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Unformatted text preview: 1 50 PART 1 STATICS Exam Ie 69' ' “ I ‘ Truss analysis by method of sections. ‘ p
7 Given: Truss conﬁguration and loads as shown. Reactions are shown on, the sketch. _ To Find: Forces‘in the truss members. Solution: A summary sketch is prepared aside. It Will be entered at the end of the 1
example. The symmetry of the truss and loads is recognized. Joint H is identiﬁed as a transfer joint. The force in member HB is found
by inspection and entered on the summary sketch: FHB = 2 kips (tension) " Section aa is taken ﬁrst. The free body to the left of section a—a is shown
in the following sketch. ‘ ’ a ,08  . FA: Isl
A — oH ' to 2m ! Summing moments about B eliminates F AB: I ME = 0; 9(10)  FAH(5) :0
PM = 18 kips (tension)  Summing forces vertically eliminates F A”: ZFY = 0; 9 + FMS/11.13) = 0
FAB = —20.12 kips (compression) These forces are entered immediately on the summary sketch. 'lfa
)ra
iin ose
we
are
illy the 6.3 . CHAPTER 6 SIMPLE TRUSSES 133
W The point where two or more truss members intersect is called a panel point. All mem—
bers coming into a panel point are assumed to come into a hinge at that point, though
it is obvious in Figure 63 that no such hinge exists. The members are so long and
slender, however, that very little error is incurred in making this assumption. This point
is discussed more fully in later chapters dealing with steel and timber truss designs. A truss is said to be properly formed if it can be formed by beginning with a base
triangle such as those shown in Figure 6—4, then adding two members and their included
hinge until the ﬁnal conﬁguration is obtained. The hinge points in Figure 6—4 are shown
as open circles. ‘ q)EASE 1— us: mum muuvrs i" b) £ASE 11— in mm; or us:
suvrnnr AS rut rum  lRlAlltxlE ARE truss
muss MEMBER  MEMBERS Figure 64 Properly formed trusses. The trusses of Figure 6—4, for example, are formed by beginning with the base triangle
ABH,‘ then adding members HG arﬂ GB with their hinge at G, then adding BF and FG
with their hinge at F, then adding BC and CF with their hinge at C, and so on ‘until the
entire truss is formed. Such a truss is statically determinate and it will be a stable truss
under load. In all schematic diagrams of trusses in 'this text, a hinged truss point is shown as an
open circle such 'as those of Figure 64. If two members are shown crossing each other ,without such a hinge point, they are assumed to pass each other freely at that point. As an example of a truss that is not properly formed, consider the truss of Figure
65. Beginning with the base triangle ABN, the truss can be properly formed to the right
only as far as panel DELM, then it can no longer be properly formed. Or, beginning with
base triangle IHJ, the truss can’be properly formed to the left only as far as panel EFLK,
then it can no longer be properly formed. ‘ I Figure 65 Improperly formed truss. Examination of'the truss of Figure 65 indicates that the panel DELM has an extra _ member, While the panel EFLK lacks one member. So, while the total 'number’of mem bers may' be correct, and the total number of hinges may be correct, the truss is not
properly formed. (As one means to correct the conﬁguration of this particular. truss, the
eXtra member ME could simply be moved. to position LF.) Loads on Truss Members A truss member may only be loaded axially. Any transverse loading on one of these long
slender members would create bending in the member, which would severely limit its
capacity to take axial force. For maximum efﬁciency, a truss member may be loaded only
in tension or in compression, as shown in Figure 66. ‘ 1 3 4 PA RT 1 STATICS ' 6.4 Figure 66 Loads on truss members. In Figure 6—6, membersAC and IE have been removed as free bodies and shown sepa
rately. Note that the only load that can come to these members is at a hinge' point.
