Eng. 6002 Ship Structures 1 Hull Girder Response Analysis Lecture 9: Review of Indeterminate Beams

Overview The internal forces in indeterminate structures cannot be obtained by statics alone. This is most easily understood by considering a similar statically determinate structure and then adding extra supports This way also suggests a general technique for analyzing elastic statically indeterminate structures

Statically Indeterminate Beams A uniformly loaded beam is shown with three simple supports. If there had been only two supports the beam would have been statically determinate So we imagine the same beam with one of the supports removed and replaced by some unknown force X

Statically Indeterminate Beams If the center support were removed the beam would sag as illustrated The sag at the centre is counteracted by the reaction force X1, providing an upward displacement Note: the subscript 0 is used to denote displacements generated by the original external load on the statically determinate structure

Statically Indeterminate Beams In the original statically indeterminate structure there is no vertical displacement of the centre due to the support Thus, the force X1 must have a magnitude that exactly counteracts the sag of the beam without the centre support

Statically Indeterminate Beams There are two approaches for solving indeterminate systems. Both approaches use the principle of superposition, by dividing the problem into two simpler problems, soling the simpler problems and adding the two solutions. The first method is called the Force Method (also called the Flexibility Method). The idea for the force method is;

Statically Indeterminate Beams The idea for the force method is: Step 1. Reduce the structure to a statically determinate structure. This step allows the structure to displace where it was formerly fixed. Step 2. solve each determinate system, to find all reactions and deflections. Note all incompatible deflections Step 3. re-solve the determinate structures with only a set of self-balancing internal unit forces at removed reactions. This solves the system for the internal or external forces removed in 1. Step 4. scale the unit forces to cause the opposite of the incompatible deflections Step 5. Add solutions (everything: loads, reactions, deflections…) from 2 and 4.