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Axial Members WORKSHEET 11 to answer just click on the button or image related to the answer

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Question 1a are tension structures or compression structures more efficient when loaded axially? compression structures a tension structures b

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Question 1b why are tension structures more efficient? they are not subject to buckling a they are usually made of steel b they pull straight c they are stronger d

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Question 2a what is meant by a short column? a column which looks short a a fat column b a column which will fail in true compression c

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Question 2b what is meant by a long column? a column which looks long a a thin column b a column which buckles before its full compressive strength is reached c

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Question 3 what affects the buckling load? the maximum allowable compressive strength a the slenderness ratio b the moment of inertia, I c the modulus of elasticity, E d the end fixing conditions e b, c and d f b, d and e g

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Question 4a what are good sections for columns? I-sections a sections with similar radii of gyration in all directions b rectangular sections c sections in which the major part of the material is as far from the Centre of Gravity as possible d b and d e

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Question 4b why are these good sections for columns? so that they do not buckle in the weak direction a they are cheaper b they use the material more efficiently c a and c d a, b and c e similar radii of gyration in all directions and material far away from the Centre of Gravity

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Question 5 what are two effects which can cause a pier to overturn? bad construction a a horizontal load b an eccentric vertical load c b and c d

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Question 6a does the middle third rule deal with? the force acting on the pier being in the middle third of the pier a the resultant reaction being in the middle third of the base b

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Question 6b if the middle third rule holds, what does that tell you? the pier will not overturn a no tension will develop in the base of the pier b the pier will not lift off its base c b and c d a and b e

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Question 6c if the middle third rule holds, what does that tell you about the safety factor? the safety factor is > 3 a the safety factor is 3 b the safety factor is 2 c

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Question 7a what is the overturning moment? 1.6 kNm a 1.0 kNm b 1.3 kNm c Pier 600 x 600mm 8kN 1kN a heavy steel gate is hung from a hollow brick pier as shown in the diagram

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Question 7b what is the stabilizing moment? 2.4 kNm a 4.8 kNm b 8 kNm c Pier 600 x 600mm 8kN 1kN a heavy steel gate is hung from a hollow brick pier as shown in the diagram

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Question 7c will the pier overturn and why or why not? yes, the weight of the gate is eccentric a no, the weight of the pier is greater than the weight of the gate b yes, the overturning moment is greater than the stabilizing moment c Pier 600 x 600mm 8kN 1kN a heavy steel gate is hung from a hollow brick pier as shown in the diagram no, the stabilizing moment is greater than the overturning moment d

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Question 7d what is the margin of safety? 2 : 1 a 2.4 : 1 b 3 : 1 c a heavy steel gate is hung from a hollow brick pier as shown in the diagram Pier 600 x 600mm 8kN 1kN > 3 : 1 d

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Question 7e what is the value of the vertical reaction, R? 8 kN a 9 kN b 2.4 kN c a heavy steel gate is hung from a hollow brick pier as shown in the diagram Pier 600 x 600mm 8kN 1kN X R

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Question 7f what distance is the vertical reaction from X? h =? mm a mm b mm c Pier 600 x 600mm 8kN 1kN X h a heavy steel gate is hung from a hollow brick pier as shown in the diagram R = 9 kN

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Question 7g is the reaction in the middle third of the base? yes a no b Pier 600 x 600mm 8kN 1kN X h = mm a heavy steel gate is hung from a hollow brick pier as shown in the diagram R = 9 kN mm

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Question 7h is this what we would expect from our previous observations? you can’t tell a no b yes c Pier 600 x 600mm 8kN 1kN X h = mm a heavy steel gate is hung from a hollow brick pier as shown in the diagram. The reaction is outside the middle third. R = 9 kN mm

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Question 7i what is the stress distribution under the pier? ( / ) MPa a ( / ) MPa b ( / ) MPa c Pier 600 x 600mm 8kN 1kN X h = mm a heavy steel gate is hung from a hollow brick pier as shown in the diagram. The reaction is outside the middle third. stress = P/A ±Pe/Z Z = bd 2 /6 R = 9 kN mm

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Question 7j is this what you would expect? yes a no b not possible to tell c Pier 600 x 600mm 8kN 1kN X h = mm a heavy steel gate is hung from a hollow brick pier as shown in the diagram. The pier is developing tension on the left-hand side. R = 9 kN mm

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Question 7k will it be safer? not possible to know a no b yes c Pier 600 x 600mm 8kN 1kN X h = mm a heavy steel gate is hung from a hollow brick pier as shown in the diagram. The pier is developing tension on the left-hand side. If you make the pier solid, i.e. it will be about twice as heavy R = 9 kN mm

