Civil Engineering (CE) Exam > Civil Engineering (CE) Videos > Method of Sections for Truss Analysis Example
Method of Sections for Truss Analysis Example Video Lecture - Civil Engineering
Text Transcript from Video
alright so in this video I'm going to go
over the method of sections by an
example problem here and in this method
of sections problem I have a truss here
this roof truss that's very large it's
quite gotta fight quite a few members
it's every Bay of this trust has about
five feet of spacing and it's like 1 5
10 15 20 25 30 35 40 feet long and it's
a roof truss it's 12 feet tall all the
way up to point E and I am I want to
know in this problem I want to know the
internal forces of members de and E and
nm which are here de n E and nm here
okay and the way that we're going to go
about doing this as with any method of
sections problem on a truss is to
calculate the support reactions of the
truss here these are the three support
reactions we have this a y and an ax at
this support and an I Y here a vertical
reaction here on that roller and then
you're going to isolate and cut and it's
important to isolate and cut through
three elements the reason you have to
cut through three elements is because
when you take that Freebody diagram
you're only going to have three
equations available to you in 2d so
you've got you've got you can only cut
through three elements okay and a lot of
times people look at a truss problem
like this in our attempts to go joint by
joint and that's just going to take you
forever for whatever the problem is
asking you okay unless it's asking you
for every force in every member of the
truss then there's no point in doing
method of joints okay so here we're
going to do method of sections and and
we're going to and then once you make
that cut once we do isolate and cut here
it's like something you might do in
basketball but here you can isolate and
cut and then you're going to apply the
equilibrium equations to that free body
diagram and I'm going to and that's what
we're going to do right now okay so here
let's take this let's take this right
here so let's calculate the reactions
first so let's do the first thing here
one calculating reactions and you know
I'm not going to apply the equations
here I've got a symmetric structure with
a symmetric loading
have a total loading of 10 20 30 40 50
60 70 Kip's okay 70 tips it's symmetric
that means each vertical reaction here
takes half of it so I have seven kids so
this one is 35 Kip's and this is 35 kids
and by some of the forces in the
x-direction some of the forces in the
x-direction equal to zero I know that ax
also equals zero and again by symmetry I
have a y equals iy which equals 35 Kip's
okay so that's that's great now I want
to isolate and cut through the members
that I'm interested in okay and here in
this question is asking me for members
de and E and nm over here so that means
I'm going to make a cut through through
some members okay and here I'm going to
cut through three and we're always going
to cut through three okay it's you just
can't apply the free you can't apply the
equilibrium equations if you cut through
more members in that and and here we're
going to apply the equilibrium equations
to one side of the free body diagram
okay and one rule of thumb to know
whether or not you're making the right
cut is that you should be able to take
either side of the cut to analyze your
structure okay if you can't do that then
really the cut is no good so here I
would let's say I take the left side of
this drawing here so I'm going to take
this blue line this isolation okay so
isolate this this left side of the free
body diagram I've cut it away and and
now really the you know unless you have
a computer program that can erase all
this or something right here you
probably want to redraw it and
thankfully I redrew this in advance here
but let's before we redraw it let's make
sure that we understand some of the
geometry here so I have here we know the
height is 12 feet this member right here
okay this member or this angle right
here this is a 5 12 13 triangle okay
that's a 5 12 13 triangle and then here
this this member de is essentially a I
say 5 10 15 20 20
by twelve by whatever this hypotenuse is
triangle okay
and that's going to be important because
we need these angles when we do whatever
we're going to do and when we apply
these equilibrium equations alright so
so now the thing to do is redraw the
free body diagram of the cut the
isolated portion this blue zone okay and
and then make sure you and one thing
that happens is that people forget the
loading sometimes they forget the
loading go like well what's going on
here okay
make sure you have to include them
loading on the on the cut that you're
interested in so I'm going to take that
I'm gonna redraw it okay so thankfully I
redrew it here and I blew it up a little
bit okay because hey my computer program
lets me do it okay and I have here what
I want to do is from that cut so my cut
zone was here I'm going to draw from
each joint the force that would have
been their intention okay so this is
force let me draw that and put that on
top over here this force de right here
and here kind of get rid of this right
here this would be force force and M and
here oops that's not force n em that's
force n E and here would be force and M
okay and here just to make sure we know
that this line of action of the force
goes through point e this line of action
goes through point he here this line of
