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University of Sydney – BDes Design Studies 1A - Structures SYSTEMS Peter Smith & Mike Rosenman EQUILIBRIUM As a wholeA system of elements.

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Presentation on theme: "University of Sydney – BDes Design Studies 1A - Structures SYSTEMS Peter Smith & Mike Rosenman EQUILIBRIUM As a wholeA system of elements."— Presentation transcript:

1 University of Sydney – BDes Design Studies 1A - Structures SYSTEMS Peter Smith & Mike Rosenman EQUILIBRIUM As a wholeA system of elements

2 University of Sydney – BDes Design Studies 1A - Structures SYSTEMS Peter Smith & Mike Rosenman Structural Systems The building must be stable as a whole There must be enough structural elements They must be in suitable places They must be strong and stiff enough, and suitably connected together As a wholeA system of elements 1/26

3 University of Sydney – BDes Design Studies 1A - Structures SYSTEMS Peter Smith & Mike Rosenman Stability Structure may be in equilibrium but not stable Arrangement of parts critical 2/26

4 University of Sydney – BDes Design Studies 1A - Structures SYSTEMS Peter Smith & Mike Rosenman Qualitative and Quantitative Understanding First we will try to understand the structural system qualitatively If we don’t have the right kind of supports in the right kind of places, it won’t work When we have a credible system, we can use precedent and simple rules to get credible sizes We need to understand the quantitative basis of a final structural design 3/26

5 University of Sydney – BDes Design Studies 1A - Structures SYSTEMS Peter Smith & Mike Rosenman Equilibrium of the Whole Building The building supports its loads All loads are finally resisted by the ground Parts of the building intervene between loads and foundation But first, let’s make sure the building as a whole is stable Loads Reactions from Foundation ? Building 4/26

6 University of Sydney – BDes Design Studies 1A - Structures SYSTEMS Peter Smith & Mike Rosenman Overall Equilibrium What do we Need to Know? all the possible loads Loads ? Building Reactions from Foundation where the building can be supported how big the reactions have to be for equilibrium 5/26

7 University of Sydney – BDes Design Studies 1A - Structures SYSTEMS Peter Smith & Mike Rosenman Overall Equilibrium What are the Problems? downward loads just need big enough footings Reactions from Foundation Building Loads Bigger building 6/26

8 University of Sydney – BDes Design Studies 1A - Structures SYSTEMS Peter Smith & Mike Rosenman Overall Equilibrium What are the Problems? (cont.) horizontal loads might overturn a tall building Reactions from Foundation Taller building Building Loads wider base and heavy building are more stable 7/26

9 University of Sydney – BDes Design Studies 1A - Structures SYSTEMS Peter Smith & Mike Rosenman Equilibrium of Forces Newton’s Third law Reactions from Foundation ‘to every force there is an equal and opposite reaction’ 8/26

10 University of Sydney – BDes Design Studies 1A - Structures SYSTEMS Peter Smith & Mike Rosenman Equilibrium (forces in line) Every force is resisted by an equal and opposite one, exactly in line 9/26

11 University of Sydney – BDes Design Studies 1A - Structures SYSTEMS Peter Smith & Mike Rosenman Equilibrium (forces out of line) Tendency to overturn The turning effect is a moment Can be resisted by other out-of-line forces Maintains equilibrium Tries to overturn Tries to restrain 10/26

12 University of Sydney – BDes Design Studies 1A - Structures SYSTEMS Peter Smith & Mike Rosenman Equilibrium l forces on whole structure and each part, just balance (sum equal zero) doesn’t move l not up-and-down, sideways or spin  V = 0  H = 0  M = 0 12/26

13 University of Sydney – BDes Design Studies 1A - Structures SYSTEMS Peter Smith & Mike Rosenman EQUILIBRIUM (cont. 1) wwwwww H  V = 0  H = 0  M = 0  w = 6w F = 5kN 13/26

14 University of Sydney – BDes Design Studies 1A - Structures SYSTEMS Peter Smith & Mike Rosenman EQUILIBRIUM (cont. 2) wwwwww H  w  V = 0  H = 0  M = 0 H 60kN 14/26

15 University of Sydney – BDes Design Studies 1A - Structures SYSTEMS Peter Smith & Mike Rosenman EQUILIBRIUM (cont. 3) wwwwww H M  w H  V = 0  H = 0  M = 0 M 15/26

16 University of Sydney – BDes Design Studies 1A - Structures SYSTEMS Peter Smith & Mike Rosenman Equilibrium of Elements total downward load = total upward reaction if the load is symmetrical, so are the reactions beam 5 units 2.5 units 16/26

17 University of Sydney – BDes Design Studies 1A - Structures SYSTEMS Peter Smith & Mike Rosenman Equilibrium of Framework Total downward load is carried down by columns beam 5 units 2.5 units We can follow the ‘Load Path’ 17/26 2.5 units

18 University of Sydney – BDes Design Studies 1A - Structures SYSTEMS Peter Smith & Mike Rosenman Equilibrium of Framework (cont.) If the load is off-centre, so are the reactions 5 units beam 4L/5L/5 1 unit 4 units 18/26

19 University of Sydney – BDes Design Studies 1A - Structures SYSTEMS Peter Smith & Mike Rosenman Work smarter, not harder Most beams are symmetrical If the reactions are equal, don’t make hard work of it W R1R1R2R2 Everything is symmetrical R1 = R2 = W/2 R2R2R1 W is off centre R1, R2 must be calculated W (central) 19/26

20 University of Sydney – BDes Design Studies 1A - Structures SYSTEMS Peter Smith & Mike Rosenman Work smarter, not harder (cont.) The  M condition says the sum of moments about any point is zero Pick a point that eliminates one of the unknowns, to make it easy W (known) R1R1R2R2 Only W and R2 have a moment about the dot 20/26 R2R2R1 W (known) R1 and R2 have a moment about the dot

21 University of Sydney – BDes Design Studies 1A - Structures SYSTEMS Peter Smith & Mike Rosenman Finding the Reactions of a Cantilever A cantilever has one V, one H, and one M reaction M R H=0 W1W2 d2 d1 Vertically: R = W1 + W2 Horizontally: H = 0 unless there is a horizontal load Moments: M = W1.d1 + W2.d2 22/26 d1 d2

22 University of Sydney – BDes Design Studies 1A - Structures SYSTEMS Peter Smith & Mike Rosenman Free-bodies (Looking Inside the Elements) l can isolate any member or part of it to study it l must put back artificial forces to replace whatever supports were cut away 23/26

23 University of Sydney – BDes Design Studies 1A - Structures SYSTEMS Peter Smith & Mike Rosenman Free-bodies 100kg 1kN can ‘cut’ the wire at any point 1kN 24/26

24 University of Sydney – BDes Design Studies 1A - Structures SYSTEMS Peter Smith & Mike Rosenman Free-bodies (cont1.) 20kg 100kg 1kN 0.2kN 1kN 0.2kN 2kN 1kN 0.2kN 2kN 1.4kN 2kN 0.2kN 1kN 0.2kN 1.8kN 25/26

25 University of Sydney – BDes Design Studies 1A - Structures SYSTEMS Peter Smith & Mike Rosenman Free-bodies (cont2.) 26/26


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