Lecture 61 Introduction to Structural Analysis STUDENT MASTER COPY
Structural Analysis:
1. Previous chapters have dealt with:
a. equilibrium of a single rigid body
b. external forces acting on the rigid body
2. This chapter will deal with:
a. the equilibrium of individual structural members
b. the evaluation of internal forces between members
3. Given the crane of Fig. 6.1a) with freebody diagram of Fig. 6.1b):
a. consists of three beams AD, CF, and BE
b. connections are frictionless pins
c. held in place by cable DG with external tensile force T
d. supports external load W
e. two external reactions Ax and Ay at point A
4. The internal forces applied to each of the three beam members are shown
in Fig. 6.1c):
a. vertical and horizontal forces as point C which are equal and
opposite for members AD and CF
b. a single diagonal force at point B which is equal and opposite for
members AD and BE
c. a single diagonal force at point E which is equal and opposite for
members CF and BE
5. These internal forces follow Newton's Third Law which states that the
forces of action and reaction between bodies in contact have the same
magnitude, the same line of action, and opposite sense
6. Newton's Third Law is one of the six basic fundamental principles of
elementary mechanics and is based on experimental evidence
7. The three types of structures that will be considered in this chapter
are as follows: a. trusses, b. frames, and c. machines
8. Truss structures:
a. are designed to support loads
b. are usually stationary and fully constrained
c. consist of straight members that are connected at joints located at
the ends of each member
d. members are "twoforce members" which are acted upon by two equal
and opposite forces applied at each end and directed along the
member centerline
9. Frame structures:
a. are also designed to support loads
b. are usually stationary and fully constrained
c. contain at least one "multiforce member" which is acted upon by
three or more forces that are, in general, not directed along the
member centerline
10. Machine structures:
a. are designed to transmit and modify forces
b. are structures containing moving parts
c. contain at least one "multiforce member" which is acted upon by
three or more forces that are, in general, not directed along the
member centerline
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Trusses:
1. One of the major types of engineering structures
2. Provides a practical and economical solution to many engineering
situations, especially in the areas of bridge and building design
3. A truss consists of straight members connected at joints at each end
a. no members are continuous through a joint
b. thus in Fig. 6.2, members AD and DB are two separate members, i.e.
there is no member AB
4. Most trusses consist of two or more twodimensional trusses joined
together to form a threedimensional structure:
a. each twodimensional truss can then be analyzed as a separate
twodimensional system
b. the bridge in Fig. 6.3 consists of two twodimensional trusses which
each support loads only at the truss joints
5. Bridge loads are transferred to truss joints by a twopart floor system
that consists of:
a. longitudinal stringers
b. lateral floorbeams
6. The weight of each truss member is assumed to be divided equally between
the joints
7. Actual trusses are connected by:
a. riveted, bolted, or welded joints in general
b. ballandsocket joints occasionally
c. all truss joints are treated as pin connections
d. thus there are no moments or couples in the analysis of a truss
8. Each member can be acted upon by either:
a. tension as shown in Fig. 6.4a) or
b. compression as shown in Fig. 6.4b)
9. Typical trusses, which are shown in Fig. 6.5, are as follows:
a. roof trusses such as Pratt, Howe, or Fink
b. singlespan bridge trusses such as Pratt, Howe, Warren, Baltimore,
and KTruss
c. multispan bridge trusses such as cantilever trusses
d. moveable bridge trusses which include bascule or draw bridges
Simple Trusses:
1. Consider the truss of Fig. 6.6a):
a. which is composed of four members connected by pins at A, B, C, and D