action goes through point M now the one
of the things that happens here is that
when you apply the equilibrium equations
you know everyone's tempted to go
straight you know we're going to apply
these equilibrium equations so here is
going to be one now you're going to
apply the equilibrium equations to solve
for these forces and this is part three
really okay so here you know people are
tempted to apply some of the forces in
the x equals zero because they're so
familiar with it and some of the forces
in the Y equals zero and this being the
vertical here and look let's just let's
just see what this does so let's call
this angle before we figure it out let's
call this alpha call this angle theta
okay alpha and theta there and so if I
do sum of the forces in the x equals
zero here okay if I apply some of the
forces in the X here I would get like I
would get F and M this free body this
whole body deal
okay FN m + f n e cosine of theta plus
FD e cosine of alpha equals to zero so I
have a you know I have kind of a
complicated equation a lot of you know
each variable appears here then I do sum
of the forces in Y I get F M isn't there
so I got F n e sine of theta plus F de
cosine of alpha equals to zero and then
I have to do sum of the moments about a
point and to get that you know it's just
it's gonna and I'm gonna have to solve
I'm gonna have to go through a lot of
algebra to solve for these you know with
this right here if you can take moments
about the right point you can get one at
a time one equation want to know you can
set it up with one equation one unknown
okay so really the power here in
applying the equilibrium equations take
sum of moments all right so let's take
let's do the first thing let's take
summer moments about a point let's start
with point let's say I want to isolate
points I want to isolate a specific
force so let's take here I have two
forces running through this line of
action over here through point point e
so if I take moments about point E I'm
going to have F M all by itself some
moments about point a here some moments
about point e equal to 0 I here's my
point E over here I'll take moments
counterclockwise is positive right here
I always like to draw write this out the
sum of moments equal to 0 and in this
direction here it just it's a reminder
to me what I'm doing okay it's just a
good technique to have and here I would
have so here let's see I know I have no
contribution of moment about point E due
to F de because the line of action goes
through point E same thing with F any
all I have is f NM and the perpendicular
distance to that point or the shortest
distance is 12 feet okay then I have in
addition to that about point e I have
plus
ten Kip's which is a distance these are
all five feet five feet five feet and
five feet here plus ten Kip's times five
feet okay plus ten more tips times ten
feet plus ten tips times fifteen feet
minus ay Y which is 35 Kip's times 5 10
15 20 feet all equal to zero okay and so
here great I have one equation one
unknown and I and I work out the algebra
and we could get geeky the up F n M
equals thirty three point three three
Kip's okay thirty three point three
three tips I got a positive result here
this thirty three point tips that
implies that this if any is in the
correct direction which SN m is in the
correct direction which means that this
member nm is in tension okay so this
sometimes you write this as thirty three
point three K and then you'll just write
the letter T to indicate tension okay
now and now we can have we can take
moments about two other locations to
calculate those forces directly and
another one is I look so now I'm
interested in n e if I look at nm I have
a line of action that goes like this
and de goes like this and where
intersects I find this point a here this
point a so now I'm going to take moments
about point a okay so let's let's take
moments about point a and move down low
okay Liz I'm going all over the place
right here so if I take moments about
point a right here some moments about
point a equal to zero I have this is
positive right here let's see so for
moments about point a I've got let's see
I have F and E okay and really if I if I
break up F and E right here so let me
use let me use black here let me use
orange right here FN e will be broken up
into a horizontal
a vertical component okay a horizontal
and vertical component right here this
this component right here this component
is the only one that causes a moment
about point a so this would be FN e sine
of this angle theta times the arm which
is 5 10 15 15 feet 15 feet then and
everything so that causes a positive
moment about point a and then plus oops
not plus if I once look at the loading
here I have everything else the loading
- 10 Kip's times 15 feet - 10 Kip's
times 10 feet - 10 Kip's times 5 feet
equals zero and I can't include a Y
anymore because a Y again goes through
point a and so here and sine theta is
really just sine theta this triangle if
you recall this was a 5 12 13 triangle
so sine theta is just equal to 12
divided by 13 and if I plug that into
there I'm going to get F and E is equal
to a twenty one point six seven Kip's
this is going to be positive indicating
tension okay and now for the last one
okay I have to find the force and de
right here I got to find this F de going
on right here and and to do that you
know I look at again where does FN e f
and f NM these two members coincide in
the cosine a point n so I'll just take
