The Warren Truss uses equilateral triangles to spread out the loads on the bridge. Examples of it can be found everywhere in the world. Regardless, the Warren Truss has been around a while and has been very popular. His patent was more about the methodology of building rather than a “design”. James Warren patented a design in 1848 (in England), which many attribute the name “Warren Truss”. It’s exact history and origination is a little muddled, however. They can be placed either horizontally or inclined.The Warren Truss is a very common design for both real and model bridges. Σ = F Ed /(200 x 300) = (600 x 10 3)/(200 x 300) = 10 N/mm 2 300 mm, when the area of the primary horizontal tie A s is such that (where A c denotes the sectional area of the concrete in the corbel at the column), then closed stirrups, having a total area not less than |0.4|As, should be distributed over the effective depth d in order to cater for splitting stresses in the concrete strut. The node 2 is a tied-compressed node, where the main reinforcement is anchored the compressive stress below the load plate is Node 2 verification, below the load plate Trying 2 legs of H10 mm, Provide 4 number of H10 mm stirrups (A sw,prov = 628 mm 2) Strut-and-tie model resolution in two elementary beams and partition of the diagonal stress Fdiag.į wd = x F c = x 335.93 = 180.525 kNĪ sw = F wd/f yd = (180.525 x 10 3)/435 = 415 mm 2 > k 1A s,prov = 0.25 x 1005 = 251.25 mm 2 The beam proposed in EC2 is indeterminate, then it is not possible to evaluate the stresses for each single bar by equilibrium equations only, but we need to know the stiffness of the two elementary beams shown below in order to make the partition of the diagonal stress between them.īased on the trend of main compressive stresses resulting from linear elastic analysis at finite elements, some researchers from Stuttgart have determined the two rates in which F diag is divided, and they have provided the following expression of stress in the secondary reinforcement. The distance y 1 of the node 1 from the lower border is evaluated setting the internal drive arm z equal to 0.8⋅d (z = 0.8 x 407 = 325.6 mm): The compression stress in each strut is restricted to a maximum value of 0.75(1 – f ck/250)f cd. (c) CTT node: If two tensile forces at a compressive force meet at a node it is called a CTT node. According to Eurocode 2 equation (6.61), the compression stress in each strut is restricted to a maximum value of 0.85(1 – f ck/250) f cd. (b) CCT node: If two compressive forces and a tie force anchored in the node through bond meet, it is called a CCT node. According to Eurocode 2 equation (6.60), the compression stress in each strut is restricted to a maximum value of 1.0(1 – f ck/250)f cd. (a) CCC Node: If three compressive forces meet at a node, it is called a CCC node. These nodal zones need to be carefully designed and detailed. When using the strut−and-tie method of design, because the given structure is replaced by a pin-jointed truss, different types of nodes occur where members meet. Then the structure is safe under the given system of external loads.
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