Transcript WORKSHEET 8 Beams to answer just click on the button or image related to the answer Question 1 floor joists are at 600mm centres.

WORKSHEET 8
Beams
to answer just click on the button or image related to the answer
Question 1
floor joists are at 600mm centres and span 2.0m between bearers,
draw the configuration
what is the tributary area for one joist?
a
12 m2
b
1.2 m2
c
1,200,000 mm2
Question 2
given a floor 18 m x 18 m with columns on a 6m x 6m grid,
draw the configuration
what is the tributary area for an internal column
a
36 m2
b
18 m2
c
324 m2
Question 3
given a floor 18 m x 18 m with columns on a 6m x 6m grid,
draw the configuration
what is the tributary area for an edge column
a
36 m2
b
18 m2
c
324 m2
Question 4
given a floor 18 m x 18 m with columns on a 6m x 6m grid,
draw the configuration
what is the tributary area for a corner column
a
36 m2
b
18 m2
c
9 m2
Question 5
a roof weighing 0.4 kPa spans between roof trusses
which are at 2.5 m centres and span 10m. We want to
determine the total load on a truss.
what is the first thing we need to do?
a
determine the load per metre on a truss
b
determine the tributary area for a truss
c
determine the bending moment on the truss
Question 6
a roof weighing 0.4 kPa spans between roof trusses
which are at 2.5 m centres and span 10m. We want to
determine the total load on a truss.
what is the tributary area for a truss?
a
2.5 m2
b
10 m2
c
25 m2
Question 7
a roof weighing 0.4 kPa spans between roof trusses
which are at 2.5 m centres and span 10m. We want to
determine the total load on a truss.
what is the total load on a truss?
(neglecting the self-weight)
a
10 kPa
b
10 kN
c
100 kN
Question 8
a roof weighing 0.4 kPa spans between roof trusses
which are at 2.5 m centres and span 10m. We want to
determine the total load on a truss.
is this a UDL or a point load?
a
UDL
b
point load
Question 9
a roof weighing 0.4 kPa spans between roof trusses
which are at 2.5 m centres and span 10m. We want to
determine the total load on a truss.
what is the load per metre on a truss?
(neglecting the self-weight)
a
1 kPa
b
1 kNm
c
1 kN/m
Question 10
what are the two main types of stress
involved in beam action?
a
buckling and shear
b
tension and compression
c
bending and shear
d
bending and buckling
Question 11
in buildings which is more important?
a
shear
b
bending
Question 12
in buildings bending is more important than shear
why?
a
spans are large
b
we design for bending and check for shear
c
loads are light
d
spans are large relative to loads
Question 13
what is the sign convention
for BMD for sagging?
a
positive
b
negative
Question 14
what is the sign convention
for BMD for hogging?
a
positive
b
negative
Question 15
what does a Shear Force Diagram tell you?
a
where the maximum shear force occurs
b
where the maximum shear stress occurs
c
the values of the shear force along the beam
d
a and c
e
b and c
Question 16
what does a Bending Moment Diagram tell you?
a
where the maximum bending moment occurs
b
where the maximum load occurs
c
the values of the bending moment along the beam
d
a and c
e
a, b and c
Question 17a
given the beam loaded as shown
2m
16 kN
4m
what’s the first thing we do?
a
calculate the maximum bending moment
b
calculate the maximum shear force
c
calculate the reactions
Question 17b
2m
4m
given the beam loaded as shown
RL
what are the reactions?
a
RL = 10 kN, RR = 6 kN
b
RL = 16 kN, RR = 16 kN
c
RL = 8 kN, RR = 8 kN
16 kN
RR
Question 17c
given the beam loaded as shown
draw the deflected shape
does the beam sag or hog?
a
sag
b
hog
2m
16 kN
4m
RL= 8 kN
RR = 8 kN
Question 17d
given the beam loaded as shown
draw the deflected shape
is the Bending Moment?