b. if a load is applied a point B, the truss will greatly deform and
will loose its original shape
2. Consider the truss of Fig. 6.6b):
a. which is composed of three members connected by pins at A, B, and C
b. if a load is applied a point B, the truss will retain its original
shape with only minor changes in the lengths of its members
c. this truss is said to be a "rigid truss" to indicate that the truss
will not collapse under normal working loads
3. Consider the truss of Fig. 6.6c):
a. obtained by adding two members and one joint to the rigid truss of
Fig. 6.6b)
b. the result is another rigid truss
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c. this process of adding two members and one joint can be repeated as
many times as desired, with the result being another rigid truss
d. a truss that is constructed in this manner is called a "simple truss"
Comments on Simple Trusses:
1. As illustrated in Fig. 6.6d), simple trusses are not necessarily
composed of only triangles
2. Rigid trusses are not always simple trusses:
a. the Fink and Baltimore trusses in Fig. 6.5 are not simple trusses
b. all of the other trusses in Fig. 6.5 are simple trusses
3. The necessary condition for simple trusses is that: m = 2n 3
where: m = the total number of members
n = the total number of joints
a. for the truss of Fig. 6.6b), n = 3 joints and m = 2 (3) 3 = 3
members
b. for the truss of Fig. 6.6c), n = 4 joints and m = 2 (4) 3 = 5
members
c. for the truss of Fig. 6.6d), n = 7 joints and m = 2 (7) 3 = 11
members
d. this formula does not guarantee that a truss is simple, however
Lecture 62 Truss Analysis STUDENT MASTER COPY
Truss Analysis by the Method of Joints:
1. A truss is essentially a group of joints and twoforce members
2. The truss of Fig. 6.7a) can be be broken down into the freebody
diagrams for each member and joint as shown in Fig. 6.7b)
3. Each member is acted upon by two equal and opposite forces at each end
4. The reactions between the end of a member and the corresponding joint
must be equal and opposite
5. The magnitude of the force in a truss member is often referred to as the
"force in the member" even though this magnitude is a scalar quantity
6. Since the lines of action of the forces in a truss are known, the only
unknowns are:
a. the magnitudes of the forces in each member
b. whether the member is in tension or compression
7. Since the entire truss is in equilibrium:
a. each pin must be in equilibrium
b. each member must be in equilibrium
8. If a truss has "n" number of pins, then:
a. there will be 2n equations available
b. which can be solved for 2n unknowns
9. If the truss is a simple truss:
a. since the number of members m = 2n 3
b. 2n = the number of equations = m + 3
c. the unknowns that can be solved will include the force in each member
and three reactions
10. The reactions can be determined separately using:
a. the three equations of equilibrium
b. the freebody diagram for the entire structure
11. The arrangement of pins and members in a simple truss is such that:
a. it is always possible to find one joint that involves only two
unknown forces
b. these forces can be determined by using the equations of equilibrium
for a point (particle)
c. the values of member force determined can then be transferred to
joints at the opposite ends of each member
d. the process is then repeated until all of the unknown member forces
are known
Example Figs. 6.7 and 6.8:
1. By analyzing the truss structure in Fig. 6.7, we can see that at point A
(Fig. 6.8a):
a. there two unknown forces in members AC and AD
b. there are two known forces which are the rectangular coordinates of
force Ra
c. thus the forces in members AC and AD can be found using the equations
of equilibrium at a point: ΣFx = 0 and ΣFy = 0
2. Next we can move to point D (Fig. 6.8b) where:
a. there two unknown forces in members DB and DC
b. there are two known forces which are the applied load P and the force
in member AC
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c. thus the forces in members DB and DC can be found using the equations