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Question 8a what‘s the weight of the wall (for 1 m length)? 5.24 kN a 52.4 kN b 52.4 kPa c A freestanding garden brick wall is 230 mm thick and 1200 mm high. The density of brick is 19 kN/m 3. The wind load is 0.5 kPa 1200mm self weight wind 0.5kPa R x A 115 h 600

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Question 8b what‘s the total wind force (for 1 m length)? 0.5 kN a 0.6 kN b 0.6 kPa c A freestanding garden brick wall is 230 mm thick and 1200 mm high. The density of brick is 19 kN/m 3. The wind load is 0.5 kPa 1200mm self weight wind 0.5kPa R x A 115 h 600

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Question 8c what‘s the value of the reaction (for 1 m length)? 5.30 kN a 5.24 kN b 5.74 kN c 1200mm wind 0.5kPa R x A 115 h 600 A freestanding garden brick wall is 230 mm thick and 1200 mm high. The density of brick is 19 kN/m 3. The wind load is 0.5 kPa W = 5.24 kN

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Question 8d what‘s the distance, h, of the reaction from the centre of the pier? 68.7 mm a 46.3 mm b mm c 1200mm wind 0.5kPa R = 5.24 kN x A 115 h 600 A freestanding garden brick wall is 230 mm thick and 1200 mm high. The density of brick is 19 kN/m 3. The wind load is 0.5 kPa W = 5.24 kN

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h Question 8e is the reaction within the base? maybe a yes b no c 1200mm W = 5.24 kN wind 0.6 kN R = 5.24 kN x A A freestanding garden brick wall is 230 mm thick and 1200 mm high. The density of brick is 19 kN/m 3. The wind load is 0.5 kPa 1200mm x A 115 h = 46.4 mm 600

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h Question 8f what does that mean? tension develops on the RHS a the pier will not overturn b no tension develops on the RHS c 1200mm W = 5.24 kN wind 0.6 kN R = 5.24 kN x A A freestanding garden brick wall is 230 mm thick and 1200 mm high. The density of brick is 19 kN/m 3. The wind load is 0.5 kPa The reaction is within the base 1200mm x A 115 h = 46.4 mm 600 the pier will overturn d

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h Question 8g is the reaction within the middle third of the base? no a maybe b yes c 1200mm W = 5.24 kN wind 0.6 kN R = 5.24 kN x A A freestanding garden brick wall is 230 mm thick and 1200 mm high. The density of brick is 19 kN/m 3. The wind load is 0.5 kPa 1200mm x A 115 h = 46.4 mm 600

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h Question 8h what does that mean? tension develops on the RHS a the pier lifts off its base b a and b c 1200mm W = 5.24 kN wind 0.6 kN R = 5.24 kN x A A freestanding garden brick wall is 230 mm thick and 1200 mm high. The density of brick is 19 kN/m 3. The wind load is 0.5 kPa The reaction is not in the middle third 1200mm x A 115 h = 46.4 mm 600 the pier overturns d

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h Question 8i how wide would the footing have to be for the reaction to fall within the middle third of the base? 400 mm a 450 mm b 410 mm c 1200mm W = 5.24 kN wind 0.6 kN x A A freestanding garden brick wall is 230 mm thick and 1200 mm high. The density of brick is 19 kN/m 3. The wind load is 0.5 kPa 1200mm x A 115 h = 46.4 mm 600 R = 5.24 kN

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h Question 8j how else could we increase the stability of the wall? no idea a make the wall thicker b show me c 1200mm W = 5.24 kN wind 0.6 kN x A A freestanding garden brick wall is 230 mm thick and 1200 mm high. The density of brick is 19 kN/m 3. The wind load is 0.5 kPa 1200mm x A 115 h = 46.4 mm 600 R = 5.24 kN

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next question enough ! tension structures are more efficient

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let me try again let me out of here what’s the problem with compression structures?

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next question enough ! Yep! No buckling! Can use all its strength !!

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let me try again they may be made of steel but then again they may be made of other materials let me out of here

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let me try again they do pull straight but so what? let me out of here

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let me try again some materials are stronger in compression some materials are equally strong in compression and tension so what is it about compression that may lead to early failure? let me out of here

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next question enough ! it does probably looks short and fat but the technical answer is that it fails in true compression

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but it’s not really the answer let me try again let me out of here

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next question enough ! it does probably looks long and thin but the technical answer is that it fails in buckling before true compression

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but it’s not really the answer let me try again let me out of here

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the ease of buckling is a function of the slenderness ratio, the modulus of elasticity and the end restraints the slenderness ratio takes into account the effective length and the radius of gyration which takes into account the Moment of Inertia. The greater the slenderness ratio, the more likely buckling will occur. Materials with a higher Modulus of Elasticity, E, will less likely buckle. The more restrained the ends, the less likely to buckle. next question enough !

