worksheet 6 trusses. q1 when would we use a truss? (a) long spans, loads not too heavy (b) when want...

WORKSHEET 6

TRUSSES

Q1

When would we use a truss?

(a) long spans, loads not too heavy

(b) when want to save weight

(d) when want light appearance

(c) when have plenty of depth

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Q2

When would we not use a truss?

(a) don’t have the depth

(b) very large loads

(d) can’t provide lateral support if needed

(c) don’t want fussy appearance

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Q3

What are the main characteristics of trusses?

(a) assembly of (short) linear members

(b) members connected to form triangles

(c) joints pinned

(d) loads applied at panel points (joints)

(d) members carry only tension or compression forces

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Q4

What materials are trusses most commonly made of?

(a) steel

(b) timber

(c) concrete (very occasionally)

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Q5

How are the joints commonly made?

(a) in timber trusses

(b) in steel trusses

(i) gusset plates

(i) using gangnail joints - light timber

(ii) using gusset plates

(iii) overlapping/double members and bolts

(iv) concealed plates - bolts

(ii) welded joints

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Q6

What do the following do?

(a) the chords

(b) the web members

(i) resist the shear forces

(ii) top chords take the compressive forces (in a triangular truss, the top chord also resists shear)

(iii) bottom chords take the tensile forces

(i) chords resist the bending moment

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Q7

What are the common span-to-depth ratios for:

(a) timber beams?

(b) trusses?

5-10:1

18-20:1

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Q8

In the truss below show where:

(a) the maximum compressive force in a chord occurs

(b) the maximum tensile force in a chord occurs

(c) the maximum shear force in a web member occurs

(c)(c)

(a)

(b)

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Q9

In the truss below show where:

(a) the maximum compressive force in a chord occurs

(b) the maximum tensile force in a chord occurs

(c) the maximum shear force in a web member occurs

(c)

(a)

(b) 10/24

Q10 (a)

(i) the horizontal component?

H

H = 10 cos 45

= 7.07 kN to the right V

(ii) the vertical component?

V = 10 sin 45

= 7.07 kN down

(a) given the force shown, what is

45o

10kN

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Q10 (b)

(i) the horizontal component?

H

H = 6 cos 30

= 5.2 kN to the right

V

(ii) the vertical component?

V = 6 sin 30

= 3.0 kN up

(b) given the force shown, what is

30o

6kN

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Q10 (c)

R = (62 + 82)

8kN

6kN(c) given the two forces shown, use the parallelogram of forces or the triangle of forces to find the resultant force

8kN

6kN 6kN

8kN

R

Øo

Tan Ø = 6 / 8 = 0.75

Ø = 36.87o

R

Øo

= 10 kN at 36.87o

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Q11

Analyze the truss shown below using the Methods of Joints

First find the reactions. Using symmetry R1 = R2 = 8/2 = 4kN

Assume all forces are in tension - i.e. away from the joint

1kN 2kN 2kN 2kN 1kN

45o

A

B C

D

E

F

R1=4kN R2=4kN

4 bays @ 3m

3m

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Q11 (cont1.)start at joint A(support)A

AB

AD

4

AB = -4.0 (compression)

AD = 0

B

BA4

BC

BD

1

vertically: 4 + AB = 0

horizontally:

next at joint B.

BD = +4.24 (tension)

BC = -3.0 (compression)

vertically: 1 - 4 + BDsin45o = 0

horizontally: BC +4.24 cos45o = 0

next at joint D.

DC = -3.0 (compression)

DF = +3.0 (tension)

vertically: 4.24sin45o + DC = 0

horizontally: 0 +DF -4.24 cos45o = 0D

DC

DFDA(0)

DB(4.24)

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Q11(cont2.)next at joint C

CF = +1.41 (tension)vertically: 2 - 3 + CFsin45o = 0

horizontally: 3 +CE +1.41cos45o = 0CE = -4.0 (compression)

next at joint E.

EF = -2.0 (compression)

vertically: 2 + EF = 0

C

CB3

CE

CFCD3

2kN

E

EC4

EG4

EF

2kN

That completes half the truss. The other half is the same by symmetry

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Q11 (cont3.)

1kN 2kN 2kN 2kN 1kN

A

B C

D

E

F

4 4.24

3 4 4 3

44.2433 1.41 1.412

3 3 00

Note that the biggest chord forces are near the middleand the biggest web forces are near the ends

Always true for parallel chord trusses with fairly uniform loading

17/24

Q12

For the same truss use the Methods of Sectionsto find the forces in members BD, CE and DF

First find the reactions. Using symmetry R1 = R2 = 8/2 = 4kN

Assume all forces are in tension - i.e. away from the joint

1kN 2kN 2kN 2kN 1kN

45o

A

B C

D

E

F

R1=4kN R2=4kN

4 bays @ 3m

3m

18/24

Q12(cont1.)

Make a cut through the truss passing through the member wanted, BD

1kN

A

B C

D

4

Consider the left part of the cut as a freebodyMark all the forces on it - including those members cut off

the freebody is in equilibrium

Using V = 0

4 - 1 -BD sin45o = 0 BD = 4.24 (tension)

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Q12(cont2.)

Make another cut through the truss passing through the other members wanted, CE and DF

Mark all the forces on it - including those members cut off

the freebody is in equilibrium

Using M = 0 about F - consider all the forces on the left4 x 6 - 1x 6 - 2 x 3 +CE x 3 = 0 CE = -4.0 (compression)

1kN 2kN

A

B C

D

E

F4

3m

3m 3mUsing M = 0 about C4 x 3 - 1x 3 - DF x 3 = 0 DF = +3.0 (tension)

Note that we have found only the members we wanted

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Q13

For the same truss use the Graphical Methodto construct a Maxwell diagram and find the forces in the members

First find the reactions. Using symmetry R1 = R2 = 8/2 = 4kN

1kN 2kN 2kN 2kN 1kN

45o

R1=4kN R2=4kN

4 bays @ 3m

3m

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Q13(cont1.)

First annotate using Bow’s Notation (label spaces between members and forces)

First find the reactions. Using symmetry R1 = R2 = 8/2 = 4kN

1kN 2kN 2kN 2kN 1kN

a

b c ed

f

R1=4kN R2=4kN

4 bays @ 3m

3m

g

hi

jk l

mn

0

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Q13(cont2.)Next select a scale and draw a line representing all the loads and reactions

0

1

2

3

4

scaleforforces

c

d

f

a

b

g

all the loads are vertical - so is the linethe line is the line a,b,c,d,e,f (ag, gf)ab = 1, bc = 2,...etc ag = 4, gf = 4

e

begin with a zone near a reaction, e.g hah is vertical and gh is horizontal - meet at g (h & g at same point) hg is 0

h

take next zone - i

bi is horizontal and ih is at 45o - draw these lines they meet at i

i

Now ij is vertical and jg is horizontal. This locates j

j

Proceed to k in same way and half the truss is solved

k

n

l

mo

Measure all lines - this gives the force in each memberneed to use a special convention to determine tension or compression 23/24

Q14

When would you use:

(a) the Method of joints?

(b) the Method of Sections?

When want to know only a few member forcese.g. end diagonals and middle chords

When want to do detailed design and need to know all the forces in all members

(b) the Graphical Method

When don’t have a calculator or don’t want to calculate

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