1. WORKSHEET 7 AXIALLY LOADED MEMBERS

2. Q1 Are tension structures or compression structures more efficient when loaded axially ? tension structures tension structures are not subject to buckling they pull straight - can use thin cables can use: stress = Force / Area Q2 why ?

3. Q3 What is meant by: (a) a short column a column which will fail in true compression (b) a long column a column which buckles before full compressive strength is reached

4. Q4 What affects the buckling load (or buckling stress)? (a) the slenderness ratio - l/r (or l/B rect. sections) (b) the modulus of elasticity, E materials with higher E will buckle less (c) end fixing conditions - restraints but included in slenderness ratio - effective length free end Eff Length = 2 x l, fixed ends Eff Length = 0.5 x l the slenderness ratio takes into account the effective length, l and the stiffness (radius of gyration) of the X-section r =  (I/A) buckling load inversely proportional to slenderness ratio i.e. greater the slenderness ratio - more will tend to buckle

5. Q5 What are good sections for columns? (i) sections which have similar radii of gyration in all directions Q6 (i) so that they do not buckle in a weak directions (ii) sections in which the major part of the material is a far from the Centre of Gravity as possible (ii) to use the material efficiently Why?

6. Q7 What are two effects which can cause a pier to overturn? (a) a horizontal load (b) an eccentric vertical load

7. Q8 What is the middle-third rule? the middle-third rule tells you that as long as the resultant reaction falls in the middle-third of the base of the pier then no tension will develop in the pier. there will be a factor of safety of >3 if the middle-third rule holds if the middle-third rule holds the pier will not lift off its base

8. Q9 (a) The diagram shows a heavy steel gate hung from a hollow brick pier which weighs 8kN. Investigate the stability of the pier. (a) assume the pier is sitting on (but not stuck to) a strong concrete footing. take moments about the point X (i) will the pier overturn?No Overturning Moment (clockwise): = 1 kNm Potential Stabilizing Moment (anticlockwise): = 2.4 kNm (ii) what is the margin of safety?2.4:1 OTM can increase up to 2.4kNm before overturning occurs M =1 x 1 M =8 x 0.3 Pier 600 x 600mm 8kN 1kN 3001000 stress = P/A ±Pe/Z Z = bd 2 /6 X R=9 h

9. Q9 (b) (b) where is the resultant of the two loads? (i) is it within the middle third of the base?No In (a) we assumed that there would be no crushing of the leading edge of the pier. Under this condition the middle-third rule gives a factor of safety of >3. For a factor of safety of 2.4 we would expect the reaction to be just outside the middle-third 155.6mm from X 144.4mm from centre (ii) is this what you would expect from (a)? 1 x 1000 + R x h = 8 x 300 9 x h = 2400 -1000 h = 1400 / 9= 155.6

10. Q9 (c) (c) calculate the stress distribution under the pier (i) is it trying to develop tension on one side? since the reaction is outside the middle third you would expect tension to tend to develop (ii) is this what you would expect from (b)? stress = P/A (compressive part) ± M/Z (bending part) M = P x e = 1 x 1300 (or 9 x 144.4)= 1300kNmm Z = bd 2 /6 = 600 x 600 2 / 6= 36 x 10 6 mm 3 stress = 9000 / (600 x 600) ± 1300000 / 36 x 10 6 = 0.025 ± 0.036 MPa on the gate side the stress is 0.061 MPa (61kPa) - compression on the other side it would be -0.011 MPa (11kPa) (tension) yes since the interface cannot develop tension, a redistribution would occur a different analysis is required to find that the maximum compressive stress would increase to 0.096 MPa yes

11. Q9 (d) (d) if you make the pier solid, it will be about twice as heavy. Will this make it safer against overturning? The potential stabilising moment would double while the overturning moment remains the same. No tension would develop (the reaction would be within the middle third) and it would be safer against overturning yes

12. Q10(a) (a) Find the location of the reaction on the base weight of wall (of length 1m) = 1.2 x 0.23 x 1 x 19 A freestanding garden wall is 230mm thick and 1200mm high. The density of brick is 19kN/m3. The wind load in this location is 0.5 kPa = 5.244 kN = 0.6 kN taking moments about A: overturning moment = 0.6 x 600 = 360 kNmm restraining moment = 5.244 x 115 = 603.1 kNmm 5.244 x h + 360 = 603.1 = 46.4 mm (i) is it within the base? no(i) is it within the middle third? the reaction is 115 - 46.4 = 68.6mm from the centre of the wall. the middle third is 38.3mm from the centre of the wall. so the reaction is well outside the middle third. yes wind load on 1m length of wall = 0.5 x 1 x 1.2 h = 243.1/5.244 1200mm self weight wind 0.5kPa R = weight h A 115 x = 115-h 600

13. Q10(a) (a) Find the location of the reaction on the base weight of wall (of length 1m) = 1.2 x 0.23 x 1 x 19 A freestanding garden wall is 230mm thick and 1200mm high. The density of brick is 19kN/m3. The wind load in this location is 0.5 kPa = 5.244kN = 0.6kN GRAPHICAL METHOD the reaction is 68.6mm from the centre of the wall. wind load on 1m length of wall = 0.5 x 1 x 1.2 1200mm self weight wind 0.5kPa R = weight h A 115 x 5.244kN 0.6kN x 600 X/600 = 0.6 / 5.244 X = 0.6 x 600 / 5.244 X = 68.6

14. Q10(b, c) (b) How wide would the footing have to be to bring the reaction within the middle third? Forgetting the weight and depth of the footing, width of footing would have to be 6 x 68.6 = = 412mm (c) What other options are there for increasing the stability of the wall? (i) put a heavy coping on top (has to be wide rather than high otherwise subject to wind) (ii) make the wall thicker (expensive) (iii) attached piers 5.44kN 68.6 We would probably make it 450mm wide. This is the width of a common backhoe bucket. (iv) zigzag plan (similar to (iii) but more interesting 2 x 68.6 6 x 68.6