Consequently, the only load that can exist is direct tensiOn or compression. As noted in
earlier chapters, compressive forces “push” toward a hinge, and tension forces “pu ”
away from a hinge.   Since the members in a truss may only be loaded axially, all external loads on a truSS
can be introduced only at hinge points. Such control over the loading of a truss is shown
in Figure 6—7, where beams are used to transmit the loads from the bridge deck to the panel points of the truss. BRlDl'xl WUSS TYDHZAL BRlDGE TRUSS muss“ EMT.th ll
DEM“ 10 TR SSES DEW. SUPWRI BUM
TYPICAL SE CTlON Figure 67 Typical loading of a truss. Under certain circumstances some loads (such as a concrete deck) may be placed along
the length of a truss member, violating the rule that loads may only be introduced at panel
points. In such cases, the truss member must be designed to sustain both the axial loads
and the bending loads. Such practices introduce inefﬁciency into the truss design, but
such designs are sometimes the least costly alternative. External Reactions on Trusses External reactions on trusses are determined in much the same way as they are deter—
mined for beams. At times, however, their locations can produce some confusing conﬁgu
rations. Some examples Will illustrate thesolution for some typical truss reactions. Exam le 61 CHAPTER 6 SIMPLE TRUSSES 135 W Solution for truss reactions. Given: Truss with loads and reactions as shown. To Find: Magnitude and direction of the reactions. Solution: The entire truss istaken ash freesbody with all forces and reactions
shown. ' ‘ in,“ It is noted that there are three unknown reactions in this planar system; _
an equilibrium solution is therefore possible. Moments are summed about point A, eliminating all unknoWn's except
RCX: EMA = o; — 5(5) 4 10(20) — chuo') = 0
ch = 17.5 kips in direction shown Forces are now summed vertically, eliminating all unknowns except RAY: *ZFY=0;+RAY10=*0 . .
RAY = 10 kips in direction shown " '_ Moments are now summed about point C: _ g 0; R440)  RAan) 4 5(5) — 10(20) = 0. _ 2
: RAx = 22.5 kips in direction shown ' As a check, forces are summed horiiOntally: 2Fx=0;+RAx—5RCX=0
"22.5;5'—17.5=0
‘ O=0(OK) _ PA RT 1 STATlCS 134 6.4 10k 10k 1 Wk Figure 66 Loads on truss members. r In Figure 66, members AC and IE have been removed as free bodies and shown sepa
rately. Note that the only load that can come to these members is at a hinge point. '
Consequently, the only load that can exist is direct tension or compression. As noted in
earlier chapters, compressive forces “push” toward a hinge, and tension forces “pull”
away from a hinge. a  Since the members in a truss may only be loaded axially, all external loads on a truSs
can be introduced only at hinge points. Such control over the loading of a truss is shown
in Figure 6—7, where beams are used to transmit the loads from the bridge deck to the panel points of the truss. BRlDlil’ “illSS ', nrttsunun Bull _
TYDlCAl. SECTlON l
Figure 67 Typical loading oi a truss. Under certain circumstances some/loads (such as a concrete deck) may be placed along
the length of a truss member, violating the rule that loads may only be introduced at panel
points. In such cases, the truss member must be designed to sustain both the axial loads
and the bending loads. Such practices introduce inefﬁciency into the truss design, but
such designs are sometimes the least costly alternative. External Reactions on Trusses External reactions on trusses are determined in much the same way as they are deter—
mined for beams.,At times, however, their locations can produce some confusing conﬁgu—
rations. Some examples Will illustrate the solution for some typical truss reactions. CHAPTER 6 SIMPLE TRUSSES 135 Exam le 61
Solution for truss reactions.
Given: Truss with loads and reactions as shown. To Find: Magnitude and direction of the reactions. l El . f iﬂk
Solution: The entire truss is taken at a free body with all forces and reactions
shown. ‘
tin‘ ~ in“
4—8“ A
El 
Rm Wk
It is noted that there are three unknown reactions in planar system;
an equilibrium solution is therefore, possible. Moments are summed about point A, eliminating all unkn'oWn's except
Rex: EMA = 0;  5(5);r 10(20)  chao‘) = o _
Rex = 17.5 kips in direction shown Forces are now summed vertically, eliminating all unknowns except RAY: ZFY=O;+RAy~10=OI . . RAY = 10 kips in direction shown Moments are now summed about point C: ZMC = 0; RA;(4O)  RAXUO) + 5(5) ~ l0(20) = 0,—
RAx = 22.5 kips in directionshown A7 As a check, forces are summed horizontally: 22.55'17.5=0 4
‘ 0 = 0 (OK) 136 PAFiT1 STATICS  _ I It should be noted that the conﬁguration of the truss never entered the solution to l3); ample _61. Onlylthe locations and direCtions of the forces were used. The rigid body _ could haVe been missed 'or solid; insofar as external reactions are oncerned, it would have
" . made no difference. * r j Exam e_ 62 Reactions on __a truss. _' Vi ' ' " ' V Given: Truss with loads and reactions as shown. To Find: Magnitude and direction of the reactions. Solutioni The entire truss is taken as a free body, with all loads and reactions
Shown. Moments are sumined about H to ﬁnd RM}: EMH = 0; RAY(18)  10(12)‘ — 10(6) + 10(6) + 5(6) = 0
RM = 5 kN in direction shown Forces are summed horizontally: 2Fx=0; +5 ~RHX=0
RHx = 5 kN in direction shown fiorces are surnmed vertically:
