Sections WORKSHEET 9c to answer just click on the button or image related to the answer.

Sections

WORKSHEET 9c

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Question 1awhen we want to design a beam

its shape and dimensionsa

its materialb

its Section Modulusc

what do we need to determine?

a and cd

a and be

Question 1bwhen we want to design a beam

the span and support typesa

the total load on the beamb

the shape and dimensions of the beamc

what do we need to know to start?

the maximum bending momentd

the moment of inertia and section moduluse

a and bf

a, b and eg

Question 1cwhen we want to design a beam

the maximum allowable bending stressa

the Modulus of Elasticityb

the maximum allowable deflectionc

what else do we need to know?

whether it is elastic, plastic or brittle d

all the abovee

a, b and cf

Question 2aIn a two-story house, we are spanning across a double garage 7m wide with steel beams at3.6 m centres the floor joists in the upper floor of a house.

The timber joist system in Q 19 spans between these steel beams (200 x 50 mm softwood

joists @ 600 mm centres spanning 3.6 m.) The live load is 1.5 kPa and the dead load is 0.4

kPa (including the self-weight of the joist)

what is the tributary area?

4.2 m2a

49.0 m2b

25.2 m2c

Question 2bIn a two-story house, we are spanning across a double garage 7m wide with steel beams at3.6 m centres the floor joists in the upper floor of a house.

The timber joist system in the previous tutorial spans between these steel beams

(200 x 50 mm softwood joists @ 600 mm centres spanning 3.6 m.) The live load is 1.5 kPa and

the dead load is 0.4 kPa (including the self-weight of the joist)

what is the total load on a beam?

47.9 kNa

100.8 kNb

47.9 kPac

Question 2c

is this load a?

uniformly distributed load (UDL)a

a point loadb

In a two-story house, we are spanning across a double garage 7m wide with steel beams at3.6 m centres the floor joists in the upper floor of a house.




Question 2d

do we design for strength or for stiffness

stiffnessa

strengthb





Question 2e

in order to get the dimensions of the beamwhat do we need to find?

the stress in the beama

the minimum required Section Modulusb

the minimum required Moment of Inertiac





Question 2f

in order to find the minimum requiredSection Modulus, what do we need to find?

the maximum Bending Momenta

the stress in the beamb

the maximum Shear Forcec





Question 2g

what is the maximum Bending Moment?

293.3 kNma

83.8 kNmb

41.9 kNmc









Question 2h

we want to find a Universal Beam which is strong enough. The maximum allowable stress of grade 300 steel is 200 MPa

what is the minimum required Section Modulus?

209.5 x 103 mm3a

477.3 x 103 mm3b

4.8 x 103 mm3c

Question 2iwe want to find a Universal Beam which is strong enough. Given the we need a Section

Modulus of at least 209.5 x 103 mm3, we can look up a Table of Universal Beams.

which beam would we select as being strong enough?

200 UB 29.8a

200 UB 25.4b

180 UB 22.2c

click here to see the Table of Universal Beams

250 UB 31.4d

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back toQ2n

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Question 2jhaving found a beam section 200UB 25.4 (with a depth of 203 mm) as being strong enough

what do we need to do now?

use ita

check it for strengthb

check it for stiffnessc

Question 2kto check the stiffness of a beam

what must we do?

check its widtha

check its deflectionb

check its span-to-depth ratioc

Question 2lto check the deflection of a beam

what must we do?

compare the actual deflection with the maximum allowable deflection

a

check that it doesn’t deflect more than thespan-to-depth ratio

b

make the beam twice as deep c

Question 2mto check the deflection of the 200 UB 25.4 beam

what else do we need to know?

the total load and the span a

the Moment of Inertia (I) and the Modulus of Elasticity, E

b

the maximum allowable deflectionc

b and cd

a, b and ce

Question 2ngiven that the Modulus of Elasticity of structural steel is 200000 MPa and the maximum allowable deflection is span / 500, we need to find the Moment of Inertia of the200 UB 25.4 beam.

We can do that by going back to the Table of Properties

what is the Moment of Inertia of the beam?

