ENGIN2203 Structural Analysis Topic 4
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ENGIN2203 Structural Analysis
Topic 4: Analysis of statically determinate structures and trusses
1. Analysis of Structure
One of the initial and important difficulties of structural analysis the accuracy of modelling the structure. Many estimations such as loadings, points of application of loadings, strength of material, etc. need to be made on the structure in order for it to be analysed. The structural engineer needs to develop a simplified and idealized model so that the practical force analysis can be performed from there. Some of these idealizations include the supports and connection of structure, and loadings.
Supports and connections
Depending on how each member is designed to transfer the loads to grounds, different supports may be used. The three basic and well-known types of supports are roller, pin and fixed connections. Examples of two common metal and concrete connections are shown below.
Typical pin support of a metal connection Typical fixed support of a concrete connection
Idealized model:
Idealized model:
In steel connections, the gusset plates, bolts or welds can change the connection from pinned to fixed. In reinforced concrete members, the connectivity of reinforcements and the concrete from one member to the other governs how the loads are transformed. For most timber structures, the members are assumed to be pin connected, since bolting or nailing them will provide sufficient restrain to avoid rotations completely and act as a fixed connection. Another simplification for beam analysis is that beam thickness can be neglected since it is small compared to the length of the beam.
Idealized structure
The structural components of a system and the applied loadings are usually represented in a simple way for analysis. Beam and column elements can be represented as lines and
the concentrated loads as point loads on these lines.
actual structure idealized structure
In case of slabs which are supported by beams or joists, the beams and joists can also be represented as lines on the plan. Depending on how the beams are connected to the system, supports are modelled at both ends of the system. The roof deck shown below is supported by wood joists. These joists are idealized as lines which deliver loads to the masonry walls that can be assumed as simply supported connection.
Loadings
When plane surfaces are supported by structural frames, the load transition on each element becomes necessary. In one-way slabs, which rest on a support system and the loads are transmitted in a one way fashion. This is where the supporting beams or walls are parallel and the floor simply spans one way between them. One-way floors are typically constructed of either reinforced concrete or timber. The loading on the supporting members is a UDL and is typically determined based on the assumption that the member supports the floor to half-way to the adjacent members (the shaded rectangular area in the figure below). The applied uniform loads acting on each rectangular area can be considered as a line load. These line loads are then transferred onto the supporting girders as point loads.
One-way slab: load distribution on beam
Two-way slab systems are cases where the load is transferred to the supporting beams and girders in two directions. In order to transfer the uniform load of the slab, 45 degree lines are used on the slab to form the areas that each beam is withholding the load. Depending on the dimensions of the slab, the areas and thus the load distribution may be different. The uniform line loads applied on each beam in case of a square slab is
triangular while that of other types of rectangular slabs will have triangular and trapezoidal shapes. This assumed distribution indicates that the supporting perimeter members (walls, in this case) are supporting varying widths of the floor. Recalling the rule from previous lectures (UDL in kN/m = UDL in kN/m2 × width of load supported in m.), this means that the distributed loads on the supporting members will also be varying. This loading distribution is assumed to apply also to ‘multi-panel’ floors; i.e. where the slab is continuous over the supporting members into other spans.
Two-way squsre and rectangular slab: load distribution on beam (loading on long and short side)
One-way or two-way slab?
According to the American Concrete Institute (ACI 318 code) and also the Australian Standard (AS3600) if the ratio of dimensions of a slab (L1/L2) exceeds 2, the slab is considered to be a one-way slab while less than 2 acts as two-way slab.
2. Analysis of Trusses
Trusses are specific types of structures composed of slender members which are connected together at their ends. The connections are normally formed using bolts or welds at the member ends which can be idealised as a pinned connection for each member. These members are assumed to be joint together by smooth pins. This happens in cases where the truss members are connected at their centrelines. In some occasions, a gusset pate is used to connect all the members which are intersecting at one joint. Another important assumption for truss structures is that all loads are applied on the joints rather than the element itself, which is true in most cases. It should also be noted
that the self-weight can be neglected since its value is small compared to the applied forces. Having all the forces applied only on ends, no bending occurs in truss members and therefore, the loads are categorised axial forces - either tension or compression. Trusses have various applications in buildings and structures including roofs, bridges, etc.
There are two main methods to analyse a truss which are known as the method of joints and method of sections. These methods have already been covered in Engineering Statics unit. As a reminder the general procedure adopted in each of these methods are summarised below.
Method of joints:
Draw the free body diagram > calculate the external reactions if required > assume the coordinates in a way that the unknowns can easily be resolved > solve each joint for force equilibrium equations in the two main directions > try to start with joints with the least
number of unknowns using member forces previously calculated.
Method of sections:
Make a decision on how to cut the truss to have the least unknowns and forces on it > determine the reaction forces if necessary > draw the free body diagram of the cut section > take moments at a point with the most unknowns > use the force equilibrium equations to calculate the remaining unknowns.
Complex trusses:
These trusses are a category of trusses that can’t be analysed simply by the method of joints or sections described previously, although they are statically determinate. The complexity with these types of trusses is that the member forces can’t be obtained at one joint or at one section only. The equilibrium equations in these cases need to be solved simultaneously. An example of a complex truss is shown below.
Space trusses:
Due to their special arrangement, these trusses can’t be subdivided into individual plane trusses therefore, they need to be analysed at their three-dimensional nature. All the assumption which were made in 2D trusses such as connection of members on centrelines or load applied only on joints are also applicable to space trusses. To analyse space trusses, six equilibrium equations are solved to calculate the three unknown
reaction forces in case of a statically determinate structure.
Three support types (pin, roller and fixed supports) known in 2D structures can be extended to a space form with 1 up to 6 unknown reactions based on the support category. Member forcers in space trusses obviously have three components in the x, y
and z direction.
1. The steel framework is used to support the reinforced stone concrete slab that is used for an office. Sketch the loading that acts along members BE and FED. Take a=3m and b=4m. (concrete slab is 200mm thick and unit weight: 23.6 kN/m3)
Take the live load as 2.4 kN/m2 for this question.
2. Determine the horizontal and vertical components of reactions at supports.
Consider joint C as a pin.
3. Determine the force in members AB, CD and DG of the compound truss below.
4. Determine the forces in member AF of the truss below.
lk
5. Determine the force in members BC, AB, AG and BE of the space truss below and state whether each member is in tension or compression. This truss is indeterminate (which will be discussed in future topics of this course), however, using symmetry of the geometry and loading, you are able to analyse the structure for member forces.
6. Determine the components of the reaction acting at the smooth journal bearings A,B and C. (Hint: You will need to decide whether you can ignore the moment components of the reactions)
Answers:
1. :
2. NA =12kN, Cy = 12kN, Cx=0, Bx =0, By =30 kN, Mb =84 kN.m
3. FAB = 22.5 k (T), FCD = 10 k, FDG =20.62 k (T)
4. FAF =646 lb (C)
5. FAB = 2.4 kN (C), FAG =1.01 kN (T), FBE =1.80 kN (T)
6. Cy=-318N, Cz= 500N, Bz=-273N, Bx = 239 N, Ax=-239N, Az=90.9 N
2023-02-19
Analysis of statically determinate structures and trusses