# WORKSHEET 9c let's go !! Sections

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WORKSHEET 9c let's go !! Sections
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what do we need to determine?
Question 1a when we want to design a beam what do we need to determine? its shape and dimensions a its material b its Section Modulus c a and c d a and b e

what do we need to know to start?
Question 1b when we want to design a beam what do we need to know to start? the span and support types a the total load on the beam b the shape and dimensions of the beam c the maximum bending moment d the moment of inertia and section modulus e a and b f a, b and e g

what else do we need to know?
Question 1c when we want to design a beam what else do we need to know? the maximum allowable bending stress a the Modulus of Elasticity b the maximum allowable deflection c whether it is elastic, plastic or brittle d all the above e a, b and c f

what is the tributary area?
Question 2a In a two-story house, we are spanning across a double garage 7m wide with steel beams at 3.6 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 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 m2 a 49.0 m2 b 25.2 m2 c

what is the total load on a beam?
Question 2b In a two-story house, we are spanning across a double garage 7m wide with steel beams at 3.6 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 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 kN a 100.8 kN b 47.9 kPa c

In a two-story house, we are spanning across a double garage 7m wide with steel beams at 3.6 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 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) is this load a? uniformly distributed load (UDL) a a point load b

do we design for strength or for stiffness
Question 2d In a two-story house, we are spanning across a double garage 7m wide with steel beams at 3.6 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 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) do we design for strength or for stiffness stiffness a strength b

in order to get the dimensions of the beam
Question 2e In a two-story house, we are spanning across a double garage 7m wide with steel beams at 3.6 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 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) in order to get the dimensions of the beam what do we need to find? the stress in the beam a the minimum required Section Modulus b the minimum required Moment of Inertia c

Question 2f in order to find the minimum required
In a two-story house, we are spanning across a double garage 7m wide with steel beams at 3.6 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 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) in order to find the minimum required Section Modulus, what do we need to find? the maximum Bending Moment a the stress in the beam b the maximum Shear Force c

what is the maximum Bending Moment?
Question 2g In a two-story house, we are spanning across a double garage 7m wide with steel beams at 3.6 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 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 maximum Bending Moment? 293.3 kNm a 83.8 kNm b 41.9 kNm c

what is the minimum required Section Modulus?
Question 2h In a two-story house, we are spanning across a double garage 7m wide with steel beams at 3.6 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 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) 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 mm3 a 477.3 x 103 mm3 b 4.8 x 103 mm3 c

Question 2i we want to find a Universal Beam which is strong enough. Given the we need a Section Modulus of at least x 103 mm3, we can look up a Table of Universal Beams. click here to see the Table of Universal Beams which beam would we select as being strong enough? 200 UB 29.8 a 200 UB 25.4 b 180 UB 22.2 c 250 UB 31.4 d

back to Q2i back to Q2n back to Q2t

Question 2j what do we need to do now? a use it b
having 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 it a check it for strength b check it for stiffness c

Question 2k what must we do? a check its width b check its deflection
to check the stiffness of a beam what must we do? check its width a check its deflection b check its span-to-depth ratio c

Question 2l what must we do? a b c
to 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 the span-to-depth ratio b make the beam twice as deep c

what else do we need to know?
Question 2m to 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 deflection c b and c d a, b and c e

Question 2n what is the Moment of Inertia of the beam? a b c
given that the Modulus of Elasticity of structural steel is 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 click here to see the Table of Universal Beams 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 mm4 c

what is the deflection of the beam?
Question 2o Given that the Moment of Inertia of the beam is 23.6 x 106 mm4 and that the Modulus of Elasticity of structural steel is MPa, we need to find the deflection The deflection formula for a simply supported beam with a uniformly distributed load is 5 x WL3 / 384 E I (remember the total load is kN, the span is 7 m) what is the deflection of the beam? 45.3 mm a 453.0 mm b 35.5 mm c

is the section stiff enough?
Question 2p Given that the deflection of the beam is 45.3 mm and the maximum allowable deflection is span / 500 is the section stiff enough? yes a no b

Question 2q so the section is not stiff enough what do we need to do?
upsize it by increasing its Section Modulus a upsize it by increasing its Moment of Inertia b increase its Modulus of Elasticity c

how much stiffer must the section be? By how much must we increase
Question 2r so 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 3.24 a by a factor of 1.92 b by a factor of 1.69 c

what value of the Moment of Inertia do we need?
Question 2s so 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 mm4 a 45.3 x 106 mm4 b 76.4 x 106 mm4 c

Question 2t what section has a satisfactory Moment of Inertia
So 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 click here to see the Table of Universal Beams what section has a satisfactory Moment of Inertia and, therefore is stiff enough? 250UB 37.3 a 310UB 46.2 b 310UB 40.4 c

a Great Start next question enough !
when we design anything what we are doing is determining its material, shape and dimensions next question enough !