5 ZFy=O;RAY—10—10+‘RHY10=0 R'HY = 25 kN in direction shown Exam le 63 ' ,si.
Reactions on a truss. Given: Truss with lOads and reactions as shown. To Find: Magnitude and direction of reactions. CHAPTER 6 SIMPLE TRUSSES 137 Solution: The entire truss is taken as a free body with all forces and unknown
reactions shown. ‘ Moments are summed about point A to ﬁnd REY:
EMA _= 0; 9(6) + 3(2) £902) + 3(4) .
+ 9(18) + 3(6)  REV(18) 1,115,, = 20 kips in direction shown Forces are summed horizontally to ﬁnd REX: ZFX = 0; 3 + 3 + 35,3“ :0
REX = in direction shown Moments are summed about E to ﬁnd RAY: ' t V ZME = 0; RAY(18) + 3(2)  9(12) + 3(4)  9(6) + 3(6) =' 0 I ' RAY = 7 kips in direction shown As a check, sum forces vertically: 'zFY=o;RAY—99—9+REY=0
799—9+20=_Q
' O=Q(OK) ; Exam le 64
Reactions on a truss. Given: Water tower with loads and reactions as shown. To Find: Magnitude and direction of reactions. “ mm (mm mm
' m; Ems SIDE) Solution: The entire truss is taken as a
reactions shown. 6.5 Member Forces by Method of Joints Forces in indivi
approaches in common use for ﬁnding Eon“
t 7.1L“. E “m 1000 ——.> m: i m“ F—P m 1°“
100°, Kit] .4000 ‘ 4000 1 Moments are summed about point A to ‘ﬁnd R N: EMA = 0; 2(8) + 1(6) + 1(4) + 1(2) 1 190(4) — RN(8) = 0
RN = 53.5 kN in direction shown 2M, = O; RAY(8) + 1(2) + 1(4) + 1(6) + 2(8)  100(4) = 0
RM = 46.5 kN in direction shown Sign forces horizontally to ﬁnd RAX: 'ZFX=0;2+1+1+1 —R,,X=0
RM: 5 kN in direction shown Ks a check, sum forces vertically: , 100+46.5+53.5=0 0 = 0 (OK) 1 dual truss members are found readily by
these forces: 1) Method of joints,
2) Method of sections. The ﬁrst of these approaches,
In the method of joints, each acting on the hinge are found by summing
progresses from joint to joint, utilizing required. Since there is no resultan
yields nothing of interest, leaving 0
ZFY = 0). There can therefore be no more solution is to be possible.
of joints is often called th the method of joints,
hinge is removed as t moment at a joint, the equation
nly the two equations of forc e method of force equilibrium. free body with all loads and unknown simple statics. There are two Since the solution involves only force equilibrium, is presented in this section.
a free body. Any unknown forces forces vertically and horizontally. The solution
the member forces from earlier sol utions as of moment equilibrium
e equilibrium (ZFX = 0,
than two unknown forces at each joint if a the method Exam le 65
Truss analysis by method of joints. Given: 1) 2) 3) i.  j 4) CHAPTER 6 SIMPLE TRUSSES 139 ""' ' __ ‘.' ———————————————————————————————— Since no more than two unknowns may occur at any joint, it becomes necessary'to
plan the sequence in which the solution will be performed such that no more than two
 unknown forces will occur. Member forces used in earlier solutions must necessarily be
used in follbwing solutions, which means that numerical errors made in one solution will
be propagated into all subsequent calculations. Extreme care must be exercised to avoid numerical errors and the resulting timeconsuming tracking to correct such errors.
