknowns will be
m + 3. A problem can be analysed if there are as many equations as there are unknowns.
Hence, a frame analysis problem is determinate if:
2j = m + 3
(1.2)
This equation is the same as Equation 4.1 which was derived on the consideration of
a perfect frame. Hence, a perfect frame is determinate. If m > 2j – 3, then the number of
unknowns is more than the number of equations. Hence, a redundant frame is indeterminate.
If m < 2j – 3, then the number of equations is more than the number of unknowns. Since
a set of solutions can satisfy such equations, it shows instability of the structure. Hence, a
deficient frame is not stable.
The method of joints is illustrated with the examples 1.1 to 1.5.
Example 1.1. Find the forces in all the members of the truss shown
in Fig. 1.8(a). Tabulate the results.
Solution.
Step 1: Determine the inclinations of all inclined members. In
this case,
A
B
3m
q
D
q
C
3
E
3m
3m
=1
3
40 kN
40 kN
∴
θ = 45°
Figure 1.8(a)
Step 2: Look for a joint at which there are only two unknowns. If
such a joint is not available, determine the reactions at the
supports, and then at the supports the unknowns may reduce to only two.
Now at joints C, there are only two unknowns, i.e., forces in members CB and CD, say FCB and
FCD.
FCB
Note:
Usually in cantilever type frames, we find such joints without the
need to find reactions.
45°
C
Step 3: Now there are two equations of equilibrium for the forces
FCD
meeting at the joints and two unknown forces. Hence, the unknown
forces can be determined.
At joint C [Ref: Fig. 1.8(b)] ΣV = 0 condition shows that the force FCB
40 kN
should act away from the joint C so that its vertical component
Figure 1.8(b)
balances the vertical downward load at C.
tan θ =
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7
Stability and Equilibrium of Plane Frames
∴
FCB = 40
Chapter 1
FCB sin 45° = 40
kN
Now ΣH = 0 indicates that FCD should act towards C.
FCD – FCB cos 45° = 0
FCD = FCB cos 45° = 40 2 ×
1
2
= 40 kN
Note:
If the assumed direction of unknown forces is opposite, the value will be negative. Then reverse
the direction and proceed.
Step 4: On the diagram of the truss, mark arrow on the member near the joint analysed to indicate
the force on the joint. At the other end, mark the arrow in the opposite direction.
In the present case, near the joint C, the arrows are marked on the members CB and CD to
indicate forces FCB and FCD directions as found in the analysis of joint C. Then opposite directions
are marked in the members CB and CD near joints B and D, respectively.
Step 5: Look for the next joint where there are only two unknown forces
and analyse that joint.
In this case, there are only two unknown forces at joint D as shown in
Fig. 1.8(c).
FDB
FDE D
40 kN
ΣV = 0
FDB = 40 kN
ΣH = 0
40 kN
2
Figure 1.8(c)
FDE = 40 kN
Step 6: Repeat steps 4 and 5 till forces in all the members are found.
In the present case, after marking the forces in the members DB and DE, we find that analysis of
joint B can be taken up. Referring to Fig. 1.8(d).
FBA
ΣV = 0
FBE sin 45° – 40 – 40
∴
B
45°
× sin 45° = 0
45°
FBE
FBE = 80 2 kN
40 2
40 kN
ΣH = 0
Figure 1.8(d)
FBA – FBE cos 45° – 40 2 × cos 45° = 0
FBA = 80 2 ×
∴
1
2
+ 40
×
1
2
FBA = 120 kN
The directions of these forces are marked on the diagram. Now the analysis
is complete since the forces in all the members are determined.
Step 7: Determine the nature of forces in each member and tabulate the
results. Note that if the arrow marks on a member are towards joints,
then the member is in tension and if the arrow marks are away from
joints, the member is in compression [Ref. Fig. 1.8(e)] as discussed in 1.3.
