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Re“ 2.12) 1.13) ,1 for em "iAMPLE. 203 a ' A bar AB of length L and uniform cross section is attached to
rigid supports at A and B before being loaded. What are the
stresses in portions AC and BC due to the application of a load
P at point C (Fig. 2.2661)? Drawing the freebody diagram of the bar (Fig. 2.2617).
we obtain the equilibrium equation RA + RB = P (2.14) Since this equation is not sufﬁcient to determine the two un
known reactions RA and R3, the problem is statically indeter
minate. However, the reactions may be determined if we observe
from the geometry that the total elongation 5 of the bar must
be zero. Denoting by 61 and 52, respectively, the elongations
of the portions AC and BC , we write 5=5l+52=0 or, expressing 51 and 82 in terms of the corresponding internal
forces PI and P2: 5—P‘L1+P2L3—0 715
_ AE AE _ (“‘ ) But we note from the free—body diagrams shown respectively
in parts b and c of Fig. 2.27 that P1 = RA and P2 = —R3. Car«
rying these values into (2.15), we write RALI ~ RBLZ = 0 (2.16) Equations (2.14) and (2.16) can be solved simultaneously for
RA and RB; we obtain RA = PIC/L and RB = PLl/L. The
desired stresses 0'l in AC and 02 in BC are obtained by
dividing, respectively, P1 = RA and P2 = —RB by the cross~
sectional area of the bar: PL2 PLl
= W. a? _ _
AL  AL (71 M Superposition Method. We observe that a structure is statically in—
determinate whenever it is held by more supports than are required to
maintain its equilibrium. This results in more unknown reactions than
available equilibrium equations. It is often found convenient to desig—
nate one of the reactions as redundant and to eliminate the correspon—
ding support. Since the stated conditions of the problem cannot be ar—
bitrarily changed, the redundant reaction must be maintained in the
solution. But it will be treated as an unknown load which, together with
the other loads, must produce deformations that are compatible with the
original constraints. The actual solution of the problem is carried out
by considering separately the deformations caused by the given loads and by the redundant reaction, and by adding—or superposing~the re—
sults obtainedt i‘The general conditions under which the combined effect of several loads can be obtained
in this way are discussed in Sec. 2.12. 1‘ ' ummemwwmiwwu SEC. 4.4 STATICALLY INDETERMINATE AXIALLY LOADED MEMBER 139 p, ,. ”= g,“
ﬁﬁ’ii “be “r The steel rod shown in Fig. 4—1251 has a diameter of 5 mm. It is
attached to the fixed wall at A, and before it is loaded there is a gap
between the wall at B’ and the rod of 1 mm. Determine the reactions
at A and B’ if the rod is subjected to an axial force of P = 20 kN as
shown. Neglect the size of the collar at C. Take Est = 200 GPa. As shown on the free—body diagram, Fig. 4—12b, we
will assume that the force P is large enough to cause the rod’s end B to contact the wall at B’. The problem is statically indeterminate since FA there are two unknowns and only one equation of equilibrium.
’ Equilibrium of the rod requires 23 2F, = 0; —FA — FB + 20(103) N = 0
' " The compatibility condition for the rod is
53/A = 0.001 m This displacement can be expressed in terms of the unknown
reactions by using the load—displacement relationship, Eq. 4—2,
applied to segments AC and CB, Fig. 4~12c. Working in units of
newtons and meters, we have _ _ FALAC _ FBLCB
5B/A — In — AE AE
w m)
0001 m _ 710.0025 m)2[200(109) N/mz]
FB(O.8 m) 710.0025 m)2[200(109) N/mz] FA(O.4 m) — FB(O.8 m) = 3927.0 N ~ in (2)
Solving Eqs. 1 and 2 yields
FA = 16.6 kN FB = 3.40 kN Since the answer for F B is positive, indeed the end B contacts the wall
at B’ as originally assumed. On the other hand, if F B were a negative
quantity, the problem would be statically determinate, so that F B = 0
and F A = 20 kN. Ehi—Ehiiﬁi §~7 The three A36 steel bars shown in Fig. 414a are pin—connected to
a rigid member. If the applied load on the member is 15 kN, deter—
mine the force developed in each bar. Bars AB and EF each have a
cross—sectional area of 25 m2, and bar CD has a crosssectional area
of 15 m2. ESL? Eiiii‘é ifn‘igmz. The freebody diagram of the rigid member is shown
in Fig. 4—14b. This problem is statically indeterminate since there are
three unknowns and only two available equilibrium equations. These
equations are +1‘ 2Fy = 0; FA + FC + FE— 15 kN = 0 (1)
1+ ZMC = 0; —FA(O.4 m) + 15 kN(0.2 m) + FE(0.4 m) = 0 (2) if ‘umpmibiiiz}: Due to the displacements at the ends of each bar, line ACE shown in Fig. 4—14c will move to the inclined points A’C’E’.
