Transcript WORKSHEET 2 FORCES, MOMENTS, LOADS & SUPPORTS

WORKSHEET 4
BEAMS
Q1
Given that floor joists are at 600mm centres and span 2.0m
between bearers, what is the tributary area for one joist?
tributary area
600mm
600mm
tributary area for joist = 2 x 0.6 = 1.2 m2
Q2
(ii) a column on the edge
6 x 3 = 18 m2
(ii) a corner column
3 x 3 = 9 m2
6m 6m 6m
6 x 6 = 36 m2
6m 6m 6m
(i) an internal column
6m 6m 6m
Given a floor 18 m x 18 m with columns on a 6m x 6m grid,
what is the tributary area for:
6m 6m 6m
Q3
Given the values in the Building Principles Notes for the
Dead Loads of materials (P17), determine the dead load of the
roof/ceiling construction shown below
6mm corrugated fibre cement sheet - 0.11kN/m2
100 x 50 hardwood rafters @ 600mm crs - 11 kN /m3
13mm plasterboard ceiling - 0.22kN/m2
Another way of doing this is to say that 1sqm can be achieved
by an area 0.6 wide x 1.67 long (1 / 0.6).
Weight of rafter 1.67m long = 0.1 x 0.05 x 1.67 x 11 = 0.09 kN
Weight of 1 sq m fibre cement = 1 x 0.11
= 0.11 kN
Weight of 1 sq m plasterboard = 1 x 0.22
= 0.22 kN
Total weight of roof/ceiling per sq m
= 0.42 kN / m2
0.6
1.67
We are after a 1sq m of roof, but the rafters are at 600mm centres
so that 1m width of roof will contain 1.67 rafters (1 / 0.6).
1.67 x 0.6 = 1.0
= 0.42 kPa
Q4
The roof above spans between roof trusses which are
at 2.5 m centres and span 10m
2.5m
a) sketch the layout described and
indicate the tributary area for
one truss
b) what is the total load
on one truss? (neglecting the
self-weight of the truss)
Tributary area
= 2.5 x 10
2.5m
2.5m
= 25 m2
Total load
= 25 x 0.42
= 10.5 kN
c) what is load per metre on one truss?
Load per metre
= 10.5 / 10
=1.05kN / m
Note:
We have neglected the
self-weight of the truss
Q5
What are the two main types of stress
involved in beam action?
a) bending
b) shear
Q6
In buildings:
a) which of the above two (bending & shear)
is more important?
bending
b) why?
have bigger spans relative to loads.
In the design of machines have short spans
with heavy loads and shear more important.
Q7
What does shear force do to:
a) timber beams?
can cause horizontal splitting along grain
b) steel beams?
not so critical - make sure don’t exceed
allowable shear stress
c) concrete?
tends to cause diagonal tension cracks
near supports
Q8
How is shear resisted in concrete beams:
a) steel reinforcement at 450
b) stirrups
Q9
What is the sign convention for
Bending Moment Diagrams for:
a) sagging?
+
positive
b) hogging?
negative
-
Q10
a) What does a Shear Force Diagram tell you?
the values of the shear force along the beam
you can see where the maximum shear force
occurs
b) What does a Bending Moment Diagram tell you?