In this case,
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Tension
Compression
Figure 1.8(e)
8
Solid and Fluid Mechanics
Member
Magnitude of force in kN
Nature
AB
120
Tension
BC
CD
40
40
Tension
Compression
DE
40
Compression
BE
BD
80 2
40
Compression
Tension
Example 1.2. Determine the forces in all the members of the truss shown in Fig. 1.9(a) and indicate
the magnitude and nature of forces on the diagram of the truss.
40 kN
50 kN
All inclined members are at 60° to horizontal and length of
each member is 2 m.
B
Solution.
C
Now, we cannot find a joint with only two unknown forces
without finding reactions.
60°
A
Consider the equilibrium of the entire frame.
2m
RD × 4 – 40 × 1 – 60 × 2 – 50 × 3 = 0
RD = 77.5 kN
60°
D
RA
2m
RD
60 kN
Figure 1.9(a)
ΣH=0
∴
60°
E
ΣMA = 0
∴
60°
2
HA = 0
∴ Reaction at A is vertical only
ΣV = 0
RA + RD = 40 + 60 + 50
∴
FAB
RA = 72.5 kN, since RD = 77.5 kN
Joint A:
ΣV = 0
60°
A
FAE
FAB sin 60° = RA = 72.5
FAB = 83.716 kN (Comp)
ΣH = 0
RA
Figure 1.9(b)
FAE = 83.716 cos 60° = 0
FAE = 41.858 kN (Tension)
FDC
Joint D:
ΣV = 0
60°
FDC sin 60° = RD = 77.5
∴
FDC = 89.489 kN (Comp)
ΣH = 0
FDE = 89.489 cos 60° = 0
∴
FDE = 44.745 kN (Tension)
Joint B:
ΣV = 0
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D
FDE
RD
Figure 1.9(c)
9
Stability and Equilibrium of Plane Frames
∴
83.716 sin 60 ° − 40
= 37.528 (Tension)
sin 60 °
ΣH = 0
FBE =
FBC = FAB cos 60° – FBE cos 60° = 0
FBC = (83.716 + 37.527) × 0.5
FBC
B
60°
60°
60°
FBE
FAB
FBC = 60.622 kN (Comp)
Figure 1.9(d)
Joint C:
ΣV = 0
50 kN
FCE sin 60° + 50 – FDC sin 60° = 0
FCE =
89.489 sin 60 ° − 50 °
= 31.754 kN (Tension)
sin 60 °
Now the forces in all the members are known. If joint E is analysed, it will
give the check for the analysis. The results are shown on the diagram of
the truss in Fig. 1.9(f).
40 kN
754
31.
489
716
89.
528
83.
Compression
C
Tension
37.
A
D
E 44.745
41.858
60 kN
Figure 1.9(f)
Example 1.3. Analyse the truss shown in Fig. 1.10(a).
All inclined members have the same inclination to horizontal.
4
3
θ = 53.13°
tan θ =
∴
C
4m
B
q
D
q
4m
HA
A
q
3m
q
F
Figure 1.10(a)
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E
3m
VA
C
60°
60°
FCE
FDC
Figure 1.9(e)
50 kN
60.622
B
FBC
20 kN
Chapter 1
40 kN
FBE sin 60° – FAB sin 60° + 40 = 0
10
Solid and Fluid Mechanics
As soon as a joint is analysed, the forces on the joint are marked
on members [Fig. 1.10(b)]
kN
RC
20 kN
25
Joint E:
ΣV=0
q 15 kN
20 kN
FED = 25 kN (Tension)
ΣH=0
FEF = 25 × cos 53.13 = 15 kN (Comp)
O
q
HA
FEF – FED cos θ = 0
O
q
15 kN
15 kN
VA
20 kN
At this stage as, no other joint is having only two unknowns,
Figure 1.10(b)
no further progress is possible. Let us find the reactions at the
supports considering the whole structure. Let the reaction be as shown in Fig. 1.10(b).
