tic equations (compatibility) and Hooke’s law) are coupled equa- tions.
Let us consider the symmetrical truss shown in Fig. 1.16a. It is stress-free before the load is applied. The axial rigiditiesEA1, EA2, EA3 =EA1 are given; the forces in the members are un- known. The system is statically indeterminate to the first degree (the decomposition of a force into three directions cannot be done uniquely in a coplanar problem, see Volume 1, Section 2.2). The two equilibrium conditions applied to the free-body diagram of pinK (Fig. 1.16b) yield
→: −S1sinα+S3sinα= 0 → S1=S3,
↑: S1cosα+S2+S3cosα−F = 0→ S1=S3=F−S2
2 cosα . (a)
The elongations of the bars are given by
∆l1= ∆l3= S1l1
EA1
, ∆l2= S2l EA2
. (b)
To derive the compatibility condition we sketch a displacement diagram (Fig. 1.16c) from which we find
a b c
= +
d
"1"− System
"0"− System
000000 000000 111111 111111 00000
00000 11111 11111
00 00 11 11 000000 111111 0000000
1111111
1 2 3
α α
α
1 2 3
1 3
α
α
K
K S1 S3=S1
F
∆l1
∆l3
∆l2
K′ S2
l
F
F X
X
K
F
Fig. 1.16
1.6 Statically Indeterminate Systems of Bars 35
∆l1= ∆l2cosα . (c)
With (a), (b) andl1=l/cosαwe obtain from (c) (F−S2)l
2EA1cos2α = S2l EA2cosα which leads to
S2= F
1 + 2EA1
EA2
cos3α .
The remaining two forces in the bars follow from (a):
S1=S3=
EA1
EA2
cos2α 1 + 2EA1
EA2
cos3α F .
Note that now the vertical displacementv of pinKcan also be written down:
v= ∆l2= S2l EA2 =
F l EA2
1 + 2EA1
EA2
cos3α .
The problem may also be solved using the method of super- position. In a first step we remove bar 2 to obtain a statically determinate system, the “0“-system. It consists of the two bars 1 and 3 and it is subjected to the given force F (Fig. 1.16d).
The forcesS1(0)andS3(0)in these bars follow from the equilibrium conditions as
S1(0)=S3(0)= F 2 cosα.
The corresponding elongations are obtained withl1=l/cosα:
∆l(0)1 = ∆l(0)3 =S1(0)l1
EA1 = F l
2EA1cos2α. (d)
In a second step we consider the statically determinate system under the action of an unknown force X (“1“-system, see also Fig. 1.16d). Note that this force acts in the opposite direction on bar 2 (actio = reactio). Now we get
S1(1) =S(1)3 =− X
2 cosα, S2(1)=X ,
∆l1(1)= ∆l(1)3 =− Xl
2 EA1cos2α, ∆l(1)2 = Xl EA2
.
(e)
The total elongation of the bars is obtained through superposition of the systems “0“ and “1“:
∆l1= ∆l3= ∆l1(0)+ ∆l(1)1 , ∆l2= ∆l2(1). (f) The compatibility condition (c) is again taken from the displace- ment diagram (Fig. 1.16c). It leads with (d) - (f) to the unknown forceX =S2(1)=S2:
F l
2 EA1cos2α− X l
2EA1cos2α= X l EA2
cosα
→X =S2= F 1 + 2EA1
EA2
cos3α .
The forcesS1 andS3 follow from superposition:
S1=S3=S1(0)+S1(1)=
EA1
EA2cos2 α 1 + 2EA1
EA2
cos3α F .
A system of bars is statically indeterminate of degree nif the number of the unknowns exceeds the number of the equilibrium conditions by n. In order to determine the forces in the bars of such a system,ncompatibility conditions are needed in addition to the equilibrium conditions. Solving this system of equations yields the unknown forces in the bars.