of equilibrium at a point
3. Next we can move to point C (Fig. 6.8c) where:
a. there there is one unknown force in member CB
b. there are two known forces which are the forces in members AC and CD
c. thus the force in member CB can be determined using one of the
equations of equilibrium at a point, while the other equation can be
used as a check
4. Next we can move to point B (Fig. 6.8d) where:
a. there are no unknown forces
b. there are three known forces which are the reaction Rb and the forces
in members BD and BC
c. thus we can check our solution by making sure that the equations of
equilibrium at this point are fully satisfied
5. Note that using the equations of equilibrium for a point is equivalent
to drawing the corresponding parallelogram or triangle
6. Other triangles, parallelograms, or diagrams can be drawn to determine
the unknowns at one or more joints
7. The single diagram for the entire truss system, which is shown in Fig.
6.10, is referred to as a Maxwell's Diagram
Joints Under Special Loading Conditions:
1. Consider the joint shown in Fig. 6.11a):
a. which connects four members lying along two intersecting lines of
action
b. the freebody diagram shown in Fig. 6.11b) shows that the joint is
subjected to two pairs of forces which must be equal and opposite
c. the corresponding force polygon will be a parallelogram and thus the
forces in opposite members must be equal
2. Consider the joint shown in Fig. 6.12a):
a. which connects three members and supports a vertical load P
b. the freebody diagram of Fig. 6.11b) is still applicable with Rae
replaced by P
c. thus the forces in the two opposite members must be equal and the
force in the other member must be equal to P
3. Consider the joint shown in Fig. 6.12b):
a. where the force P from Fig. 6.12a) is zero
b. thus the forces in the two opposite members must be equal and the
force in the other member must be zero
4. Consider the joint shown in Fig. 6.13a):
a. which connects two members along the same line
b. the forces in the two members must be equal
5. Consider the joint shown in Fig. 6.13b):
a. which connects two members along different lines
b. the force in both members must be zero
6. Consider the truss structure of Fig. 6.14:
a. member BC will be a zeroforce member because of the conditions at
joint C
b. the forces in members AC and CE will be equal
c. member JK will be a zeroforce member because of the conditions at
joint K
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d. the forces in members IK and KL will be equal
e. member IJ will be a zeroforce member because of the conditions at
joint J
f. the forces in members HJ and JL will be equal
g. member DE does NOT have a force of 24 kN because of the existence of
member DG
h. the forces in members BD and DF are not equal because of the
existence of member DG
7. Until these special cases become easily recognized:
a. assume all member forces are nonzero
b. until proven otherwise by application of the equations of equilibrium
at each joint in each truss structure
8. Note that zeroforce members are not useless:
a. they can be nonzero under other load conditions
b. they maintain the shape of the truss
c. they help to support the selfweight of the truss
Space Trusses:
1. A threedimensional truss is referred to as a space truss
2. The most elementary rigid space truss is the tetrahedron ABCD
as shown in Fig. 6.15 with:
a. six members
b. four joints
3. As shown in Fig. 6.15, this basic tetrahedron can be expanded to form
other rigid space trusses by:
a. adding three members (AE, BE, and CE)
b. adding one joint (E)
4. A space truss formed in this way is called a simple space truss
5. The joints of a space truss can be formed by:
a. riveted joints
b. bolted joints
c. welded joints
d. ballandsocket joints
6. The members of a space truss are twoforce members
7. The necessary condition for a simple space truss is: m = 3n 6
where: m = the total number of members
n = the total number of joints
8. This is not a sufficient condition to guarantee that a truss is a simple
space truss, however
Sample Problem 6.1 - Using the method of joints, determine the force in each member of the truss shown.
Solution:
Step #1 Equilibrium of Entire Truss Structure:
Σ Fx = 0 = Cx
Cx = 0
Σ Mc = 0 = (2000 lb) (24 ft) + (1000 lb) (12 ft) E (6 ft)
E = (48,000 lbft + 12,000 lbft) / 6 ft = 10,000 lb
Σ Fy = 0 = 2000 lb 1000 lb + Cy + E
Cy = 2000 lb + 1000 lb 10,000 lb = 7000 lb
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Σ Me = 0 = (2000 lb)(18 ft) + (1000 lb)(6 ft) (7000 lb)(6 ft)
= 36,000 lbft + 6000 lbft 42,000 lbft = 0
Step #2 Analysis of Joint A:
Fadx = (3/5) Fad
Fady = (4/5) Fad
Σ Fy = 0 = 2000 lb + Fady = 2000 lb + (4/5) Fad
Fad = (5/4) (2000 lb) = 2500 lb (compression)