 An example will illustrate the method of joints, or the method of force equilibrium, in ﬁnding the forces in truss members. / Truss conﬁguration‘and loads as shown. Reactions at A and E are given. 3 To Find: Force in each member. ‘_ Solution: A separate summary sketch of the truss is prepared and kept aside; in this I
example, this summary sketch is shown at the end of the example. When .
a force is determined, its value is entered immediately on the summary
sketch. ‘ The sequence of joints to beused in the solution is established: Begin at joint A. There are two members and therefore two unknown
forces on the joint at A; joint A can therefore be solved. A single tally
mark is placed on members AB and AH since they can be solved. The
number 1 is encircled at this joint. Proceed to joint B. With the force in member AB known from the
solution of joint A, there are still three unknown member forces at
joint B. Joint B cannot yet be solved. Proceed to joint H. With the force in member AH known frOm the
solution of joint A, there are only tWo unknown member forces re—
maining; joint H can now be solved. A double tally mark is placed on
members HB and HG to indicate that they are solved second. The
number 2 is encircled at joint H. PmCeed to joint B With the force in members AB and HB now known, there remains only two unknowns; jointB can now be solved. A triple tally mark is placed on members BG and BC to indicate that they are
solved third. The number 3 is encircled at joint B. 140 PART 1 STATICS 5) Thesymmetry of the truss is noted. Loads in symmetrical members
will of course be equal, leaving only the force in member CG_as_an
unknown. Joint C will be solved to ﬁnd the force in member CG; a
tally mark is placed on member CG to indicate that it is the fourth , solution. The‘number 4 is encircled at this joint. ' 6) The joint G reTnains the only joint that has not been solved. All forces coming into joint G have already been computed, however. Joint G
therefore becomes the “checking” joint for all earlier calculations; if
joint G is notin equilibrium, there is an error somewhere in the earlier
solutions. Joint G is checked for equilibrium. _ The solution can now begin, with joint A taken ﬁrst. The joint at A is removed as a free body with all forces shown on the joint. Known forces
are shown in their correct directions; unknown forces are shown in tension. . \\'\% 5
_ ' F 10
/ AB ~ A°'—_"FAH lg 2F, =0; + 9 4 FAB(5/ll.18) = o _
' FAB =  20.12 kips (compression) 2F“: 0; FAH+ Paco/11.13) = o
‘ FAH = + 18.0 kips (tension) TheSe values are entered immediately on the summary sketch (shown at
the end of the solution). ' Joint H is. solved next. F M is entered in its correct direction, pulling on
thca joint. All unknown forces are shown in tension. ‘ ’ Fits
11.
‘ / 18 
I <———l‘‘o————> p“
‘ 'L ">:Fv=o; —2+FH,,=0
FHB = 2 kips (tension) 21F1H =0; —18 +FHG= 0
. FHG = 18 kips (tension) These values are entered immediately on the summary sketch (shown at
the end of the solution). . D‘NSUS at In at CHAPTER 6 SIMPLE ,TRUSSES 141 Joint B is solved next. Forces AB and HB are entered in their correct
directions. F AB is in compression, pushing on the joint, and F "B is in 
tension, pulling on the joint. The unknown forces are shown in tension. ZFV = O; 20.12(5/11.1s). 4 — 2 + FROG/11.18)  FBG(5/ll.18) = 0
ZFH = 0; 20.1200/111? + FBC(10/Il.18) + FBG(10/ll.l8) = 0
Solve simultaneously to ﬁnd: FBC =;l3.42rikips (compression)
FBG = — 6.72 kips (compression) , These values are entered...immediately on the summary sketch (shown at
the end of the solution). Joint C is solved next. Forces BC and DC are equal, and both are in compression. Force CG is shown in tension. 4. . C 13.42 ° 13,42
 10 ;\
% 1 ca $0815 EFV = 0; 13.42(5/u.18)  4  FCG + 13.42(5/ll.18)=0
' ‘FCG = 8.00 (tension)
This value is entered immediately on the summary sketch (shown at. the. '
end of the solution). “  ’ Joint G‘is now checked to see that it is in equilibrium. All forces at joint
G are now known and are shown on the free body in their correct directions. ‘0 6.1216312 ID : [3<——é0——>13
12 ZFV = O;  6.72(5/11.ls) + 8 — 6.72(5/i1..18) — 2 = 0
0 = 0 (OK) ~142 PART 1 STATICS The check shows that there are no numerical errors. The ﬁnal solution is
shown in the following sketch. Joint H in Example 65 is a special type of joint. Refer to the solution of joint H in the
example. All external loads coming into the joint are seen to be transmitted unchanged
to joint B. Joint H is identiﬁed as a transfer joint and member H3 is identiﬁed as a
7 transfer member. In future solutions, such members will be identiﬁed and their forces
solved by simple inspection._ _ '
' The major points in the procedure of Example 6—5 are summarized:  Plan a sequence of joints to be solved such that no more than two unknowns
occur in any solution. ‘ — ' Remove the joints in the planned sequence, showing each joint as a free body.