A statically indeterminate system of degreencan also be solved with the method of superposition. Then n bars are removed in order to obtain a statically determinate system. The action of the
1.6 Statically Indeterminate Systems of Bars 37 bars which are removed is replaced by the action of the static redundants Si = Xi. Next n+ 1 different auxiliary systems are considered. The given load acts in the “0“-system, whereas the
“i“-system (i = 1,2, ..., n) is subjected only to the force Xi. In each of the statically determinate auxiliary problems the forces in the bars and thus the elongations can be calculated. Applying the ncompatibility conditions yields a system of equations for then unknown forcesXi. The forces in the other bars can subsequently be determined through superposition.
Example 1.7 A rigid beam (weight negligible) is suspended from E1.7 three vertical bars (axial rigidityEA) as shown in Fig. 1.17a.
Determine the forces in the originally stress-free bars if a) the beam is subjected to a forceF (∆T = 0),
b) the temperature of bar 1 is changed by ∆T (F = 0).
a b c
000000000 111111111
2
1 F 3 l
∆T
∆l1 ∆l2 ∆l3
A
S1 F S2 S3
a/2a/2 a Fig. 1.17
Solution The system is statically indeterminate to the first degree:
there are only two equilibrium conditions for the three unknown forcesSj (Fig. 1.17b). a) If the structure is subjected to forceF the equilibrium conditions are
↑ : S1+S2+S3−F = 0,
A: −a
2 F+a S2+ 2a S3= 0.
(a)
The elongations of the bars are given by (∆T = 0)
∆l1=S1l
EA, ∆l2=S2l
EA, ∆l3= S3l
EA. (b)
We sketch a displacement diagram (Fig. 1.17c) and find the com- patibility condition
∆l2= ∆l1+ ∆l3
2 . (c)
Now we have six equations for the three forcesSj and the three elongations ∆lj . Solving for the forces yields
S1= 7
12F , S2= 1
3F , S3= 1 12F .
b) If bar 1 is heated (F = 0), the equilibrium conditions are
↑ : S1+S2+S3= 0,
A : aS2+ 2aS3= 0,
(a′)
and the elongations are given by
∆l1= S1l
EA + αT∆T l , ∆l2= S2l
EA, ∆l3=S3l
EA. (b′) The compatibility condition (c) is still valid. Solving (a′), (b′) and (c) yields
S1=S3=−1
6EA αT∆T , S2=1
3EA αT∆T .
E1.8 Example 1.8 To assemble the truss in Fig. 1.18a, the free end of bar 3 (lengthl−δ,δ≪l) has to be connected with pinC.
a) Determine the necessary forceF acting at pinC (Fig. 1.18b).
b) Calculate the forces in the bars after the truss has been assem- bled and forceF has been removed.
a
0000000 0000000 0000000 0000000 0000000
b c1111111 1111111 1111111 1111111 1111111
0000000 0000000 0000000 0000000 0000000 1111111 1111111 1111111 1111111 1111111
1 2
3 δ
1
δ
2 3
l l
C F
C∗ l
v∗
∆l3
Fig. 1.18
1.6 Statically Indeterminate Systems of Bars 39 Solution a) The force F causes a displacement of pin C. The horizontal componentvof this displacement has to be equal toδ to allow assembly. The required force follows with α= 45◦ from (1.21):
v= F l EA
1 +√
√ 2/4
2/4 =δ → F= EA δ
(2√
2 + 1)l.
b) The force F is removed after the truss has been assembled.
Then pin C undergoes another displacement. Since now a force S3 in bar 3 is generated, pin C does not return to its original position: it is displaced to positionC∗ (Fig. 1.18c). The distance between pointsC andC∗ is given by
v∗=S3l EA
1 +√
√ 2/4 2/4 . The compatibility condition
v∗+ ∆l3=δ
can be taken from Fig. 1.18c. With the elongation
∆l3=S3(l−δ) EA ≈ S3l
EA of bar 3 we reach
S3l EA
1 +√
√ 2/4
2/4 +S3l
EA =δ → S3= EA δ 2(√
2 + 1)l. The other two forces follow from the equilibrium condition at pin C:
S1=√
2 S3, S2=−S3.