moments about point n I'm going to zoom
in a little bit on the screen so you can
see a little bit better okay so here and
and I have I will maybe right over here
I'll take moments about point n okay
about point n equal to zero and I'll say
here this is positive right here and
again I will have again if I break up
this de I'll have a
a horizontal component and a vertical
component of de and the only component
that induces a moment about point n is
this component right here so I will have
minus F de cosine of alpha times the
distance here okay
the distance here this distance here and
by similar triangles I know that this
height is nine feet okay so I would
compare really to determine this height
so let's say this is is y here so I'm
going to show you similar triangles real
fast I've got here this is five ten
fifteen so I've got Y divided by fifteen
feet is equal to twelve feet divided by
five ten fifteen twenty feet okay and if
I solve for this right here just by
ratios here y equals 15 over twenty
which is 3/4 of twelve which is thirty
nine feet so y equals nine feet and
that's how I know that's nine feet here
so this distance is nine feet so cosine
alpha times nine feet and here I have
plus so at point n here this ten Kip's
does nothing in terms of deucing moment
plus I have ten Kip's loading times five
feet plus ten Kip's times fifteen feet
and then I have the a Y inducing a
negative moment - thirty-five Kip's
times twenty feet
so like 20 feet no five ten fifteen
should be 15 feet 15 feet equal to zero
all right and and now this now that I'm
pretty happy with my equilibrium
equation here just I just gotta figure
out what this cosine alpha is and and
here just to let's see clarify again
here this this point right here is you
know this this angle right here this
angle alpha is the same as this angle
right here okay
and so really I can there's I have a
number of ways right I can find the
hypotenuse calculate cosine alpha from
this you know just using the ratio of
triangles or I just calculate this angle
alpha and and really what I find is that
in a in this triangle that is 20 20 by
12 this is from you know 20 squared this
20 squared plus 12 squared square root
that number is just twenty three point
three two and that makes cosine alpha
right here alpha if this is twenty three
point three two cosine alpha is 20/20
three point three two
so all I've got to do put in here is 20
divided by twenty three point three to
solve for fde okay so solve for F de
which is I'm going to get a negative
result negative forty eight point five
eight kits which is forty eight point
five eight kits in compression so
element de is in compression okay
and so here these one two and three are
my final results okay
hope that was helpful let me know if you
have any questions like the video if you
liked it subscribe you know do your
thing
good luck luck
over the method of sections by an
example problem here and in this method
of sections problem I have a truss here
this roof truss that's very large it's
quite gotta fight quite a few members
it's every Bay of this trust has about
five feet of spacing and it's like 1 5
10 15 20 25 30 35 40 feet long and it's
a roof truss it's 12 feet tall all the
way up to point E and I am I want to
know in this problem I want to know the
internal forces of members de and E and
nm which are here de n E and nm here
okay and the way that we're going to go
about doing this as with any method of
sections problem on a truss is to
calculate the support reactions of the
truss here these are the three support
reactions we have this a y and an ax at
this support and an I Y here a vertical
reaction here on that roller and then
you're going to isolate and cut and it's
important to isolate and cut through
three elements the reason you have to
cut through three elements is because
when you take that Freebody diagram
you're only going to have three
equations available to you in 2d so
you've got you've got you can only cut
through three elements okay and a lot of
times people look at a truss problem
like this in our attempts to go joint by
joint and that's just going to take you
forever for whatever the problem is
asking you okay unless it's asking you
for every force in every member of the
truss then there's no point in doing
method of joints okay so here we're
going to do method of sections and and
we're going to and then once you make
that cut once we do isolate and cut here
it's like something you might do in
basketball but here you can isolate and
cut and then you're going to apply the
equilibrium equations to that free body
diagram and I'm going to and that's what
we're going to do right now okay so here
let's take this let's take this right
here so let's calculate the reactions
first so let's do the first thing here
one calculating reactions and you know
I'm not going to apply the equations
here I've got a symmetric structure with
a symmetric loading
have a total loading of 10 20 30 40 50
60 70 Kip's okay 70 tips it's symmetric
that means each vertical reaction here
takes half of it so I have seven kids so
this one is 35 Kip's and this is 35 kids
and by some of the forces in the
x-direction some of the forces in the
x-direction equal to zero I know that ax
also equals zero and again by symmetry I
have a y equals iy which equals 35 Kip's
okay so that's that's great now I want
to isolate and cut through the members
that I'm interested in okay and here in
this question is asking me for members
de and E and nm over here so that means