Σ Fx = 0 = Fab Fadx = Fab (3/5) Fad
Fab = (3/5) (2500 lb) = 1500 lb (tension)
Step #3 Analysis of Joint D:
Fdbx = (3/5) Fdb
Fdby = (4/5) Fdb
Fdax = (3/5) Fda = (3/5) (2500 lb) = 1500 lb
Fday = (4/5) Fda = (4/5) (2500 lb) = 2000 lb
Σ Fy = 0 = 2000 lb + Fdby = 2000 lb + (4/5) Fdb
Fdb = (5/4) (2000) = 2500 lb (tension)
Σ Fx = 0 = Fde + Fdbx + Fdax = Fde + (3/5) Fdb + 1500
Fde = (3/5) (2500) + 1500 = 3000 lb (compression)
Step #4 Analysis of Joint B:
Fbex = (3/5) Fbe
Fbey = (4/5) Fbe
Fbdx = (3/5) Fbd = (3/5) (2500 lb) = 1500 lb
Fbdy = (4/5) Fda = (4/5) (2500 lb) = 2000 lb
Σ Fy = 0 = 1000 lb Fbey Fbdy = 1000 lb (4/5) Fbe 2000 lb
Fbe = (5/4) (2000 lb + 1000 lb) = 3750 lb (compression)
Σ Fx = 0 = 1500 lb Fbdx + Fbex + Fbc
= 1500 lb 1500 lb + (3/5) ( 3750 lb) + Fbc
Fbc = 5250 lb (tension)
Step #5 Analysis of and Check at Joint E:
Fecx = (3/5) Fec
Fecy = (4/5) Fec
Febx = (3/5) Feb = (3/5) (3750 lb) = 2250 lb
Feby = (4/5) Feb = (4/5) (3750 lb) = 3000 lb
Σ Fy = 0 = E Feby + Fecy = 10,000 lb 3000 lb + (4/5) Fec
Fec = (5/4) (3000 lb 10,000 lb) = 8750 lb (compression)
Σ Fx = 0 = Fed + Febx + Fecx = 3000 lb + 2250 lb + (3/5) Fec
= 5250 lb + (3/5) ( 8750 lb) = 5250 lb 5250 lb = 0
Step #6 Checks at Joint C:
Fcex = (3/5) Feb = (3/5) (8750 lb) = 5250 lb
Fcey = (4/5) Feb = (4/5) (8750 lb) = 7000 lb
Σ Fy = 0 = Cy + Fcey = 7000 lb + 7000 lb = 0 lb OK
Σ Fx = 0 = Fcb + Cx + Fcex = 5250 lb + 0 lb + 5250 lb = 0 lb OK
Analysis of Trusses by Method of Sections:
1. The method of joints is most effective if all of the member forces are
required
2. The method of sections is more effective if only a few member forces are
required
3. Taking the truss in Fig. 6.16a):
a. if we need to calculate the force only in member BD for example
1
b. we can choose the freebody derived by cutting the truss along line
nn as shown in Fig. 6.16b)
c. the cut must cross the member BD whose force we wish to calculate
d. the cut should also be chosen so that the other cut members BE and CE
are concurrent at some point such as point E
4. By taking moments about the concurrent point E:
a. the forces acting through the concurrent point are eliminated
b. we can solve for the only unknown force which is in the remaining cut
member BD
5. If we also needed the force in another cut member such as member CE:
a. we can take moments about point B
b. such that the forces in the remaining two cut members are eliminated
6. Summing forces in y direction may be necessary, however, if:
a. cutting to leave only one unknown is not possible
b. the unknown force desired, such as the force in member BE, passes
through two concurrent points such as points B and E
7. If all of the forces in the cut members are calculated:
a. then checks can be made of the final answers
b. these checks would use the remaining (unused) equations of
equilibrium
Trusses Made of Several Simple Trusses:
1. Simple trusses can be combined to form a rigid truss that is not a
simple truss:
a. the two simple trusses ABC and DEF are combined in Figs. 6.17a) and
6.17b)
b. this type of truss is called a compound trusses
c. the resulting trusses satisfy the following equation for
twodimensional trusses: m = 2n 3
d. the number of unknowns is m + 3 = 29 + 3 = 32
e. the number of equations is 2 n = 2 (16) = 32
2. Consider the truss in Fig. 6.18:
a. the number of members m = 30 is larger than:
2n 3 = 2 (16) 3 = 29
b. this truss is said to be over-rigid
c. members BE and CD are said to be redundant, i.e. only one member is
required for rigidity
d. the number of unknowns is m + 3 = 33 which is larger than the number
of equations which is 2 n = 2 (16) = 32
e. the truss is statically indeterminate
3. Compound trusses that are supported by one pin and one roller, or by an
equivalent system of supports are statically, rigid, and completely
constrained
4. Consider the truss of Fig. 6.19a):
a. the number of members m = 26 is less than:
2n 3 = 2 (15) 3 = 27
b. the number of unknowns is m + 3 = 29 which is larger than the number
of equations which is 2 n = 2 (15) = 30
c. the truss is said to be nonrigid and will collapse under very little
load
1
5. Consider the truss of Fig. 6.19b):
a. the number of members m = 26 is less than:
2n 3 = 2 (15) 3 = 27
b. the number of unknowns is m + 4 = 30 which is equal to the number of
equations which is 2 n = 2 (15) = 30
c. thus the truss is rigid and statically determinate
6. Given that "r" is the number of support reactions for a truss:
a. the necessary condition for a compound truss to be statically
determinate, rigid, and completely constrained is: m + r = 2n
b. this condition is not sufficient to guarantee that a truss will be in
equilibrium
Sample Problem 6.2 - Determine the force in members EF and IG of the truss shown.
