BUCKLING

The Hong Kong Polytechnic University Dept. of building and Real Estate

Department of Building and Real Estate

The Hong Kong Polytechnic University

BRE 204 Structure I

Laboratory #1Intended learning outcomes:

1. apply the concepts of structural mechanics to solve structural problems involving statically determinate trusses, and;

2. quantify and analyse the internal forces of a determinate truss.Experiment on SIMPLE TRUSS WITH SIMPLE SUPPORTS1. Introduction

For simple structures, it is assumed that all the connection between individual members will be considered as pin-jointed structure, i.e. there is no bending moment built up at all the pin-joints.

The term simple structure applies to those trusses, which are made up of individual triangular elements, and all these elements are connected by pin-jointed method.

Fig.1 Fig. 2

Since each individual triangle is pin-jointed at its corners, the truss is capable of carrying load until the load carrying capacity of any one of its members are exceeded.

2. Structural equilibrium for simple truss

For a simple structure in equilibrium, each individual pin-joint will be in equilibrium under the action of forces which act on the joint i.e. the external forces applied to any member must be balanced by internal forces created within that member if equilibrium is to be maintained. When the external forces acting on a member tend to increase its length then the internal forces will counteract these external forces as shown below. The member is then in tension and the member is known as a TIE.

Fig. 3

When the external forces tend to shorten the length of the member then internal forces will counteract these external forces as shown below. The member is then in compression and the member is known as a STRUT.

Fig. 4

Thus a member is a TIE (tension) if the internal forces pull on a joint.

A member is a STRUT (compression) if the internal forces push on a joint.

In this experiment, we use strain gauges bonding to each members of the truss to determine the internal forces in each truss members of a simple 2-D pin-jointed structure as shown below: Fig. 5

*** All members of the truss are pin-jointed

All pin to pin distance = 250mmThe equilibrium condition is that the resultant forces at each pin- joint are zero. It is usual to resolve the forces at each pin-joint into vertical and horizontal components. The equilibrium condition at each joint requires that

vertical force components = 0

horizontal force components = 0

As there are only two simultaneous equations for each joint, no more than two unknown forces are allowed. When determining the forces at a particular pin joint, the initial assumption is made that the unknown force are positive tensile forces. If the result is negative, the forces are in compression. The resolution must be commenced at a joint where there are not more than two unknown forces i.e. usually start at the two supports of known reaction forces.Experimental procedure:

Under the loading condition as in Fig 5 above

1. Using the dummy truss frame determining the compression or tension state in each of the truss members by removing one of the dummy truss members at each time.

2. Record your observations for each truss members and fill into the following table:

*** a + symbol represents that the member is a TIE, i.e. in Tension

a symbol represents that the member is a STRUT, i.e. in Compression

3. Using the testing truss frame set the initial strain gauge meter reading to zero. Hang a dead weight W=40N to the loading point carefully as shown in Fig. 5 above and then record the force readings in each of the truss members.

*** Be aware of the +/- sign before the strain gauge force reading.

+ means tension state and - means compression state

4. Calculate the reaction forces R1 and R2 at the testing from frame supports when W=40N.

Discussion and calculation:

1. What is meant by simple frame and simple support?

2.Calculate step by step clearly all the internal forces in the truss members from no.0 to no.10 by using any method you are most familiar with.

No. 0No. 1No. 2No. 3No. 4No. 5No. 6No. 7No. 8No. 9No. 10

By experiment(N)

By theory(N)

3. Compare the results obtained by calculation and by experiment.4. Conclusion.End of Experiment

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