I'm going to make a cut through through
some members okay and here I'm going to
cut through three and we're always going
to cut through three okay it's you just
can't apply the free you can't apply the
equilibrium equations if you cut through
more members in that and and here we're
going to apply the equilibrium equations
to one side of the free body diagram
okay and one rule of thumb to know
whether or not you're making the right
cut is that you should be able to take
either side of the cut to analyze your
structure okay if you can't do that then
really the cut is no good so here I
would let's say I take the left side of
this drawing here so I'm going to take
this blue line this isolation okay so
isolate this this left side of the free
body diagram I've cut it away and and
now really the you know unless you have
a computer program that can erase all
this or something right here you
probably want to redraw it and
thankfully I redrew this in advance here
but let's before we redraw it let's make
sure that we understand some of the
geometry here so I have here we know the
height is 12 feet this member right here
okay this member or this angle right
here this is a 5 12 13 triangle okay
that's a 5 12 13 triangle and then here
this this member de is essentially a I
say 5 10 15 20 20
by twelve by whatever this hypotenuse is
triangle okay
and that's going to be important because
we need these angles when we do whatever
we're going to do and when we apply
these equilibrium equations alright so
so now the thing to do is redraw the
free body diagram of the cut the
isolated portion this blue zone okay and
and then make sure you and one thing
that happens is that people forget the
loading sometimes they forget the
loading go like well what's going on
here okay
make sure you have to include them
loading on the on the cut that you're
interested in so I'm going to take that
I'm gonna redraw it okay so thankfully I
redrew it here and I blew it up a little
bit okay because hey my computer program
lets me do it okay and I have here what
I want to do is from that cut so my cut
zone was here I'm going to draw from
each joint the force that would have
been their intention okay so this is
force let me draw that and put that on
top over here this force de right here
and here kind of get rid of this right
here this would be force force and M and
here oops that's not force n em that's
force n E and here would be force and M
okay and here just to make sure we know
that this line of action of the force
goes through point e this line of action
goes through point he here this line of
action goes through point M now the one
of the things that happens here is that
when you apply the equilibrium equations
you know everyone's tempted to go
straight you know we're going to apply
these equilibrium equations so here is
going to be one now you're going to
apply the equilibrium equations to solve
for these forces and this is part three
really okay so here you know people are
tempted to apply some of the forces in
the x equals zero because they're so
familiar with it and some of the forces
in the Y equals zero and this being the
vertical here and look let's just let's
just see what this does so let's call
this angle before we figure it out let's
call this alpha call this angle theta
okay alpha and theta there and so if I
do sum of the forces in the x equals
zero here okay if I apply some of the
forces in the X here I would get like I
would get F and M this free body this
whole body deal
okay FN m + f n e cosine of theta plus
FD e cosine of alpha equals to zero so I
have a you know I have kind of a
complicated equation a lot of you know
each variable appears here then I do sum
of the forces in Y I get F M isn't there
so I got F n e sine of theta plus F de
cosine of alpha equals to zero and then
I have to do sum of the moments about a
point and to get that you know it's just
it's gonna and I'm gonna have to solve
I'm gonna have to go through a lot of
algebra to solve for these you know with
this right here if you can take moments
about the right point you can get one at
a time one equation want to know you can
set it up with one equation one unknown
okay so really the power here in
applying the equilibrium equations take
sum of moments all right so let's take
let's do the first thing let's take
summer moments about a point let's start
with point let's say I want to isolate
points I want to isolate a specific
force so let's take here I have two
forces running through this line of
action over here through point point e
so if I take moments about point E I'm
going to have F M all by itself some
moments about point a here some moments
about point e equal to 0 I here's my
point E over here I'll take moments
counterclockwise is positive right here
I always like to draw write this out the
sum of moments equal to 0 and in this
direction here it just it's a reminder
to me what I'm doing okay it's just a
good technique to have and here I would
have so here let's see I know I have no
contribution of moment about point E due
to F de because the line of action goes
through point E same thing with F any
all I have is f NM and the perpendicular
distance to that point or the shortest
distance is 12 feet okay then I have in
addition to that about point e I have
plus
ten Kip's which is a distance these are
all five feet five feet five feet and
five feet here plus ten Kip's times five
feet okay plus ten more tips times ten
feet plus ten tips times fifteen feet