Partly correct let me try again let me out of here
that’s only part of it let me try again let me out of here

No! let me try again let me out of here
we will need to use the Section Modulus but that’s not what we are aiming at let me try again let me out of here

You Bewdy ! next question enough !
yes, we always need to know what the span, support type and loads are next question enough !

Partly correct let me try again let me out of here
that’s only part of it let me try again let me out of here

isn’t that what we are trying to find?
Oh my gosh!! isn’t that what we are trying to find? let me try again let me out of here

No! let me try again let me out of here
we will need to calculate the maximum Bending Moment but we can do that from the information we need to know let me try again let me out of here

No! let me try again let me out of here
we will need to find the minimum required I and Z but we can calculate that from the information we will have let me try again let me out of here

you've got it it!! next question enough !
in order to calculate some of the things that we will have to we 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 next question enough !

Partly correct let me try again let me out of here
that’s only part of it let me try again let me out of here

Oh my gosh!! let me try again let me out of here
Why would we want to know whether the beam will be of an elastic, plastic or brittle material at this stage? What would that give us? let me try again let me out of here

you've got it it!! next question enough ! 3.6 m 3.6 m 7 m steel beams
timber joists @ 600mm crs tributary area for beam = 7 x 3.6 = 25.2 m2 3.6 m 7 m next question enough !

The beams span 7m and they are 3.6m centres
Oh my gosh!! How did you get that? The beams span 7m and they are 3.6m centres let me try again let me out of here

Fantastic next question enough !
The load on the beam (neglecting the weight of the joists) is the sum of the live load (1.5kPa) and the dead load (0.4kPa) 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) next question enough !

What is the total load per sq m? What is the tributary area?
Not right !! How did you get that? What is the total load per sq m? What is the tributary area? let me try again let me out of here

Oh my Goodness!! let me try again let me out of here you should learn
We are talking about total load. So what are the units for a load (force)? let me try again let me out of here

the load is distributed over the length of the beam
Yeaaahh! that’s right the load is distributed over the length of the beam next question enough !

it's enough to make one cry
is the load acting at just one point or is it acting all along the length of the beam? let me try again let me out of here

we design for strength and check for stiffness
terriffic !! we design for strength and check for stiffness next question enough !

Uhhhh ??? No, not really let me try again let me out of here

Fantastic next question enough !
once we have the Z value we can find the dimensions since we are designing for strength we need to find the minimum required Section Modulus since that is the property of a beam’s section which determines its strength i.e. a beam with the minimum required Z will be strong enough. next question enough !

we already have that given as the maximum allowable bending stress
Uhhhh ??? we already have that given as the maximum allowable bending stress let me try again let me out of here

The Moment of Inertia of a beam’s X-section determines its stiffness
it's enough to make one cry The Moment of Inertia of a beam’s X-section determines its stiffness NOT its strength let me try again let me out of here

Yipee !! next question enough !
the formula for bending stress is f = M / Z (stress = BM / Section Modulus) so Z = M / f since we know f (max allowable bending stress) we need to first find the maximum BM next question enough !

we already have that given as the maximum allowable bending stress
Uhhhh ??? we already have that given as the maximum allowable bending stress let me try again let me out of here

what has the shear force to do with
No, No, No !! what has the shear force to do with the Section Modulus? let me try again let me out of here

Yipee !! next question enough !
since this is a simply supported beam with a UDL Max BM = W L / 8 (where W = total load) = x 7 / 8 = 41.9 kNm next question enough !

Not right!! let me try again let me out of here
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? let me try again let me out of here

What is the max BM formula for a simply supported beam with a UDL?
No, No, No !! What is the max BM formula for a simply supported beam with a UDL? let me try again let me out of here

yes, yes, yes !! next question enough ! Z = M / f
= 41.9 / 200 (since M is in kNm bring to N and mm) = 41.9 x 106 / 2 x 102 = x 103 mm3 next question enough !