a
negative
b
positive
c
positive and negative
2m
16 kN
4m
RL= 8 kN
RR = 8 kN
Question 17e
given the beam loaded as shown
draw the Shear Force Diagram (SFD)
is the SFD?
a
block shaped
b
triangular shaped
2m
16 kN
4m
RL= 8 kN
RR = 8 kN
Question 17f
given the beam loaded as shown
draw the Shear Force Diagram (SFD)
2m
4m
RL= 8 kN
what is the maximum Shear Force?
a
16 kNm
b
16 kN
c
8 kN
d
8 kNm
16 kN
RR = 8 kN
Question 17g
given the beam loaded as shown
draw the Shear Force Diagram (SFD)
2m
16 kN
4m
RL= 8 kN
RR = 8 kN
where does the maximum Shear Force occur?
a
at the centre of the beam
b
at the ends of the beam
c
all along the beam
Question 17h
given the beam loaded as shown
draw the Bending Moment Diagram (BMD)
is the BMD?
a
trapezoidal / triangular
b
parabolic
2m
16 kN
4m
RL= 8 kN
RR = 8 kN
Question 17i
given the beam loaded as shown
draw the Bending Moment Diagram (BMD)
2m
4m
RL= 8 kN
what is the maximum Bending Moment?
a
16 kNm
b
32 kNm
c
8 kNm
d
16 kN/m
16 kN
RR = 8 kN
Question 17j
given the beam loaded as shown
draw the Bending Moment Diagram (BMD)
2m
16 kN
4m
RL= 8 kN
RR = 8 kN
where does the maximum Bending Moment occur?
a
at the centre of the beam
b
at the ends of the beam
c
all along the beam
Question 18a
given the beam loaded as shown
UDL 5kN/m
2m
what’s the first thing we do?
a
calculate the maximum bending moment
b
calculate the maximum shear force
c
calculate the reactions
Question 18b
UDL 5kN/m
given the beam loaded as shown
2m
what is the vertical reaction?
RV
a
5 kN
b
10 kN
c
20 kN
Question 18c
given the beam loaded as shown
draw the deflected shape
does the beam sag or hog?
a
sag
b
hog
UDL 5kN/m
2m
RV = 10 kN
Question 18d
given the beam loaded as shown
draw the deflected shape
UDL 5kN/m
2m
is the Bending Moment?
RV = 10 kN
a
negative
b
positive
c
positive and negative
Question 18e
given the beam loaded as shown
draw the Shear Force Diagram (SFD)
UDL 5kN/m
2m
is the SFD?
RV = 10 kN
a
block shaped
b
triangular shaped
Question 18f
UDL 5kN/m
2m
given the beam loaded as shown
draw the Shear Force Diagram (SFD)
RV = 10 kN
what is the maximum Shear Force?
a
5 kNm
b
10 kN
c
10 kNm
d
5 kN
Question 18g
UDL 5kN/m
2m
given the beam loaded as shown
draw the Shear Force Diagram (SFD)
RV = 10 kN
where does the maximum Shear Force occur?
a
at the centre of the beam
b
at the left end of the beam / at the support
c
at the right end of the beam
d
same all along the beam
Question 18h
UDL 5kN/m
2m
given the beam loaded as shown
draw the Bending Moment Diagram (BMD)
RV = 10 kN
is the BMD?
a
trapezoidal / triangular
b
parabolic
Question 18i
UDL 5kN/m
2m
given the beam loaded as shown
draw the Bending Moment Diagram (BMD)
RV = 10 kN
what is the maximum Bending Moment?
a
- 20 kNm
b
- 10 kNm
c
- 5 kNm
d
- 2.5 kNm
Question 18j
UDL 5kN/m
2m
given the beam loaded as shown
draw the Bending Moment Diagram (BMD)
RV = 10 kN
where does the maximum Bending Moment occur?
a
at the centre of the beam
b
at the right end of the beam
c
at the support
tributary area
600mm 600mm
tributary area = 2 m x 0.6 m = 1.2 m2
next question
enough !
how did you get that?