28.9 x 106 mm4 a

44.4 x 106 mm4 b

23.6 x 106 mm4c


Question 2oGiven that the Moment of Inertia of the beam is 23.6 x 106 mm4 and that the Modulus of Elasticity of structural steel is 200000 MPa, we need to find the deflectionThe deflection formula for a simply supported beam with a uniformly distributed load is

5 x WL3 / 384 E I

(remember the total load is 47.88 kN, the span is 7 m)

what is the deflection of the beam?

45.3 mma

453.0 mm b

35.5 mmc

Question 2pGiven that the deflection of the beam is 45.3 mm and the maximum allowable deflection is span / 500

is the section stiff enough?

yesa

no b

Question 2qso the section is not stiff enough

what do we need to do?

upsize it by increasing its Section Modulusa

upsize it by increasing its Moment of Inertia b

increase its Modulus of Elasticityc

Question 2rso the section is not stiff enough.

A Moment of Inertia of 23.6 x 106 mm4 produced a deflection of 45.3 mm The maximum allowable deflection is 14.0 mm

how much stiffer must the section be ?By how much must we increase

the Moment of Inertia?

by a factor of 3.24a

by a factor of 1.92 b

by a factor of 1.69c

Question 2sso the section is not stiff enough.

A Moment of Inertia of 23.6 x 106 mm4 produced a deflection of 45.3 mm

So we need to increase the Moment of Inertia by a factor of 3.24

what value of the Moment of Inertia do we need?

330.4 x 106 mm4a

45.3 x 106 mm4b

76.4 x 106 mm4c

Question 2tSo we need a section that has a Moment of Inertia of 76.4 x 106 mm4

We need to go back to the Table of Properties to find a satisfactory section

what section has a satisfactory Moment of Inertiaand, therefore is stiff enough?

250UB 37.3a

310UB 46.2b

310UB 40.4c


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enough !

when we design anything what we are doing is determining its material, shape and dimensions

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that’s only part of it

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we will need to use the Section Modulusbut that’s not what we are aiming at

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enough !

yes, we always need to know what the span, support type and loads are

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isn’t that what we are trying to find?

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we will need to calculate the maximum Bending Momentbut we can do that from the information we need to know

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we will need to find the minimum required I and Zbut we can calculate that from the information we will have

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enough !

in order to calculate some of the things that we will have towe need to be given the maximum allowable bending stress,

the Modulus of elasticity and the maximum allowable deflection.This we will get from codes once we decide on a material

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Why would we want to know whether the beam will be ofan elastic, plastic or brittle material at this stage?

What would that give us?

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enough !

3.6 m7

m

steel beams

timber joists@ 600mm crs

3.6 m

tributary areafor beam = 7 x 3.6 = 25.2 m2

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How did you get that? The beams span 7m and they are 3.6m centres

The load on the beam (neglecting the weight of the joists) isthe sum of the live load (1.5kPa) and the dead load (0.4kPa)

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enough !

Tributary area = 25.2 m2

load per sq m = 1.9 kPa

TOTAL LOAD = 25.2 x 1.9 = 47.9 kN (kPa x m2 = kN)

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How did you get that?What is the total load per sq m?

What is the tributary area?

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you should learnWe are talking about total load.

So what are the units for a load (force)?

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enough !

that’s rightthe load is distributed over the length of the beam

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is the load acting at just one pointor is it acting all along the length of the beam?

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enough !

we design for strength and check for stiffness

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No, not really

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enough !

once we have the Z value we can find the dimensionssince we are designing for strength we need to find the

minimum required Section Modulus since that is the propertyof a beam’s section which determines its strength

i.e. a beam with the minimum required Z will be strong enough.

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we already have that given as the maximum allowable bending stress

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The Moment of Inertia of a beam’s X-section determines its stiffnessNOT its strength

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enough !

the formula for bending stress is f = M / Z (stress = BM / Section Modulus)

so Z = M / fsince we know f (max allowable bending stress)

we need to first find the maximum BM

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we already have that given as the maximum allowable bending stress

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what has the shear force to do withthe Section Modulus?

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enough !

since this is a simply supported beam with a UDL Max BM = W L / 8 (where W = total load)

= 47.88 x 7 / 8 = 41.9 kNm

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How did you get that? Don’t guess!what is the max BM formula for a simply supported beam with a UDL?

What is the total load? (see Q2b) What is the span?