Known forces are shown in their correct sense. Unknown forces are shown in
tension. (Transfer joints may be solved visually without a free body.) ,  Solve for the unknown forces on the free bodyvby summing forces vertically and
horizontally.  When a force has been determined, its value is entered immediately on a sum
mary sketch of the truss, without algebraic sign, followed by C or T to indicate
compression or tension.  When all forces have been determined, there will be one last joint which has not
been used in a solution but for which all forces have been determined elsewhere.
Check to see that this joint is in equilibrium. If it is not, there is a numerical error
in the calculations. The emphasis on good “housekeeping” practices in Example 6—5 is essential. The greatest disadvantage in using the method of joints is that of making numerical errors. The use of good housekeeping practices will help to reduce the chance of makingsuch errors.
A second example in the method of joints will further illustrate the procedure. Exam le 66 _
Truss analysis by method of joints. . ' ~/ Given: Truss‘gonﬁguration and loads as shown. Reactions at supports are shown. To Find: Force in each member. ’. 2k“ MAN lull ' sketch: ' CHAPTER 6 SIMPLE TRUSSES 143 Solution: A summary sketch is prepared and set aside. It will be included at the end of the solution. It is noted that there are several transfer joints. Joints B, J, D, H and F
are all transfer joints. Free bodies will not be required to solve these
joints. The sequence of joints to be solved is now established. Joint B has only
two unknowns; it will be solved ﬁrst. A single tally mark is placed on the
two unknown member forces BA and BC, and an encircled 1 is placed at
the joint. Joint A is solved second. A double tally mark is placed on the two un
known member forees Aland AC and an encircled 2 is placed at the joint. Joint J is solvedﬂnext. A_' 'ple tally mark is placed on the two unknown
member forces J1 and J ' and an encircled 3 is placed at the joint. The sequence is extended through the remainder of the joints in the same
way, leaving only joint G unsolved. Joint G will be used as the checking
joint. The solution is now performed in the sequence just determined. Joint B is ﬁrst. It is identiﬁed as a transfer joint. Forces in members BC
and BA are solved by inspection and their values entered on the summary
sketch:  FBC = 10 kN (compression); FBA = 2 kN (compression) ~ Joint A is second. A free body is drawn, with known forces shown in their
correct directions and unknown forces shown in tension. 4\‘,\l\ . ZFY = 0
$ —2 + 6.17 + FAC(4/721) = 0 2/1 FM: 7 FAG = —7.52 kN (compression)
PAW—.1; '
M _ ZFX = 0
6" —20 + ace/7.21) + FA] = 0 PM = 26.26 kN (tension)
These values are entered on the summary sketch: Joint J is It is identiﬁEd as a transfer joint. Forces in members J1
and JC are solved by inspection and their values entered on the summary F" = 26.26 kN (tension); FJC = 1 kN (tension) J‘ Joint C is next. A free body is drawn, with known forces shown in their
correct directions and unknown forces shown in tension. ' 144 _ PART1 STATICS 6 i , 6 2m = o 5 .. .  2274 4 All?» 7.5261721) — 4 — i mac/7.21) =0 FCI = 1.50 kN (compression) , g, _ I m c FED , ——,—>o—————> " . " 2F :0 .