Solution:
Step #1 Equilibrium of Entire Truss:
Σ Fx = 0 = Bx + 16 kips
Bx = 16 kips
Σ Mb = 0 = (28 kips) (8 ft) (28 kips) (24 ft) (16 kips) (10 ft)
+ J (32 ft)
J = (224 kft + 672 kft + 160 kft) / (32 ft) = 33 kips
Σ Mj = 0 = By (32 ft) + (28 kips) (24 ft) + (28 kips) (8 ft)
(16 kips) (10 ft)
By = (672 kft + 224 kft 160 kft) / (32 ft) = 23 kips
Σ Fy = 0 = 28 kips + 28 kips 33 kips 23 kips = 0
Step #2 Force in Member EF:
Σ Fy = 0 = 23 kips 28 kips Fef
Fef = 5 kips (compression)
Step #3 Force in Member IG:
Σ Mh = 0 = Fig (10 ft) (16 kips) (10 ft) + (33 kips) (8 ft)
Fig = 10.4 kips (compression)
Sample Problem 6.3 - Determine the force in members HF, HG, and IG of the roof truss shown.
Solution:
Step #1 Equilibrium of Entire Truss:
Σ Fx = 0 = Ax
Ax = 0 kips
Σ Ma = 0 = (1 kN) (5 m + 10 m + 15 m + 20 m + 25 m)
(5 kN) (5 m + 10 m + 15 m) + L (30 m)
L = (75 kNm + 150 kNm) / (30 m) = 7.5 kN
Σ Ml = 0 = (1 kN) (5 m + 10 m + 15 m + 20 m + 25 m)
+ (5 kN) (15 m + 20 m + 25 m) Ay (30 m)
Ay = (75 kNm + 300 kNm) / (30 m) = 12.5 kN
Σ Fy = 0 = 5 (1 kN) + 3 (5 kN) + 7.5 kN + 12.5 kN = 0
Step #2 Force in Member IG:
Hih = height of truss from points I to H = (2/3) (8 m) = 5.33 m
Σ Mh = 0 = Fig (5.33 m) + (7.5 kN) (10 m) (1 kN) (5 m)
Fgi = (75 kNm 5 kNm) / (5.33 m) = 13.12 kN (tension)
Step #3 Force in Member HF:
α = arctan (8 m / 15 m) = 28.07
1
Move force Fhf along its line of action to point F:
Fhfx = Fhf cos (28.07) = 0.8824 Fhf
Σ Mg = 0 = Fhfx (8 m) + (7.5 kN) (15 m) (1k N) (10 m) (1 kN) (5 m)
Fhfx = (112.5 kNm 10 kNm 5 kNm) / (8 m) = 12.19 kN
Fhf = ( 12.19 kN) / (0.8824) = 13.81 kN (compression)
Step #4 Force in Member HG:
ß = arctan (5 m / 5.33 m) = 43.17
Move force Fhg to point G:
Fhgy = Fhg cos (43.17) = 0.7293 Fhg
Σ Ml = 0 = Fhgy (15 m) + (1 kN) (10 m) + (1 kN) (5 m)
Fhgy = (10 kNm + 5 kNm) / (15 m) = 1 kN
Fhg = ( 1 kN) / (0.7293) = 1.371 kN (compression)
Lecture 63 Frames and Machines STUDENT MASTER COPY
Structures Containing Multiforce Members:
1. Members acted upon by three or more forces
2. Forces are generally not directed along the member centerline
3. Force directions are unknown
4. Forces should be represented by two unknown rectangular components
5. Frame and machine structures contain multiforce members
6. Frames are designed to support loads and are usually stationary and
fully constrained
7. Machines are designed to transmit and modify forces
8. Machines may not be stationary and will always have moving parts
Analysis of a Frame:
1. Consider the crane shown in Fig. 6.20:
a. the freebody diagram of the entire frame is shown in Fig. 6.20b) and
can be used to determine external forces acting on the frame
b. summing moments about point A, we can solve for the force T
c. summing forces in the x and y directions, we can solve for the forces
Ax and Ay at point A
2. In order to determine the internal forces that are holding the frame of
Fig. 6.20a) together:
a. we draw the freebody diagrams of each member as shown in Fig. 6.20c)
b. member BE is the only twoforce member