ΣMA = 0
FED
RC × 8 – 20 × 6 = 0
q
RC = 15 kN
VA = 20 kN
20 kN
ΣH = 0
Figure 1.10(c)
HA = RC = 15 kN
Joint A:
ΣV = 0
C
FEF
ΣV = 0
15
cos 53.13
FAB
A
FAB – VA = 0
HA
FAB = 20 kN (Comp)
FAF
VA
ΣH = 0
FAF – HA = 0
Figure 1.10(d)
FAF = 15 kN (Comp)
Joint C:
ΣH = 0
FCB × cos 53.13 – RC = 0
FCB =
= 25 kN (Comp)
ΣV = 0
FCD = FCB sin θ
= 25 × sin 53.13
= 20 kN (Tension)
Joint B:
ΣV = 0
FBF × sin 53.13 – FBC × sin 53.13 + FAB = 0
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kN
∴
25
FED × sin 53.13 – 20 = 0
11
Stability and Equilibrium of Plane Frames
∴
FBF = 0
ΣH = 0
q
B
q
FBD – 25 × cos 53.13 = 0
FBD = 15 kN (Tension)
FBF
Figure 1.10(e)
ΣV = 0
FBF = 0
FFD = 0 (since FBF = 0)
Note: When three members are meeting at an unloaded joint and out of
them two are collinear, then the force in third member will be zero.
Such situations are illustrated in Fig. 1.10 (g) and (h).
A
C
FBD
FAB
Joint F:
Chapter 1
FBC
FBF × sin 53.13 = 25 × sin 53.13 – 20 = 0
FFD = 0
q
FAF
FFE
Figure 1.10(f)
A
B
D
C
B
D
Figure 1.10(g) (h)
Example 1.4. Find the forces in all the members of the truss shown in Fig. 1.11 (a).
tan θ1 =
∴
FG 8 × 1 IJ
H 3 2K
4
6
θ1 = 33.69°
2m
A
q1
12 kN
20 kN
3m
2m
2m
B
q1
D
F
q2
H
q3
C
θ2 = tan–1
θ3 = tan–1
= 53.13°
4m
E
FG 4 IJ = 53.13°
H 3K
G
Figure 1.11(a)
Joint-by-joint analysis is carried out as given below and the joint forces are marked in Fig. 1.11(b).
Then nature of the force in the members is determined.
12 kN
A 15 kN B 15 kN D 15 kN
18
.
kN028
RA
0
C
0
0
18
.
kN028
0
20 kN
F 15 kN
12
H
25 kN
E1
8.
kN028
G
RG
Figure 1.11(b)
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Compression
Tension
12
Solid and Fluid Mechanics
Joint H:
ΣV = 0
FHG sin θ3 = 20
FHG sin 53.13 = 20 = 25 kN (Comp)
ΣH = 0
FHF – FHG cos θ2 = 0
FHF = 25 × cos 53.13 = 15 kN (Tension)
ΣMG = 0
Now
RA × 6 – 20 × 3 = 0
RA = 10 kN Downward
ΣV = 0
RG – 10 – 12 – 20 = 0
RG = 42 kN
Joint A:
ΣV = 0
FAC sin θ1 – 10 = 0
FAC = 18.028 kN (Comp)
ΣH = 0
FAB – FAC cos θ1 = 0
FAB = 15 kN (Tension)
Joint B:
ΣV = 0
FBC = 0
ΣH = 0
FBD = FBA = 15 kN (Tension)
Joint C:
Σ Forces normal to AC = 0, gives
FCD = 0 since FBC = 0
Σ Forces parallel to CE = 0
FCE = FCA = 18.028 (Comp)
Joint D:
ΣV = 0
FDE = 0
ΣH = 0
FDF = FDB = 15kN (Tension)
Joint E:
Σ Forces normal to CG = 0, gives
FEF = 0 and
Σ Forces in the direction of CG = 0 gives
FEG = FCE = 18.028 kN (Comp)
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13
Stability and Equilibrium of Plane Frames
Chapter 1
Joint F:
ΣV = 0
FFG – 12 = 0
FFG = 12 kN (Comp)
Example. 1.5. Analyse the truss shown in Fig. 1.12 (a). All the members are of 3 m length.