75/ L\_‘\Fé: ' f0 + 7.5207721) + FCD + FOG/7.21) ; 0 FCD = 15 .02 kN (compression)
' These values are entered on the summary sketch. Joint D is next. It is identiﬁed as a transfer joint. Forces in members DE
, and D1 are solved by inspection and their values entered on the summary
sketch: ' t v  FDE = 15:57 kN (compression); FBI: 4 kN (compression) The solution proceeds” through all joints in the sequence. At the con”—
clusion of the sequence, the summary sketch contains the following 77 values: 10 C. 19.07. C _ 15.0212 IDT . a)
N Joint G becomes the checking joint. All forces at joint G have been found
from other solutions. ‘ £952 2 EFY = 0 \ 4952(4/721) — 2 + 12.83 = 0
' , * ' a 0:0 (OK)
[Gil 03 _ .5 . Z? ZFX _ 0
, . 12.33 4 __/ 1952(6/721) — 16.27 = 0
{1. v ‘ 6 ‘ 0 = 0 (OK) ' Joint "G is in equilibrium, indicating that the numerical calculations are
correct. ‘ There is a type of truss called a Ktruss that can sometimes cause a problem when one
is trying to establish the sequence of joints. example will illustrate a means to handle such trusses. Exam le 67 _ ‘ Truss analysis by’method of joints. 6.6 CHAPTER 6 SIMPLE TRUSSES 145 M Given: Ktruss conﬁguration and loads as shown. To Find: Sequence of joints to be used in the solution. Solution: The selection of joints tagbe used in the sequence progresses, as usual, up
to joint 6, as shOwn.» Thereafter, both jointC and joint L have three unknowns and cannot be solved next. Proceed to joint N and come backwards through joints N, E, F, D, M and
' G. Joint C can now be solved. Joint L then becomes the checking joint. ' Before leavingeithe method Of joints, it is again noted that its biggest single disadvantage is in the potential for propagating numerical errors into succeeding joint solutions. Cor— > recting suchjerrors can indeed be a frustrating and time—consuming exercise. It is an ever present hazard in any manual solution for member forces in a truss. A strong advantage of the method of joints, however, lies in its mechanical simplic—
ity and repetitiveness. For that reason, the procedure is ideally suited to computerized
solutions."And, since the'co'mputer cannot make numerical errors, the biggest single
disadvantage of the method of joints disappears.  , Alrnost certainly, a computerized solution for member forces in a truss will utilize
the method of joints. Such computerized solutions are readily available in the industry but
are not recommended for use by students. Only when the student understands the method
of Solution should'the student be permitted to use such software. Member Forces by Method of sections it was noted in earlier discussions that there are two approaches in common use for truss
analysis: ‘  \ ’
' 1) Method of joints. 2) Method of sections. The second of these approaches, the method of sections is presented here. I In the method of sections,.a section is taken of the truss, expOsing the member force
that is to be determined. A free body is then drawn for the part of the truss lying to one
side of that sectiOn, with. all unknown fOrces shown. Moments are then summed about
some point in the plane: such that all. unknowns are eliminated except the one member
force that‘is to be determined; in some'cases, forces may be summed (rather than mo
ments) to achieve the same result. Successivesections are cut, and the method is applied
until every membef force has been found. I ' r 146 PART 1 STATICS _ In the method of sections, no member force found in an earlier solution is used in
succeeding solutions. The primary disadvantage noted earlier in the method of joints, that V of propagation of error, has therefore been eliminated. The method of sections has a
disadvantage of its own, however, in that some degree of skill is required in choosing where to take the sections. ‘
The method of sections is best illustrated byan example. Discussions following the example then summarize the method. f Exam le 68 , I . __
Truss analysis by method of sections. 4 Given: Truss'conﬁguration and loads as shown. Reactions at supports are shown“. To Find: Force in each member. m1 6 4m 4m m1 m1
103i A i b i i d
wk“
41100
2m
+—
C. . . smut ' in» in“ in» mm
mm V son» . am ‘ soon q SOlution: As with the method of joints, the results of the analysis will be entered
on a summary sketch. The summary sketch for this example is kept aside
and is shown at the end' of the example. Section aa is taken in the ﬁrst panel. The free body to the left of that :2 " section is shown in the following sketch.
