WORKSHEET 4

BEAMS

Given that floor joists are at 600mm centres and span 2.0m between bearers, what is the tributary area for one joist?

Q1

tributary area for joist = 2 x 0.6 = 1.2 m2

tributary area

600mm 600mm

2m

Given a floor 18 m x 18 m with columns on a 6m x 6m grid, what is the tributary area for:

Q2

(i) an internal column6m 6m 6m

6m6m

6m

6 x 6 = 36 m2

6m6m

6m6m

6m6m

(ii) a column on the edge

6 x 3 = 18 m2

(ii) a corner column

3 x 3 = 9 m2

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

Q3

6mm corrugated fibre cement sheet - 0.11kN/m2

13mm plasterboard ceiling - 0.22kN/m2

100 x 50 hardwood rafters @ 600mm crs - 11 kN /m3

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).

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).

0.6

1.67

1.67 x 0.6 = 1.0Weight of rafter 1.67m long = 0.1 x 0.05 x 1.67 x 11Weight of 1 sq m fibre cement = 1 x 0.11Weight of 1 sq m plasterboard = 1 x 0.22

Total weight of roof/ceiling per sq m

= 0.09 kN= 0.11 kN= 0.22 kN= 0.42 kN / m2

= 0.42 kPa

The roof above spans between roof trusses which are at 2.5 m centres and span 10m

Q4

Tributary area = 2.5 x 10 = 25 m2

= 10.5 kN

=1.05kN / m

a) sketch the layout described and indicate the tributary area for one truss

2.5m 2.5m

10.0m

2.5m

b) what is the total load on one truss? (neglecting the self-weight of the truss)

Total load = 25 x 0.42

c) what is load per metre on one truss?

Load per metre = 10.5 / 10Note:We have neglected the self-weight of the truss

Q5

a) bending

What are the two main types of stress involved in beam action?

b) shear

Q6

a) which of the above two (bending & shear) is more important?

b) why?

In buildings:

bending

have bigger spans relative to loads. In the design of machines have short spans with heavy loads and shear more important.

Q7

a) timber beams?can cause horizontal splitting along grain

b) steel beams?

What does shear force do to:

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 beamyou can see where the maximum shear forceoccurs

b) What does a Bending Moment Diagram tell you?

the values of the bending moment along the beamyou can see where the maximum bending moment occurs and whether it is positive or negative

Q11For 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

A

16 kN2m

4m

+10 kN 10 kN

+10 kN

- 10 kN

B

4m

UDL 5kN/m

+8 kN 8 kN

Deflected Shape

+8 kN

- 8 kNSFD

+16 kNmWL/4 = 16 x 4 / 4 =

BMD

+10 kNmwL2/8 = 5 x 4 x 4 / 8 =

Q11 C & D

SFD+10 kN

Deflected Shape-

R =10 kN

-10 kNm

-wL2/2 = -5 x 2 x 2 / 2 =

W = w x L = 5 x 2 =

-

10 kN

UDL 5kN/m

2m

D

2m

C10 kN

BMD-WL = -20 kNm

+10 kN

Q11 E

TL = 20 + 20 = 40 kN

For reactionsMoments about ARR x 5 = 20 x 2 +20 x 4 = 120RR = 24 kNRL = 16 kN

5m

20kN2m

20kN2m 1m

A B C D

Moment at B = 16 x 2 = 32 kNmMoment at C = 24 x 1 = 24 kNm

32 kNm24 kNm

BMD

SFD

+16 kN

-24 kN

-4 kN

24 kN

Deflected Shape

+16 kN

Q11 F

TL = 10 + 5 = 15 kN

For reactionsMoment at A = 10 x 1 + 5 x 2 = 20 kNm

Moment at A = - 20 kNm Moment at B = - 5 x 1 = - 5 kNm

-20 kNm

-5 kNm

BMD

2m

10kN1m

5kN1m

A B C

SFD

+15 kN+5 kN

-

Deflected Shape15 kN

Q11 G

7.5 kN+

17.5 kN

-

Deflected Shape

TL = 20 + 5 = 25 kN

For reactionsTake Moment at C RL x 4 = 5 x 6 + 20 x 2 = 70 RL = 17.5kN RR = 7.5kN

+12.5 kN

SFD

-7.5 kN-5 kN

4m2m

20kN5kN

A CB Moment at A = -5 x 2 = -10 kNm

Moment at B = 7.5 x 2 = 15kNm

WL/4 = 20x4/4 = 20kNm

20 k

Nm

-10 kNm

BMD

+15 kNm

Q11 H

7.5 kN

10kN 20kN

+22.5 kN

-Deflected Shape

TL = 5 x 6 = 30 kN

For reactionsTake Moment at C RL x 4 = 30 x 3 = 90 RL = 22.5kN RR = 7.5kN

4m

UDL 5kN/m

2m

A B C

SFD

-10 kN

+12.5 kN

-7.5 kNMoment at A = -10 x 1 = -10 kNm

Moment at B = 7.5 x 2 - 5 x 2 x 1 = 15 - 10 = 5 kNm

WL/8 = 20x4/8 = 10kNm

-10 kNm

+5 kNm

+~5.6 kNm

10 k

Nm

BMD

30kN

Q11 H (cont.)

BMD(Comb)

~+5.6 kNm

-10 kNm

BMD(SimplySupported) +10 kNm

wL2/8 = 10 kNm

BMD (cantilever)

-10 kNm

wL2/2 = -10 kNm

Q11 I

4m

UDL 5kN/m

2m2m

+10 kN

-10 kN

SFD(Combined)

+10 kN

-10 kN

RL = 10 RR = 10

-

10 kN 10 kN

10kN -Cantilevers10kN

RL = 20 RR = 20

-20 kN 20 kN

20kN-

10kN 10kNCombined

SFD(Cantilevers)

+10 kN

-10 kN

+10 kN

-10 kN

SFD(SimplySupported)RL = 10 RR = 10 +

10 kN 10 kN

20kNSimply Supported

Q11 I (cont.)

wL2 / 2 = 5 x 2 x 2 / 2 = 10 kNm

wL2 / 8 = 5 x 4 x 4 / 8 = 10 kNm

BMD(Comb)

BMD(SimplySupported)

+10 kNm

BMD (cantilevers)

-10 kNm

-10 kNm -10 kNm