Solution.
Since all members are 3 m long, all triangles are
equilateral and hence all inclined members are at
60° to horizontal. Joint-by-joint analysis is carried
out and the forces are represented in Fig. 1.12(b).
Then nature of the force is determined.
Joint G:
40 kN
30 kN
B
D
60°
A
60°
ΣV = 0
FGF sin 60° = 20
FGF = 23.094 kN
(Tension)
ΣH = 0
FGE – FGF cos 60° = 0
FGE = 11.547 kN (Comp)
60°
C
60°
3×3=9m
F
60°
E
Figure 1.12(a)
10 kN
60°
G
20 kN
Joint F:
ΣV = 0
FFE sin 60° – FGF sin 60° = 0
FFE = FGF = 23.094 kN (Comp)
ΣH = 0
FFD + 10 – FGF cos 60° – FFE cos 60° = 0
FFD = 13.094 kN (Tension)
Now, without finding reaction we cannot proceed. Hence, consider equilibrium of the entire truss.
30 kN
10 kN
Compression
RE
4
11.547
E
.09
9.4
33
7
F
23
47
1.0566
C
VA
.07
8.3771
D 13.094
44
7
A
33
HA
9.4
38
.75
43
B 13.6603
23
.09
4
40 kN
G
20 kN
Tension
Figure 1.12(b)
∴
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Σ MA = 0
RE × 6 + 10 × 3 × sin 60° – 40 × 1.5 – 30 × 4.5 – 20 × 9 = 0
RE = 58.170 kN
ΣV = 0
VA = 40 + 30 + 20 – RE = 31.830 kN
ΣH=0
HA = 10 kN
14
Solid and Fluid Mechanics
Joint A:
ΣV = 0
FAB sin 60° – 31.830 = 0
FAB = 36.754 kN (Comp)
ΣH = 0
FAC – FAB cos 60° + 10 = 0
FAC = 8.377 kN (Tension)
Joint B:
ΣV = 0
FBC sin 60° + FAB sin 60° – 40 = 0
FBC = 9.434 kN (Comp)
ΣH = 0
FBD + FBC cos 60° – FBA cos 60° = 0
FBD = 13.660 kN (Comp)
Joint C:
ΣV = 0
FCD sin 60° – FBC sin 60° = 0
FCD = FBC = 9.434 kN (Tension)
ΣH = 0
FCE + FAC – FCD cos 60° – FBC cos 60° = 0
FCE = 2 × 9.434 ×
1
– 8.377 = 1.057 kN (Comp)
2
Joint D:
ΣV = 0
FDE sin 60° – FCD sin 60° – 30 = 0
FDE = 44.075 kN (Comp)
IMPORTANT DEFINITIONS AND FORMULAE
1. A pinjointed frame which has got just sufficient number of members to resist the loads without
undergoing appreciable deformation in shape, is called a perfect frame.
2. A frame is said to be deficient if the number of members in it, are less than that required for a
perfect frame. Such frames cannot retain their shape when loaded.
3. A frame is said to be redundant if the number of members in it, are more than that required for
a perfect frame. They are statically indeterminate.
Equations:
In a perfect frame,
m = 2j – 3
where
m = number of members
j = number of joints.
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PROBLEMS FOR EXERCISE
1.1. to 1.17. Determine the forces in all the members of the frames shown in Fig. 1.13 to 1.29. Indicate
the nature of the forces also (Tension as + ve and compression as – ve).
1.1.