the values of the bending moment along the beam
you can see where the maximum bending moment
occurs and whether it is positive or negative
Q11
For each of the Following Loading Conditions
a) Sketch the deflected shape and note where positive
and negative bending moments are expected to occur
b) Find the reactions
c) Draw the Shear Force Diagrams
d) Find the maximum bending moment(s) and
draw the Bending Moment Diagrams
draw the diagrams approximately to scale (i.e. in proportion) and mark
significant values
make use of symmetry and standard Bending Moment coefficients
where appropriate
Q11 A & B
2m
A
B
16 kN
UDL 5kN/m
4m
4m
Deflected Shape
+
8 kN
10 kN
8 kN
+8 kN
- 8 kN
+
SFD
+10 kN
BMD
WL/4 = 16 x 4 / 4 = +16 kNm
10 kN
wL2/8 = 5 x 4 x 4 / 8 = +10 kNm
- 10 kN
Q11 C & D
C
D
10 kN
UDL 5kN/m
2m
-
2m
Deflected Shape
W=wxL
=5x2
=
10 kN
SFD
+10 kN
R =10 kN
+10 kN
-WL = -20 kNm
-
BMD
-wL2/2
= -5 x 2 x 2 / 2
=
-10 kNm
Q11 E
20kN 20kN
2m
2m 1m
TL = 20 + 20
= 40 kN
+16 kN
5m
For reactions
A
Moments about A
RR x 5 = 20 x 2 +20 x 4
= 120
RR
= 24 kN
RL
= 16 kN
16 kN
Deflected Shape
B
C
D
SFD
-4 kN
-24 kN
+
Moment at B
= 16 x 2
= 32 kNm
Moment at C
= 24 x 1
24 kN
= 24 kNm
BMD
24 kNm
32 kNm
Q11 F
TL = 10 + 5
= 15 kN
1m
10kN
1m
+15 kN
5kN
+5 kN
SFD
2m
For reactions
Moment at A
= 10 x 1 + 5 x 2
= 20 kNm
A
B
C
-
Moment at A
= - 20 kNm
Moment at B
=-5x1
= - 5 kNm
-20 kNm
-5 kNm
15 kN
Deflected Shape
BMD
Q11 G
+12.5 kN
5kN
For reactions
Take Moment at C
RL x 4 = 5 x 6 + 20 x 2
= 70
RL = 17.5kN
RR = 7.5kN
SFD
2m
4m
A
B
Moment at B
= 7.5 x 2
= 15kNm
+
Deflected Shape
17.5 kN
C
Moment at A
= -5 x 2
= -10 kNm
7.5 kN WL/4 = 20x4/4
= 20kNm
BMD
-5 kN
-7.5 kN
-10 kNm
20 kNm
TL = 20 + 5
= 25 kN
20kN
+15 kNm
Q11 H
30kN
For reactions
Take Moment at C
RL x 4 = 30 x 3
= 90
RL = 22.5kN
RR = 7.5kN
UDL 5kN/m
2m
A
10kN
Deflected Shape
SFD
4m
-
B
20kN
+
22.5 kN
C
Moment at A
= -10 x 1
= -10 kNm
Moment at B
= 7.5 x 2 - 5 x 2 x 1
= 15 - 10
= 5 kNm
7.5 kN WL/8 = 20x4/8
= 10kNm
BMD
-7.5 kN
-10 kN
-10 kNm
10 kNm
TL = 5 x 6
= 30 kN
+12.5 kN
+5 kNm
+~5.6 kNm
Q11 H (cont.)
-10 kNm
BMD
(cantilever)
wL2/2 = -10 kNm
BMD
(Simply
Supported)
+10 kNm
wL2/8 = 10 kNm
BMD(Comb)
-10 kNm
~+5.6 kNm
Q11 I
UDL 5kN/m
2m
Cantilevers
4m
2m
10kN
10kN
-
-
+10 kN
SFD
(Cantilevers)
-10 kN
RL = 10 RR = 10
+10 kN
10 kN
10 kN
Simply Supported
20kN
RL = 10 RR = 10
-10 kN
+
10 kN
Combined
SFD
(Simply
Supported)
10kN
10 kN
20kN
10kN
-
-
20 kN
20 kN
RL = 20 RR = 20
+10 kN
+10 kN
-10 kN
-10 kN
SFD
(Combined)
Q11 I (cont.)
-10 kNm
BMD
(cantilevers)
BMD
(Simply
Supported)
wL2 / 2 = 5 x 2 x 2 / 2
= 10 kNm
wL2 / 8 = 5 x 4 x 4 / 8
= 10 kNm
+10 kNm
-10 kNm
BMD(Comb)
-10 kNm