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What is the max BM formula for asimply supported beam with a UDL?

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enough !

Z = M / f = 41.9 / 200 (since M is in kNm bring to N and mm)

= 41.9 x 106 / 2 x 102

= 209.5 x 103 mm3

How did you get that? What did we say the formula for Z was?

What is the max BM?

What is the stress that we can allow the beam to take?

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next question

enough !

a universal beam 200UB 25.4 has a Zx value of 232 x 103 mm3

since this > the min required Zx of 209.5 x 103 mm3 it is strong enough.A smaller beam would not be strong enough and

a larger beam would be wasteful

The Zx value is certainly adequate andthe beam would be strong enough

But it is more than necessary and would be wasteful

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The Zx value of a 180 UB 22.2 is only 171 x 103 mm3

such a beam would not be strong enough

since we need at least a Zx value of 209.5 x 103 mm3

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enough !

Yes, the beam may be strong enoughbut maybe it deflects too much

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It is strong enough but what else could go wrong?

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but we’ve already designed it for strengthwe know that it’s strong enough

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enough !

Exactly! We have to compare its actual deflection withthe maximum allowable deflection

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what will that give us?

We could do that. That will give us an indicationBut we really need to be precise.

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enough !

this will tell us if the beam is stiff enough

how do you compare a deflection with a span-to depth ratio?One is a distance, the other is a ratio

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how do you know whether that’s enough or too much?

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enough !

remembering the deflection formula d = k x WL3 / E Iwe see that we need to also know E and I

and of course we need to know the max allowable deflection

we already know the total load and the span

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It’s not the whole story

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enough !

from the Table we see that a 200UB 25.4

has a value of Ix of 23.6 x 106 mm4

check againIt’s a 200 UB 25.4 we are looking at

not a 200 UB 29.8

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check againIt’s a 200 UB 25.4 we are looking at

not a 250 UB 31.4

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enough !

= 5 x W L3 / (384 x E I)

= 5 x 47.88 x 73 / (384 x 2 x 105 x 23.6 x 106)(the load is in kN and the span is in m, so bring everything to N & mm)

= 5 x 47.88 x 103 x (7 x 103)3 / (384 x 2 x 105 x 23.6 x 106)

= 5 x 47.88 x 343 x 1012 / (384 x 2 x 23.6 x 1011) = 82114.2.84 x 10 / 18124.8 = 45.3 mm

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don’t you think that that’s a rather large deflection

check your zeros

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work it out

you have all the necessary values

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enough !

the actual deflection is 45.3 mmthe max allowable deflection is span / 500 = 7000 / 500 = 14 mm

so the actual deflection is much greater than the max allowable deflectionso the section is not stiff enough

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what is the deflection of this beam?what is the maximum allowable deflection?

is the actual deflection less or greater than the max allowable deflection?

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enough !

that’s exactly it.increasing the Moment of Inertia of a beam’s section

will increase its stiffness

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we are talking about stiffness – not strength

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theoretically we could but where would we find a materialwith such a high Modulus of Elasticity?

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enough !

the actual deflection is 45.3 mmthe deflection we want is 14.0 mmso we want the beam to be 45.3 / 14 times stifferso we must increase the Moment of Inertia by a factor of 3.24

work it out

what is the actual deflection?what is the maximum allowable deflection that we want?

so how much stiffer must the beam be?

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next question

enough !

the Moment of Inertia of the 200UB25.4 is 23.6 x 106 mm4

we need to increase that by a factor of 3.24

this gives us a required I of 23.6 x 3.24 = 76.4 x 106 mm4

work it out

what is the Moment of Inertia of the 200UB 25.4?what is the factor by which we have to increase it?

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You’ve graduated with honours!

FINISH

a 310UB 40.4 has an Ix value of 85.2 x 106 mm4 which is the lowest we can find that is

greater than the required min value of 76.4 x 106 mm4

look again

the 250UB 37.3 has an Ix of 55.6 x 106 mm4

but we need an Ix of at least 76.4 x 106 mm4

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look again

the 310UB 46.2 has an Ix of 99.5 x 106 mm4

that is greater than 76.4 x 106 mm4 but somewhat wasteful as we could use a smaller beam.

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Sections WORKSHEET 9c to answer just click on the button or image related to the answer.

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