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buckling is not a function of the maximum allowable compressive stress. Buckling doesn’t allow the element to reach its maximum allowable stress let me try again let me out of here

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but it is affected by other things too let me try again let me out of here

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the Moment of Inertia figures in it but as part of something else let me try again let me out of here

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next question enough ! you’ve got it !!

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think again – why are I-sections or rectangular sections not good for columns let me try again let me out of here

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let me try again let me out of here

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next question enough ! with equal radii of gyration, columns will not buckle in a weak direction. Remember the studs and noggings. With more material away from the Centre of Gravity, the same Radius of Gyration can be achieved with less material

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let me try again let me out of here

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Where did cheaper come from? let me try again let me out of here

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next question enough ! both an eccentric load or a horizontal load cause an overturning moment on a pier

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while bad construction may contribute it’s not the cause let me try again let me out of here

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there is something else which has the same effect let me try again let me out of here

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next question enough ! the middle third deals with the reaction

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the middle third rule does not deal with the forces acting THINK !! If there were several forces acting which one would you look at? let me try again let me out of here

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next question enough ! since no tension tends to develop, the pier will not lift off its base

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the middle third rule doesn’t deal with the overturning of the pier let me try again let me out of here

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let me try again let me out of here

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next question enough ! yes, if the middle third rule holds, the safety factor will be > 3

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let me try again let me out of here

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next question enough ! Taking moments about X 1 (kN) x 1(m) = 1 kNm Pier 600 x 600mm 8kN 1kN x

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taking moments about X What force is causing overturning? What is the distance of this force to X? Pier 600 x 600mm 8kN 1kN x let me try again let me out of here

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next question enough ! taking moments about X 8 (kN) x 0.3 (m) = 2.4 kNm Pier 600 x 600mm 8kN 1kN x the force contributing to the stabilizing moment is the weight of the pier

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taking moments about X What force is resisting overturning? What is the distance of this force to X? Pier 600 x 600mm 8kN 1kN x let me try again let me out of here

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next question enough ! the overturning moment, OTM, is 1 kNm the stabilizing moment, SM, is 2.4 kNm so SM > OTM so the pier will not overturn Pier 600 x 600mm 8kN 1kN x

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a weight is a force. Forces DO NOT overturn Moments overturn let me try again let me out of here

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WHAT????? the overturning moment is 1kNm the stabilizing moment is 2.4 kNm so which is bigger? let me try again let me out of here

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next question enough ! if the stabilizing moment is 2.4 kNm and the restraining moment is 1 kNm then the OTM can be 2.4 times greater before the pier will overturn

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how did you arrive at that? what are the overturning moment and the stabilizing moment? By how much could the OTM increase before the pier would overturn? let me try again let me out of here

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next question enough ! the total downward forces = = 9 kN so the reaction must be 9kN Pier 600 x 600mm 8kN 1kN x R = 9 kN

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what is the sum of the downward forces? let me try again let me out of here Pier 600 x 600mm 8kN 1kN x

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Taking moments about X clockwise moments = anti-clockwise moments 1 x x h = 8 x 300 9h = 1400 h = next question enough ! Pier 600 x 600mm 8kN 1kN x R = 9 kN h = mm

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try again take moments about X. what are the clockwise moments? what is the anti-clockwise moment ? let me try again let me out of here

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next question enough ! the reaction is 300 – = mm from the centre the middle third is 100 mm from the centre so the reaction is outside the middle third Pier 600 x 600mm 8kN 1kN x R = 9 kN h = mm

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try again How far is the reaction from the centre? How wide is the middle third? let me try again let me out of here

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next question enough ! If the middle third rule holds, the factor of safety would be > 3 since the factor of safety is only 2.4 we would expect the reaction to be just outside the middle third

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let me try again let me out of here THINK !! of course you can ! what would the factor of safety be if the middle third rule holds? what is the actual factor of safety?

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let me try again let me out of here THINK !! what would the factor of safety be if the middle third rule holds? what is the actual factor of safety?

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next question enough ! the compressive stress due to the weight P/A = 9000 / (600 x 600) = MPa The stress due to the eccentricity Pe/Z = 1 x 1300 / (600 x /6) = MPa Total stress = / LHS = = MPa (comp) RHS = – = MPa (tension) Pier 600 x 600mm 8kN 1kN X h = mm stress = P/A ±Pe/Z Z = bd 2 /6 R = 9 kN mm

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let me try again let me out of here stress = P/A +/- Pe/Z compression +/- compression / tension what is P (total downward load)? what is A (area of the pier)? what is P (the eccentric force)? what is e (the eccentricity)? what is Z (the Section Modulus of the pier)? Pier 600 x 600mm 8kN 1kN X h = mm stress = P/A ±Pe/Z Z = bd 2 /6 R = 9 kN mm

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next question enough ! since the reaction is outside the middle third, we would expect tension to develop on the side away from the eccentric force.