'b
1.?” ‘ . The unknown member forces FAB, F ,3 and F J, have been exposed by the
.‘éut section. They are shown on the free body in tension. Moments are taken about joint J. Forces J, and F ,3 pass through joint J
and are thus eliminated from the equation, leaving only the force F AB: 2M,= 0; 10(4) + FAB(4) = 0 .
' FAB = 10 kN (compression) Moments are now summed about joint B, eliminating forces F AB and F13,
leaving only the force F”: 2MB = 0; — 2(6) + 20(4) + 617(6) — Fn(4) = 0
‘:i‘ Fn = 26.26 kN (tension) \IHWi—riI (I) u: CHAPTER 6 SIMPLE TRUSSES 147 W Forces are now Summed vertically, eliminating the two horizontal forces
F M, and F ,1, leaving only the force F ,3: ER, = O; 6.17 — 2 + EEG/7.21) = 0
FIB = —7.52 kN (compression) These forces are entered immediately on the summary sketch. The next section is section b—b in the second panel. The free body to the
left of section brb is shown in the following sketch. 2' '14 41300 r—
20 ‘ “x
. h ' FIE“OH
, I b s
5411  1; 23M 5000 soon Summing momentsiabout joint B eliminates F BC and F 5,1, leaving only the
force F m: 4 a 4. 2M; = 0; 2(6) + 20(4) + 617(6)  Fm“) = 0 V
Fm = 26.26 kN (tens \Il) Summing moments about joint H eliminates F 3,, and Fm, leaving only the
force F BC: ‘ ZMH = 0; 10(4)  2(12)  4(6) + FBC(4)
+ 6.17(12) — 1(6)‘: 0
.‘ FBc = '15.01 kN (compression) Summing forces vertically eliminates the two horizontal forces F BC and
F I”, leaving only F B”? EFY = o; —2  4 4r 6.17  1  FBH(4/7.2x) = 0
A FBY = «1.50 kN (compression) These values are entered immediately on the summary sketch. The next section is taken as section cc. The free body to the left of
sectibn 06 is shown in the following sketch. ‘ 148 PART1 STATICS } _ ' Moments are summed about joint H, eliminating forces F HG and F "D,
leaving only the force F CD: ‘ ' ZMH = o; — 2(12) + 10(4)  4(6) + FCD(4)
+ 6.1702)  1(6) = o
FCD = 15.01 kN (compression) Moments aresummed about joint D, eliminating forces Feb and ‘FHD,
leaving only force FHG: , ZMD = 0;  2(18)  4(12) — 4(6)  FHG(4) ’ 1(6)
 1(12) + 6.1708) + 20(4) = 0
FHG = 16.27 kN (tension) Forces are summed vertically, eliminating the two horizontal forces'FCD
and F Ha, leaving only F Hp: ZFY=0;—2’4—4+6.17 1— l +FHD(4/7.21)=0
FHD = 10.51 kN (tension)
These values are entered immediately on the summary sketch. The next section is taken as section d—d; The free bodies to the left of the
section are becoming too cumbersome, so the free body for this case will be shown to the right of the section. '1
d 1
Doing” [— i0
;. E; 4000
ﬁO——.'FF& F , _ ., d 10.33 5000 Moments are summed about joint F, elirninating F FD and F F6, leaving
. only F ED: ‘ 1 EMF = 0;  FED(4) + 10(4) = O
‘ FED = 10 kN (tension) Moments are summed about joint D, eliminating F ED and F FD, leaving
only F m: ZMD = 0; 2(6) — 1283(6) + FFG(4) = O
Fm = 16.25,kN (tension) Forces are summed vertically, eliminating the two horizontal forces F ED
and F PG, leaving only F FD: ZFY = 0; — 2 + 12.83 + PEG/7.21) = 0
FFD = —19.52‘kN (compression) 1 49 CHAPTER 6 SIMPLE TRUSSES These values are entered immediately on the summary sketch. The remaining member forces are seen to be those in the vertical mem—
bers. All of these members are transfer members and may be solved by inspection: FIE = 1 kN (tension)
FGD = 1 kN (tension) FM = 2 kN (compression)
FCH = 4 kN (compression)
FEF = 2 kN (compression) / These values are entered directly on the summary sketch. The ﬁnal results are shown on the following summary sketch. They are
of course identical to those of Example 66, which is the same truss. / we mt mm tor '2C The major points in the procedure of Example 68 are summarized: Select a section that cuts the members such that the desired member forces will
be exposed. In selecting a section, try to expose no more than three forces since there are only
three equations of equilibrium. Identify some point to take moments about (if one exists) such that only the
desired member force will appear in the equation when moments are summed.