The length of a joist is 2 m and
the joists are at 0.6 m centres
let me try again
let me out of here
the number is right.
But these are stupid units for an area this size
let me try again
let me out of here
tributary area
6 m x 6 m = 36 m2
next question
enough !
6m 6m 6m
6m 6m 6m
How did you get that? The columns are on a 6 m x 6 m grid.
We are talking about an internal column
let me try again
let me out of here
6 m x 3 m = 18 m2
next question
enough !
6m 6m 6m
tributary area
How did you get that? The columns are on a 6 m x 6 m grid.
We are talking about a column on the edge.
What are the distances to the neighbouring columns?
let me try again
let me out of here
3 m x 3 m = 9 m2
next question
enough !
6m 6m 6m
tributary area
How did you get that? The columns are on a 6 m x 6 m grid.
We are talking about a column at a corner.
What are the distances to the neighbouring columns?
let me try again
let me out of here
That’s right. Before we can work out the load on a member,
We have to work out what that member is carrying
next question
enough !
We are after the TOTAL LOAD
We need to find out what the truss is carrying
let me try again
let me out of here
10 m x 2.5 m = 25 m2
next question
enough !
The length of the truss is 10 m
The trusses are at 2.5 m centres
let me try again
let me out of here
The tributary area is 25 m2 the load is 0.4kPa
25m2 x 0.4 kPa = 10 kN
(remember 1 kPa = 1 kN/m2)
next question
enough !
We are talking about the TOTAL LOAD
What are the units of a load (force)?
let me try again
let me out of here
How did you get that?
What’s the tributary area? What’s the load per sq m?
let me try again
let me out of here
If we look at the truss we see that
the load is distributed over its length
next question
enough !
How did you arrive at the conclusion that all the load
that the truss carries is concentrated at one point?
let me try again
let me out of here
you’ve got it !!
10 kN / 10 m = 1 kN /m
next question
enough !
we are talking about a force over a distance
let me try again
let me out of here
we are talking about a force over a distance
let me try again
let me out of here
you’ve got it !!
Beam action means that a beam may fail
in either bending or in shear
next question
enough !
Where did buckling come into it?
We’re talking about a beam
let me try again
let me out of here
Yes but that’s what happens internally
but it’s not the main actions that we ascribe to beams
let me try again
let me out of here
brilliant
next question
enough !
Don’t guess
let me try again
let me out of here
It’s not that loads are necessarily light or
that spans are so large.
It is relative
next question
enough !
Spans are not necessarily so large
let me try again
let me out of here
Yes, but that’s the consequence
not the reason
let me try again
let me out of here
Loads in buildings are not
necessarily light
let me try again
let me out of here
+
Yes, remember the happy smile
next question
enough !
Is it happy or sad?
let me try again
let me out of here
Yes, remember the sad face
next question
enough !
Is it happy or sad?
let me try again
let me out of here
+12.5 kN
You can see the values of the shear force
at any point along the beam and you can see
where the maximum shear force occurs
next question
enough !
-5 kN
-7.5 kN
It’s not the whole story
let me try again
let me out of here
Shear Force Diagrams show forces.
Stresses depend on the section of the beam
let me try again
let me out of here
-10 kNm
You can see the values of the bending moment
at any point along the beam and you can see
where the maximum bending moment occurs
next question
enough !
~+5.6 kNm
It’s not the whole story
let me try again
let me out of here
Bending Moment Diagrams don’t show loads
let me try again
let me out of here
2m
16 kN
4m
RL
have to determine ALL the forces and that means
determining the reactions
next question
enough !
RR
We could but that’s
not the first thing we should do
We really should determine all the forces acting
let me try again
let me out of here
How can we do that?
What do we need to know first?
let me try again
let me out of here
2m
16 kN
4m
RL= 8 kN
The system is symmetrical and so
the reactions are also symmetrical.