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try again is the reaction in the middle third? so what does that mean? let me try again let me out of here

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let me try again let me out of here THINK !! of course it is possible to tell is the reaction in or outside the middle third? SO?

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next question enough ! since it is the weight of the pier which produces the stabilizing moment, then it stands to reason that anything that increases the weight will increase the stabilizing moment and hence make the pier safer

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let me try again let me out of here THINK !! of course it is possible to tell will increasing the weight of the pier increase the stabilizing moment? SO?

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THINK!! let me try again let me out of here The greater the stabilizing moment the less likely the pier is to overturn or lift off its base. What contributes to the stabilizing moment?

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next question enough ! for 1 m length of wall the volume, V of the wall is: V = 1 x 1.2 x 0.23 m 3 V = m 3 the weight of the wall, W, is volume x density W = x 19 kN W = 5.24 KN 1200mm self Weight = 5.24 kN wind 0.5kPa R x A 115 h 600

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let me try again let me out of here there must be an error in your calculations

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a weight is a force let me try again let me out of here what are the units of force?

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next question enough ! for 1 metre length of wall, the area, A, of the wall is A = 1 x 1.2 m 2 A = 1.2 m 2 for a wind load of 0.5 kPa, the total wind load, WL, is WL = 1.2 x 0.5 kN WL = 0.6 kN 1200mm self Weight = 5.24 kN wind 0.6 kN R x A 115 h 600

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let me try again let me out of here there must be an error in your calculations

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a weight is a force let me try again let me out of here what are the units of force?

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ΣV = 0 next question enough ! if the total downward load is 5.24 kN the total upward load must equal 5.24 kN 1200mm self Weight = 5.24 kN wind 0.5kPa R = 5.24 kN x A 115 h 600

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let me try again let me out of here THINK !! ΣV = 0 what is the total downward load? So what must the total upward load be

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next question enough ! taking moments about A clockwise moments = anti-clockwise moments W * 115 = R * x x * 115 = 5.24 * x x = 5.24x x = / 5.24 x = h = 115 – x h = 68.7 mm 1200mm self Weight = 5.24 kN wind 0.6 kN R = 5.24 kN x A 115 h = 68.7 mm 600 for a graphic method of solving for h CLICK HERE

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let me try again let me out of here taking moments about A, what are the clockwise moments? What are the anti-clockwise moments? remember ΣM = 0 (you can also solve this using a graphical method)

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next question enough ! the base is 230 mm wide. So half the width is 115 mm. The reaction is 68.7 mm from the centre So the reaction is within the base.

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let me try again there’s no maybe about it it is or it isn’t let me out of here

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try again What is the width of the base? What is the half-width of the base? How far is the reaction from the centre? let me try again let me out of here 1200mm self Weight = 5.24 kN wind 0.6 kN R = 5.24 kN x A 115 h = 68.7 mm 600

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next question enough ! as the reaction is within the base the pier will not overturn

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we really don’t know yet we have to know something first let me try again let me out of here

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let me try again isn’t the reaction within the base ? let me out of here

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The base is 230 mm wide 1/3 of that is 76.7 mm so from the centre of the pier, the middle third is 38.3 mm the reaction is 68.3 mm from the centre So the reaction is outside the middle third next question enough ! 38.3 mm 68.3 mm 230 mm

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let me try again there’s no maybe about it it is or it isn’t let me out of here

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what is the distance of the middle third from the centre of the pier? what is the distance of the reaction from the centre of the pier let me try again let me out of here

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next question enough ! Yes, if the reaction is outside the middle third then tension tends to develop on the RHS (in this case) and if the pier does not stick to its base, it will lift off.

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let me try again let me out of here

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Firstly, the middle third rule doesn’t deal with overturning Secondly, didn’t we say before that the pier won’t overturn? let me try again let me out of here

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for mathematical answer enough ! This is not the mathematical answer but it is the practical answer Since it is the lowest width of a back-hoe that would do

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next question enough ! This is the mathematical answer but in practice we would make it 450 mm wide Since it is the lowest width of a back-hoe that would do distance of reaction from centre = 68.3 mm so middle third would have to be 68.3 mm from centre so width of base would have to be 6 x 68.3 = 410 mm 68.3 mm mm 410 mm

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Think of how wide the middle third would have to be so that the reaction would be within it let me try again let me out of here

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that’s the end! zig-zag plan stepped plan attached piers heavy coping on top

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you’re not trying let me try again let me out of here

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let me try again let me out of here It would be expensive

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back we draw congruent triangles. one represents the forces and the other represents distances 1200mm self Weight = 5.24 kN wind 0.6 kN R = 5.24 kN x A 115 h = 68.7 mm kN 0.6 kN 600 mm h mm then h / 600 = 0.6 / 5.24 h = (0.6 / 5.24) x 600 h = mm

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