20 kN
30 kN
6m
6m
A
C
B
4m
E
F
D
3m
6m
Figure 1.13 (Prob. 1.1)
[Ans. FAB = + 67.5 kN; FBC = + 15 kN; FCD = – 25 kN; FDE = – 30 kN; FEF = – 105 kN; FAE =
+ 62.5 kN; FBE = – 62.5 kN; FBD = + 25 kN]
1.2.
A
20 kN
B
4m
C
D
G
F
3m
E
3m
3m
20 kN
Figure 1.14 (Prob. 1.2)
[Ans. FAB = + 82.074 kN; FBC = + 73.866 kN; FCD = 49.244 kN; FDE = – 45 kN; FEF = – 45 kN; FFG
= – 67.5 kN; FBG = – 10.0 kN; FFC = + 24.622 kN; FCE = 0; FBF = 10 kN]
1.3.
A
C
E
30°
G
F
D
3m
100 kN
3m
3m
Figure 1.15 (Prob. 1.3)
[Ans. FAC = FCE = FEG = + 193.185 kN; FBD = FDE = FFG = – 193.185 kN; All others are zero
members]
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Chapter 1
15
Stability and Equilibrium of Plane Frames
16
Solid and Fluid Mechanics
1.4.
E
200 kN
2m
C
A
2m
D
B
4m
4m
Figure 1.16 (Prob. 1.4)
[Ans. FEC = + 447.214 kN; FCA = + 400 kN; FAB = – 447.214 kN; FBD = – 400 kN; FCD = 0;
FCB = – 200 kN]
1.5.
F
D
B
3m
90°
3m
3m
3m
3m
3m
3m
E
A
C
10 kN
5 kN
Figure 1.17 (Prob. 1.5)
[Ans. FDB = FBA = + 5.773 kN; FBC = FDE = – 5.774 kN; FAC = – 2.887 kN; FCE = – 14.434 kN;
FDC = + 17.320 kN; FDF = + 20.0 kN]
1.6.
160 kN
160 kN
30 kN
B
A
4m
20 kN
D
C
4m
F
E
3m
Figure 1.18 (Prob. 1.6)
[Ans. FAB = – 30 kN; FAC = – 160 kN; FBC = + 50 kN; FBD = – 200 kN; FCD = – 50 kN; FCE =
– 120 kN; FDF = – 266.67 kN; FDE = + 83.33 kN]
P-2\D:\N-fluid\ANA1-1.pm5
1.7.
100 kN
2m
200 kN
2m
A
C
B
2m
E
D
F
2m
G
H
Figure 1.19 (Prob. 1.7)
[Ans. FAB = – 200 kN; FAC = – 100 kN; FBC = FCF = 0; FBD = 141.42 kN; FBF = – 141.42 kN; FDE =
– 100 kN; FDG = 0; FEF = + 100 kN; FEH = – 141.42 kN; FEG = + 141.42 kN; FGH = + 100 kN]
A
1.8.
B
2m
1 kN
2 kN
C
D
E
F
G
H
2m
2m
2m
2m
Figure 1.20 (Prob. 1.8)
[Ans. FBD = – 2.828 kN; FBA = + 3 kN; FAC = + 4.243 kN; FAD = – 3 kN; FDC = – 2 kN; FDF = – 5 kN;
FCF = – 1.414 kN; FCE = + 6 kN; FFE = + 1 kN; FFH = – 4 kN; FEH = – 1.414 kN; FEG = + 5 kN;
FGH = + 1 kN]
100 kN
1.9.
A
B
100 kN
D
5m
C
F
E
5m
5m
5m
Figure 1.21 (Prob. 1.9)
[Ans. FAC = – 223.6 kN; FAB = + 200 kN; FBD = + 200 kN; FBC = – 100 kN; FCD = + 111.8 kN;
FCE = – 335.4 kN; FDE = + 35.0 kN; FDF = 424.3 kN; FEF = – 300 kN]
P-2\D:\N-fluid\ANA1-1.pm5
Chapter 1
17
Stability and Equilibrium of Plane Frames
18
Solid and Fluid Mechanics
D
1.10.