Since ΣV = 0 and the total downward load is 16 kN
The total upward load equals 16 kN
next question
enough !
RR = 8 kN
have a look at the system
work smarter
let me try again
let me out of here
What’s the downward load?
Think of the equations of static equilibrium
let me try again
let me out of here
16 kN
Yes, as you might expect
next question
enough !
and hogs may fly
let me try again
let me out of here
+
Yes, remember the happy smile
next question
enough !
Is it happy or sad?
let me try again
let me out of here
Yes, point loads produce block-shaped
Shear Force Diagrams
next question
enough !
try again
Remember how you draw the SFD by following the forces
let me try again
let me out of here
+8 kN
- 8 kN
you just ‘follow’ the forces
up 8, across, down 16, across, up 8.
next question
enough !
we are talking about a shear force
what are the units of force?
let me try again
let me out of here
how did you get that?
Look at your answer for Question 17 b
let me try again
let me out of here
+8 kN
- 8 kN
The signing of positive and negative shear forces is
just conventional. There is no difference in the effect.
So the shear force of 8 kN is constant all along the beam
next question
enough !
THINK !!
Look again at the Shear Force Diagram
let me try again
let me out of here
16 kN
triangular
just like a string loaded similarly
next question
let me out of here
remember the string
let me try again
let me out of here
that’s it exactly!
WL/4 = 16 x 4 / 4 = +16 kNm
next question
enough !
Not right !!
Look up the formula for a single point load at the centre
let me try again
let me out of here
THINK !!
This is not load per metre
We are talking about moments – force x distance
let me try again
let me out of here
16 kN
that’s it exactly!
next question
enough !
THINK !! GO BACK TO THE BMD
In a simply supported beam the ends cannot produce moment reactions
let me try again
let me out of here
THINK !! GO BACK TO THE BMD
let me try again
let me out of here
UDL 5kN/m
2m
have to determine ALL the forces and that means
determining the reactions
next question
enough !
We could but that’s
not the first thing we should do
We really should determine all the forces acting
let me try again
let me out of here
How can we do that?
What do we need to know first?
let me try again
let me out of here
UDL 5kN/m
2m
V = 10 kN
Since ΣV = 0 and the total downward load is 5 x 2 = 10 kN
The total upward load equals 10 kN
next question
enough !
What is the TOTAL downward load?
let me try again
let me out of here
Yes, as you would expect !
10 kN
next question
enough !
not correct
let me try again
let me out of here
Yes, remember the sad smile
next question
enough !
Is it happy or sad?
It’s not both
let me try again
let me out of here
Yes, UDLS* produce triangular-shaped
Shear Force Diagrams
*UDL = Uniformly Distributed Load
next question
enough !
try again
Remember how you draw the SFD by following the forces.
Think of a UDL as a series of little point forces
let me try again
let me out of here
+10 kN
you just ‘follow’ the forces
up 10, across, down (a little), across, down (a little), ….
next question
enough !
we are talking about a shear force
what are the units of force?
let me try again
let me out of here
how did you get that?
Look at your answer for Question 18 b
let me try again
let me out of here
+10 kN
As shown in the SFD,
the maximum shear force of 10kN occurs at the support
next question
enough !
THINK !!
Look again at the Shear Force Diagram
let me try again
let me out of here
parabolic
UDLs produce parabolic-shaped BMDs
next question
let me out of here
Think again
let me try again
let me out of here
that’s it exactly!
-wL2/2
= -5 x 2 x 2 / 2
=
-10 kNm
next question
enough !
Not right !!
Look up the formula for a UDL load on a cantilever
let me try again
let me out of here
You’ve graduated with honours!
-wL2/2
= -5 x 2 x 2 / 2
=
-10 kNm
FINISH
THINK !! GO BACK TO THE BMD
let me try again
let me out of here
THINK !! GO BACK TO THE BMD
How can a free end of a cantilever produce a moment?
let me try again
let me out of here