Sorry but, No !! let me try again let me out of here
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? let me try again let me out of here

Brilliant !! 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 x 103 mm3 it is strong enough. A smaller beam would not be strong enough and a larger beam would be wasteful next question enough !

Sorry but, No !! let me try again let me out of here
The Zx value is certainly adequate and the beam would be strong enough But it is more than necessary and would be wasteful let me try again let me out of here

Sorry but, No !! let me try again let me out of here
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 x 103 mm3 let me try again let me out of here

Yes, the beam may be strong enough but maybe it deflects too much
Cheers !! Yes, the beam may be strong enough but maybe it deflects too much next question enough !

It is strong enough but what else could go wrong?
You devil ! It is strong enough but what else could go wrong? let me try again let me out of here

You devil ! let me try again let me out of here
but we’ve already designed it for strength we know that it’s strong enough let me try again let me out of here

Cheers !! next question enough !
Exactly! We have to compare its actual deflection with the maximum allowable deflection next question enough !

You devil ! let me try again let me out of here
what will that give us? let me try again let me out of here

Yes but .... let me try again let me out of here
We could do that. That will give us an indication But we really need to be precise. let me try again let me out of here

this will tell us if the beam is stiff enough
Good on You !! this will tell us if the beam is stiff enough next question enough !

A devil of an answer ! let me try again let me out of here
how do you compare a deflection with a span-to depth ratio? One is a distance, the other is a ratio let me try again let me out of here

how do you know whether that’s enough or too much?
A devil of an answer ! how do you know whether that’s enough or too much? let me try again let me out of here

Good on You !! next question enough !
remembering the deflection formula d = k x WL3 / E I we see that we need to also know E and I and of course we need to know the max allowable deflection next question enough !

Don't cry !! we already know the total load and the span let me try again let me out of here

It’s not the whole story
Yes but .... It’s not the whole story let me try again let me out of here

from the Table we see that a 200UB 25.4
Good on You !! from the Table we see that a 200UB 25.4 has a value of Ix of 23.6 x 106 mm4 next question enough !

It’s a 200 UB 25.4 we are looking at
Not correct! check again It’s a 200 UB 25.4 we are looking at not a 200 UB 29.8 let me try again let me out of here

It’s a 200 UB 25.4 we are looking at
Not correct! check again It’s a 200 UB 25.4 we are looking at not a 250 UB 31.4 let me try again let me out of here

Good on You !! next question enough ! = 5 x W L3 / (384 x E I)
= 5 x 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 x 103 x (7 x 103)3 / (384 x 2 x 105 x 23.6 x 106) = 5 x x 343 x 1012 / (384 x 2 x 23.6 x 1011) = x 10 / = 45.3 mm next question enough !

don’t you think that that’s a rather large deflection
Not right ! don’t you think that that’s a rather large deflection check your zeros let me try again let me out of here

you have all the necessary values
Don't guess ! work it out you have all the necessary values let me try again let me out of here

so the section is not stiff enough
Cheers !! the actual deflection is 45.3 mm the max allowable deflection is span / 500 = 7000 / 500 = 14 mm so the actual deflection is much greater than the max allowable deflection so the section is not stiff enough next question enough !

You devil ! let me try again let me out of here
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? let me try again let me out of here

Cheers !! next question enough ! that’s exactly it.
increasing the Moment of Inertia of a beam’s section will increase its stiffness next question enough !

we are talking about stiffness – not strength
You devil ! we are talking about stiffness – not strength let me try again let me out of here

Yes but .... let me try again let me out of here
theoretically we could but where would we find a material with such a high Modulus of Elasticity? let me try again let me out of here

Good on You !! next question enough ! the actual deflection is 45.3 mm
the deflection we want is mm so we want the beam to be 45.3 / 14 times stiffer so we must increase the Moment of Inertia by a factor of 3.24 next question enough !

Don't guess !! let me try again let me out of here 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? let me try again let me out of here

Good on You !! 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 next question enough !

Don't guess !! let me try again let me out of here 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? let me try again let me out of here

Well Done ! FINISH You’ve graduated with honours!
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 FINISH

but we need an Ix of at least 76.4 x 106 mm4
Not Correct ! 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 let me try again let me out of here

Not Correct ! let me try again let me out of here 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. let me try again let me out of here