Such a point will occur at the point of intersection of the two forces that are to
be eliminated. '
Repeat (where possible) to ﬁnd other member forces. q
Two forces that are parallel have no point of intersection and therefore cannot be
eliminated by summing moments. In such cases, forces can be summed perpen
dicular to these forces, thereby eliminating these forces from the resulting equation.
When a force has been determined, its value is entered on a summary sketch of
the truss, without algebraic sign, followed by C or T to indicate compression or tension. A second example in the method of sections will illustrate the method when there are no
parallel members. Itwill be seen in this example that it is not always possible to eliminate
all unknowns except one. In such cases, a solution by method of joints may sometimes
be necessary to ﬁnd a particular force. H, CHAPTER 6 SIMPLE TRUSSES 151 Section bb is taken next. The free body to the left of section b—b is shown
in the following sketch. Summing moments abbut B eliminates F 35 and F 86: 2MA = 0; 4(10) + 200)? FBG(?°/11.xs)5 + FBG(5/11.18)10 = O
 _ FHG = 18 kips (tension) Summing moments about a“ eliminates F BC and F HG. The moment of the
remaining force F BGabout joint A is found by summing the moments of
its vertical and horizontal components about A: EMA = 0; 4(10) + 2(19) + FBG(l°/11.18)5 + FBG(5/11.18)10 = 0
FBG =  6.71 kips (compression) Summing moments about G eliminates F 36 and F H0. The moment of the
remaining force F BC about joint G is found by summing the moments of
its vertical and horizontal components about G: ZMG = 0; 9(20)  4(10) 2(10) + FBC(1°/u.18)5
+ FBC(5/11.18)10‘= 0
FBC = —13.42_ kips (compression) The force in the remaining member CG cannot be found by eliminating
all unknowns except Fm. The solution for FCG will be performed by the . method of joints. /.
I /. \‘b
9 L? 4 " \\;_15
10 ' l0 0 is.“ ' N42.
Er, ZFY = (l;  4 + 13.42(5/u.1s) + 13.42(5/ll.18) — FCG = 0
FCG = 8.00 kips (tension) The results of the analysis are shown on the following summary sketch.
These results are of course identical to those of Example 65 for the same
truss. ' ‘ 152 PART1 STATICS‘H _ ' H In the feregoing exainples, separate free bodies were drawn for each section. As one
becomes more familiar with the methOd, such separate sketches will probably not be
nece5sary. 'One simply selects the position where the s tion is to be taken, then covers
up the unwanted portion, of the' sketchwith apiece of so atch paper; the visible portiOn‘ will then be the free body. Summing of forces and moments then proceeds'as usual, recognizing that the unknown forces are always assumed to be in tension, that is, directed
away from the joint where they. are attached. ' ' ' 4 Review GUestions i sovsww ll. 12.
13. 14. 15. 99w What kind of loads occur in truss members? ’
‘. Why is a truss so efﬁcient in its utilization of materials? What is the differencebetween a Parker truss and a Bowstring truss? What is a panel'point amiss? ' I ‘ ' ‘ Deﬁne a properly formed truss. ‘ What end conditions are assumed for the ends of truss members? What is the difference in computing reactions for a truss and in computing reactions
for a solid beam or frame? ‘ ' ' " Naine the two methods in common use for computing member‘forces in a truss.
What is another name for the method of ferce equilibrium? . In the method of force equilibrium, what is the maximum number of unknowns that
can be found at a single panel point? What makes the method of force equilibrium a potentially timeconsuming type of
solution? > I . i ‘ What is the primary advantage of the method of sections mm the method of joints?
What disadvantage is introduced in the method of sections that did not ocur in the
method of joints? '  I "
How is it that numerical errors can propagate into succeeding calculationsin the
inede of joints but not in' the method of sections? ' Which of thetth methods would most likely be used to program a computer solution? I t ...
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 Fall '08
 Chen

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