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Outline:

Frames

Machines

Trusses

Properties and Types

Zero Force Members

Method of Joints

Method of Sections

Space Trusses

ENGR 1205 1 Chapter 9

“structures” are made up of several connected parts

we consider forces holding those parts together (in

addition to any external forces that may be present)

for the whole structure, these are internal forces, but

we can consider each of the parts as a RB and can

treat forces from connecting parts as external forces

corresponding points on different parts have equal but

opposite rxns [Newton’s 3rd Law]


in this chapter, we will look at:

Frames: they support loads, are usually

static and completely constrained, but they

have at least one multi-force member (as in

Figure on prior slide)

Machines: they transmit and modify forces -

they have moving parts and at least one

multi-force member

Trusses: they support loads, are static and

fully constrained, and are made of straight

members connected at joints - all parts are

“2 force elements”


For all types of structures:

If the whole structure is in equilibrium then

any part of the structure must also be in

equilibrium

Can separate a structure into individual parts

(all in equilibrium)

Can “cut” a structure and examine a piece of

it (also in equilibrium)


A system designed to support loads (forces and

moments)

Made up of multiple members

At least one member is acted upon by 3+ forces

(not two-force elements)

Can be internally stable or internally unstable

Stable – no matter what the external support

reactions are the frame maintains its shape

Unstable – if external support reactions are

removed the members of the frame move

relative to each other.


frames are inter-connected rigid bodies

start by analyzing the whole system to get the

reactions at the supports (if possible – frames may

be statically indeterminate)

then isolate members or parts, for each one

separately draw a FBD and apply equations of

equilibrium

Total number of unknown force components must

be less than or equal to the number of

independent equations (3 per piece – the 3

equations for the whole frame are NOT

independent of the 3 equations for each piece)


In general:

start with two force element and show their

member forces

guess at force senses, verify later by

looking at signs

then look at multi-force members

remember that Fx and Fy forces are same

magnitude but opposite sense on each side

of a connection

Write equilibrium equations for multi-force

member FBDs and back-substitute answers

to solve all of your FBDs ENGR 1205 7 Chapter 9

Example analysis:

note that pins are assumed to be part of an RB

if an external force is applied at a pin, make sure the

external force is only shown being applied to the member

with the pin in it (if the pin connects two members you can

pick which RB the external force is applied to)


Determine the

components of the

forces acting on each

member of the frame

shown.




Like frames (support loads, made of up multiple

members – at least one supports 3+ loads)

BUT, primary purpose is to transmit and modify

forces

they transform input forces (I/P) into output

forces (O/P)

Generally we want to find the magnitude of the

O/P knowing the I/P … or vice versa (ratio of

output:input is “mechanical advantage”)

Often, machines won’t be fully constrained

because they’re designed to move – but they are

in equilibrium (assume they are at rest)


We apply force P on the pliers in order for the pliers

to overcome force Q to cut a wire

pliers are not rigid, so

break them into two pieces


Use the same method as for frames

Draw FBD for each member

Put opposite forces at connections

Include input forces (act on the machine)

and the REACTIONS TO the output forces

(output forces act on something else – show

what the “something else” does to the

machine)

Often there are no external support reactions


What forces are exerted on the bolt at E as a result of

the 150 N forces on the pliers?




Typical Truss

Note:

floor beams connect at joints

Can analyse each side as a planar (2D) system


trusses are major engineering

structures (bridges & buildings)

they consist of straight (thin)

members connected at joints

the members are only connected

at the ends (not in the middle)

there’s no AB, just AD and DB

Joints are represented as pin

connections (idealizations)

All loads act at the joints

(including member weights - half

at each end)


a rigid truss is a 3 member truss (it won’t collapse,

but 4 members will collapse)

we can construct a larger truss by adding pairs of

members to existing joints, connected at a new joint


we can repeat this procedure

indefinitely

every time we get +2 members

and +1 joints

a truss constructed this way is

called a “simple truss”

the total number of members in it is: m = 2n

-3 (where n = the # of joints)

you don’t always have to have triangles in

trusses


We assume each end of a member has one force

acting on it, along the member, with no couple

(every member is a two-force element)

therefore, a truss can be regarded

as a group of pins and 2F members

where the members are either in

tension or under compression


Member in tension

Member in compression

comes from a member in compression

comes from a member in tension

For two members in tension connected with a pin:

equal & opposite


Consider 4 (2 element) members connected with a pin such

that they lie in 2 intersecting straight lines = 2 pairs of

directly opposite forces where the magnitudes of the forces in

opposing members are equal


now consider and

if 2 of 3 members are in a line and an applied force is in

line with the 3rd member, then the 3rd member’s force is

equal and opposite to the applied force

if there is no such applied force then the force in the 3rd

member = AC is zero

such a member is called a zero force member (ZFM) and

they are worth spotting early in problem solving, to

simplify/speed up the analysis


if only 2 members

connect at a joint

and they have a

common LOA, then

their forces are

equal and opposite

in sense (a), but if

they do not have a

common LOA, then

they will be zero

force members (b)