minus ay Y which is 35 Kip's times 5 10
15 20 feet all equal to zero okay and so
here great I have one equation one
unknown and I and I work out the algebra
and we could get geeky the up F n M
equals thirty three point three three
Kip's okay thirty three point three
three tips I got a positive result here
this thirty three point tips that
implies that this if any is in the
correct direction which SN m is in the
correct direction which means that this
member nm is in tension okay so this
sometimes you write this as thirty three
point three K and then you'll just write
the letter T to indicate tension okay
now and now we can have we can take
moments about two other locations to
calculate those forces directly and
another one is I look so now I'm
interested in n e if I look at nm I have
a line of action that goes like this
and de goes like this and where
intersects I find this point a here this
point a so now I'm going to take moments
about point a okay so let's let's take
moments about point a and move down low
okay Liz I'm going all over the place
right here so if I take moments about
point a right here some moments about
point a equal to zero I have this is
positive right here let's see so for
moments about point a I've got let's see
I have F and E okay and really if I if I
break up F and E right here so let me
use let me use black here let me use
orange right here FN e will be broken up
into a horizontal
a vertical component okay a horizontal
and vertical component right here this
this component right here this component
is the only one that causes a moment
about point a so this would be FN e sine
of this angle theta times the arm which
is 5 10 15 15 feet 15 feet then and
everything so that causes a positive
moment about point a and then plus oops
not plus if I once look at the loading
here I have everything else the loading
- 10 Kip's times 15 feet - 10 Kip's
times 10 feet - 10 Kip's times 5 feet
equals zero and I can't include a Y
anymore because a Y again goes through
point a and so here and sine theta is
really just sine theta this triangle if
you recall this was a 5 12 13 triangle
so sine theta is just equal to 12
divided by 13 and if I plug that into
there I'm going to get F and E is equal
to a twenty one point six seven Kip's
this is going to be positive indicating
tension okay and now for the last one
okay I have to find the force and de
right here I got to find this F de going
on right here and and to do that you
know I look at again where does FN e f
and f NM these two members coincide in
the cosine a point n so I'll just take
moments about point n I'm going to zoom
in a little bit on the screen so you can
see a little bit better okay so here and
and I have I will maybe right over here
I'll take moments about point n okay
about point n equal to zero and I'll say
here this is positive right here and
again I will have again if I break up
this de I'll have a
a horizontal component and a vertical
component of de and the only component
that induces a moment about point n is
this component right here so I will have
minus F de cosine of alpha times the
distance here okay
the distance here this distance here and
by similar triangles I know that this
height is nine feet okay so I would
compare really to determine this height
so let's say this is is y here so I'm
going to show you similar triangles real
fast I've got here this is five ten
fifteen so I've got Y divided by fifteen
feet is equal to twelve feet divided by
five ten fifteen twenty feet okay and if
I solve for this right here just by
ratios here y equals 15 over twenty
which is 3/4 of twelve which is thirty
nine feet so y equals nine feet and
that's how I know that's nine feet here
so this distance is nine feet so cosine
alpha times nine feet and here I have
plus so at point n here this ten Kip's
does nothing in terms of deucing moment
plus I have ten Kip's loading times five
feet plus ten Kip's times fifteen feet
and then I have the a Y inducing a
negative moment - thirty-five Kip's
times twenty feet
so like 20 feet no five ten fifteen
should be 15 feet 15 feet equal to zero
all right and and now this now that I'm
pretty happy with my equilibrium
equation here just I just gotta figure
out what this cosine alpha is and and
here just to let's see clarify again
here this this point right here is you
know this this angle right here this
angle alpha is the same as this angle
right here okay
and so really I can there's I have a
number of ways right I can find the
hypotenuse calculate cosine alpha from
this you know just using the ratio of
triangles or I just calculate this angle
alpha and and really what I find is that
in a in this triangle that is 20 20 by
12 this is from you know 20 squared this
20 squared plus 12 squared square root
that number is just twenty three point
three two and that makes cosine alpha
right here alpha if this is twenty three
point three two cosine alpha is 20/20
three point three two
so all I've got to do put in here is 20
divided by twenty three point three to
solve for fde okay so solve for F de
which is I'm going to get a negative
result negative forty eight point five
eight kits which is forty eight point
five eight kits in compression so
element de is in compression okay
and so here these one two and three are
my final results okay
hope that was helpful let me know if you
have any questions like the video if you
liked it subscribe you know do your
thing
good luck luck
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