From this position, the displacements of points A, C, and E can be
related by proportional triangles. Thus the compatibility equation for
these displacements is 5A_5E_ 5C—5E
0.8m “ 0.4m
1 1 5=—5 +——a
C 2A 2E Using the load—displacement relationship, Eq. 4—2, we have FCL 1[ PAL ] 1[ FEL'] = +
(15 mm2)Est 2 (25 mm2)Est 2
FC = 0.3FA ‘l‘ 0.3FE (25 mm2)Est Solving Eqs. 1—3 simultaneously yields FA = 9.52 kN
FC = 3.46 kN
FE = 2.02 kN i8 kN/m Probs. 4—64/4—65 *4—64. The horizontal beam is assumed to be rigid and suppons
the distributed load shown. Determine the vertical reactions at the
supports. Each support consists of a wooden post having a diame
ter of 120 mm and an unloaded (original) length of 1.40 m. Take EN = 12 GPn. _..__———._.—....— \ SC’OCA'
j \ \ \‘p w; L .2ng '
OCB’EK _ + AE ' .'.. .. w.aumnu.gmsu_.u+. .'... “Lungs a r—————12in—~—“I SAMPLE PROBLEM The rigid casting DEF is held by two gin—diameter bolts AD and CF against the
1.5—in.diameter rod BE. The pitch of each bolt thread is 0.]. in.. and after being
snugly ﬁtted. each nut is tightened one quarter of a turn. Knewing that for steel
E = 30,000 ksi and for aluminum E = 10.000 ksi. determine the normal stress in the rod. r'AIuminumf; 2" Statics. Considering the free body of the entire assembly, we note that the R . l \ reactions are statically indeterminate. However. We have by symmetry RA = R0,
R B and using statics we write '3
RF C El i SF = 0: —RA + RS _ RC = 0 or RB = 233 (l) Deformations. We shall use the method of superposition and consider the .
reaction RB as redundant. The connection at B is temporarily removed and, as
the nuts are tightened one quarter of a turn. the casting DEF and rod BE move to »
the left (Fig. l) a distance 51 =1(0.lin.) = 0.025 in. 1—— The reaction R3 is then applied to force the end B to move back to the right
through 8.2 (Fig. 2). Since the ﬁnal deﬂection 6 of B must be zero \Fig. 3), we have 7
82 = 51. Noting that the deﬂection 82 is the sum of the deformations of bolt AD and rod BE, we write ' R L R 1" . 8mm : 2 : L132 X 10—33;
‘ ﬁlm.) (30 x 1031611
R R 12‘ . 8m = if = , 3‘ m) = 0.679 x roe3123
‘ $1.5 in.)2(10 X 105 ksil abolt ‘l' Srod = ‘5: = ‘51
1.132 X 10—3111i + 0.679 X 10—333 = 0.025 in. (2)
Reactions. Substituting for 8,, from i into (2): V
1.132 X 10‘3RA + 0.679 iD‘TiRA) = 0.025 in.
RA = 10.04 kips RB = 234 = 2(10.04 laps) = 20.08 hips Stress in Rod ...
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 Spring '10
 Biegel
 Statics, Force, Trigraph, Bar association, available equilibrium equations

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