B
2m
F
E
A
C
2m
10 kN
G
5 kN
H
2m
2m
2m
2m
Figure 1.22 (Prob. 1.10)
[Ans. FAB = + 7.07 kN; FAC = – 5 kN; FBC = – 5 kN; FBD = + 5 kN; FCD = + 21.2 kN; FCE = – 20 kN;
FDE = – 15 kN; FDF = + 28.28 kN; FEH = – 15 kN; FEF = – 20 kN; FFG = + 42.43 kN; FFH =
+ 14.14 kN]
1.11.
40 kN
30 kN
45°
B
45°
D
2m
A
E
C
2m
2m
Figure 1.23 (Prob. 1.11)
[Ans. FAB = – 15 kN; FAC = + 16.97 kN; F BD = – 38.89 kN; F BC = – 17.68 kN; FCE = 0;
FCD = + 25 kN; FED = – 38.89 kN]
20 kN
1.12.
D
2m
20 kN
B
E
60°
2m
30°
20 kN
A
C
Figure 1.24 (Prob. 1.12)
[Ans. FAB = – 17.32 kN; FAC = + 5 kN; FBC = – 20 kN; FBD = – 17.32 kN; FCD = + 20 kN; FCE =
– 15 kN; FDE = – 30 kN]
20 kN
20 kN
10 kN
1.13.
A
D
B
20 kN
E
30°
30°
C
3m
10 kN
F
3m
Figure 1.25 (Prob. 1.13)
[Ans. FAB = 60 kN; FAC = + 51.96 kN; FBC = – 20 kN; FBD = – 40 kN; FCD = + 40 kN; symmetry]
P-2\D:\N-fluid\ANA1-1.pm5
E
1.14.
C
1m
F
2m
H
A
B
D
G
6 kN
6 kN
3 × 4 = 12 m
6 kN
Figure 1.26 (Prob. 1.14)
[Ans. FAC = – 16.22 kN; FAB = + 13.5 kN; FBC = + 6 kN; FBD = + 13.5 kN; FCD = – 1.80 kN;
FCE = – 12.65 kN; FDE = 8 kN]
1.15.
30 kN
B
15 kN
D
3m
C
A
2m
E
F
2m
2m
30 kN
Figure 1.27 (Prob. 1.15)
[Ans. FAB = + 36.06 kN; FAC = – 20 kN; FCB = – 48.75 kN; FCE = – 20 kN; FCD = – 7.5 kN;
FBE = + 22.53 kN; FDE = 18.75 kN; FDF = – 13.52 kN; FFE = – 7.5 kN]
1.16.
C
40 kN
B
D
60°
A
60° 60°
60° 60°
E
F
3m
3m
40 kN
Figure 1.28 (Prob. 1.16)
[Ans. FAB = 16.91 kN; FAF = + 31.55 kN; FBF = + 23.91 kN; FBD = – 23.91 kN; FBC = + 40 kN;
FCD = – 40 kN; FDE = – 63.1 kN; FDF = + 23.91 kN; FEF = + 31.55 kN]
P-2\D:\N-fluid\ANA1-1.pm5
Chapter 1
19
Stability and Equilibrium of Plane Frames
20
Solid and Fluid Mechanics
1.17.
E
30 kN
C
3m
1.5 m
F
30°
A
B
D
10 kN
G
3m
2m
2m
Figure 1.29 (Prob. 1.17)
[Ans. FAC = – 67.48 kN; FAB = + 53.99 kN; FBC = + 10 kN; FCD = – 8.33 kN; FCE = – 59.15 kN;
FEF = – 24.5 kN; FED = + 52.81 kN; FFD = + 47.21 kN; FFG = – 34.64 kN; FDG = + 47.32 kN]
P-2\D:\N-fluid\ANA1-1.pm5