the trick is to spot these special situations

C connects 3

members, 2 in a line

(no external F at C),

FBC=0

same at K for JK

Now J is the same

along HL, so FIJ=0


also AC=CE, HJ=JL, and IK=KL since ∑ FC = ∑ FK = ∑ FJ = 0

now look at I … FHI=20 & FGI= FIK= FKL

all of this would be very different with different external

loads (the ZFMs would not necessarily still be ZFMs)

you don’t HAVE to catch these tricks, but it helps

dismember a simple truss

each member is a 2F member and has its own FBD

The pins also have FBDs

the magnitude of the force at the end of each member

is called the force in the member (though it’s a scalar)

truss analysis is finding the magnitudes and senses

(compression or tension) of the forces in each member

Becomes


since the truss is an RB in equilibrium, so

each pin (is a particle) has 2 equilibrium equations:

Σ𝐹𝑥 = 0 and Σ𝐹𝑦 = 0

if a truss has n pins, we have 2n equations to solve for

up to 2n unknowns

for a simple truss, m = 2n-3, so 2n = m+3

we can find the forces of all members + 3 rxns (e.g.

RAx, RAy, RBy) using pins as FBDs

the 3 equations we could use for the truss as a whole

are not independent of these pin equations, but you

can use them to find the rxns at the supports


To analyse a truss using the Method of Joints

Start with the joints that only have 2 unknowns

if all joints have 3+ unknowns, then analyze the

whole structure first with the standard equations of

equilibrium (ΣFx= 0,ΣFy= 0,ΣMA=0) to get RA & RB

then do the M of J at each pin using force triangles

(or polygons), or systems of equations

For following truss assuming we know RA & RB :


Start with Joint A

Know RA & dirn of FAC

and FAD

Find magnitude and

sense of FAC and FAD

Move to Joint D

Already found FAD

given P

Know directions of FDC

& FDB

Find magnitude and

sense of FDC and FDB


Move to Joint C

Already found FCA and

FCD

Know direction of FCB

Find magnitude and

sense of FCB

Joint B

Already know all

forces acting

Can use it as a check


Determine the force in each member of the truss shown.



MofS is more efficient if you don’t need to solve for

all of the internal forces, you may only be asked for

forces for one or a few members

MofS is based on the idea that if the whole truss is in

static equilibrium then any section of the truss must

also be in equilibrium

Method:

“Cut” the truss through the members that you are

interested in solving for

Draw the FBD for the section you are left with

Ideally you have 3 unknowns (and 3 equations)

You may have to repeat the process


For example - want FBD in Figure

need effect of FBD on B or D

with MofJ, we would solve for FBD at B or D

with MofS, we pick a chunk of truss where

FBD is an external force and we choose a

chunk of truss with only 3 unknown

external forces so it’s solvable

we now have 3 equations for 3 unknowns

if we only need FBD, then we might get it

with just one equation e.g. ∑ ME = 0

as usual, a (+) sign on your answer means

that you guessed the right direction, while

(-) means you didn’t


Calculate the force in members FH, GH, and GI of the

roof truss shown.





several straight members joined

at ends form a simple 3D/space

truss

similar to 2D trusses, add sets of

3 members to 3 existing joints

and join them at one new joint

(where all 4 joints do not lie in a

plane) to expand a simple space

truss

start with 6 members and 4

joints, and then expand with +3

members and +1 joint each time

(m = 3j – 6)


for a space truss to be statically determinate and

completely constrained, use balls, rollers and ball and

socket joints to get 6 unknown rxns

(then use 6 equations of equilibrium for 3D)

for analyses, assume you have ball & socket joints

although they are often welds

therefore, each member is a 2F member and at each

joint you’ll have

∑ Fx = 0, ∑ Fy = 0, ∑ Fz = 0

in a truss with j joints, you get 3j equations and with

m = 3j-6 you get 3j = m + 6 (can solve for all the Fs in

the members + 6 rxns at supports)

start solving joints with ≤ 3 unknowns


Determine the forces acting on the members of the

space truss shown.





simple trusses simply supported (1 pin and 1 roller) are

completely constrained and determinate trusses

for any other truss, if m = the # of members, j = the # of

joints, r = the # of rxns at supports, then we have 2j

independent equations of equilibrium

if (m+r) < 2j → we have fewer unknowns than equations and the

system is partially constrained (internally unstable)

if (m+r) > 2j → we have more unknowns than equations and the

system is statically determinate (redundancies exists)

if (m+r) = 2j → we have the same # of unknowns as equations

BUT it may not be determinate since some equations may

not be able to be satisfied

ONLY if all rxns and forces in members can be determined

will it be completely constrained and determinate

(otherwise it is improperly constrained)


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Outline: Frames Machines Trusses - Weeblyroneducate.weebly.com/uploads/6/2/3/8/6238184/statics_lecture_6_f... · Outline: Frames Machines Trusses Properties and Types ... no matter