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

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Question 1a when we want to design a beam its shape and dimensions a its material b its Section Modulus c what do we need to determine? a and c d a and b e

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

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

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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 m 2 a 49.0 m 2 b 25.2 m 2 c

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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 kN b 47.9 kPa c

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Question 2c is this load a? uniformly distributed load (UDL) a a point load b 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)

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Question 2d do we design for strength or for stiffness stiffness a strength b 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)

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Question 2e 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 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)

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Question 2f 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 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)

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Question 2g what is the maximum Bending Moment? kNm a 83.8 kNm b 41.9 kNm 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)

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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) 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? x 10 3 mm 3 a x 10 3 mm 3 b 4.8 x 10 3 mm 3 c

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Question 2i we want to find a Universal Beam which is strong enough. Given the we need a Section Modulus of at least x 10 3 mm 3, we can look up a 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 click here to see the Table of Universal Beams 250 UB 31.4 d

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back to Q2i back to Q2n back to Q2t

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Question 2j 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

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Question 2k 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

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Question 2l 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

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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

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Question 2n 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 what is the Moment of Inertia of the beam? 28.9 x 10 6 mm 4 a 44.4 x 10 6 mm 4 b 23.6 x 10 6 mm 4 c click here to see the Table of Universal Beams

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Question 2o Given that the Moment of Inertia of the beam is 23.6 x 10 6 mm 4 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 WL 3 / 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 mm b 35.5 mm c

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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

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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

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Question 2r so the section is not stiff enough. A Moment of Inertia of 23.6 x 10 6 mm 4 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

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Question 2s so the section is not stiff enough. A Moment of Inertia of 23.6 x 10 6 mm 4 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? x 10 6 mm 4 a 45.3 x 10 6 mm 4 b 76.4 x 10 6 mm 4 c

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Question 2t So we need a section that has a Moment of Inertia of 76.4 x 10 6 mm 4 We need to go back to the Table of Properties to find a satisfactory section 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 click here to see the Table of Universal Beams

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

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

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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

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

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

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

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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

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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

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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

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

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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?

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next question enough ! 3.6 m 7 m steel beams timber 600mm crs 3.6 m tributary area for beam = 7 x 3.6 = 25.2 m 2

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

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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) next question enough ! Tributary area = 25.2 m 2 load per sq m = 1.9 kPa TOTAL LOAD = 25.2 x 1.9 = 47.9 kN (kPa x m 2 = kN)

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let me try again let me out of here How did you get that? What is the total load per sq m? What is the tributary area?

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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)?

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

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let me try again let me out of here is the load acting at just one point or is it acting all along the length of the beam?

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

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

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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.

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

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let me try again let me out of here The Moment of Inertia of a beam’s X-section determines its stiffness NOT its strength

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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

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

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let me try again let me out of here what has the shear force to do with the Section Modulus?

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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

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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?

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let me try again let me out of here What is the max BM formula for a simply supported beam with a UDL?

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next question enough ! Z = M / f = 41.9 / 200 (since M is in kNm bring to N and mm) = 41.9 x 10 6 / 2 x 10 2 = x 10 3 mm 3

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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

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next question enough ! a universal beam 200UB 25.4 has a Z x value of 232 x 10 3 mm 3 since this > the min required Z x of x 10 3 mm 3 it is strong enough. A smaller beam would not be strong enough and a larger beam would be wasteful

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The Z x 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

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The Z x value of a 180 UB 22.2 is only 171 x 10 3 mm 3 such a beam would not be strong enough since we need at least a Zx value of x 10 3 mm 3 let me try again let me out of here

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

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

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

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

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

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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

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

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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

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

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next question enough ! remembering the deflection formula d = k x WL 3 / E I we see that we need to also know E and I and of course we need to know the max allowable deflection

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we already know the total load and the span let me try again let me out of here

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

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next question enough ! from the Table we see that a 200UB 25.4 has a value of I x of 23.6 x 10 6 mm 4

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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

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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

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next question enough ! = 5 x W L 3 / (384 x E I) = 5 x x 7 3 / (384 x 2 x 10 5 x 23.6 x 10 6 ) (the load is in kN and the span is in m, so bring everything to N & mm) = 5 x x 10 3 x (7 x 10 3 ) 3 / (384 x 2 x 10 5 x 23.6 x 10 6 ) = 5 x x 343 x / (384 x 2 x 23.6 x ) = x 10 / = 45.3 mm

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

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let me try again let me out of here work it out you have all the necessary values

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next question enough ! 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

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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?

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

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

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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?

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next question enough ! the actual deflection is 45.3 mm the deflection we want is 14.0 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

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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

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next question enough ! the Moment of Inertia of the 200UB25.4 is 23.6 x 10 6 mm 4 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 10 6 mm 4

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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

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You’ve graduated with honours! FINISH a 310UB 40.4 has an I x value of 85.2 x 10 6 mm 4 which is the lowest we can find that is greater than the required min value of 76.4 x 10 6 mm 4

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look again the 250UB 37.3 has an I x of 55.6 x 10 6 mm 4 but we need an Ix of at least 76.4 x 10 6 mm 4 let me try again let me out of here

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look again the 310UB 46.2 has an I x of 99.5 x 10 6 mm 4 that is greater than 76.4 x 10 6 mm 4 but somewhat wasteful as we could use a smaller beam. let me try again let me out of here

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