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Numbers <span style="background-color:#F8F8F8; border-style:solid; border-width:1px; border-color:#AAAAAA;">in grey</span> are balanced moments; arrows (<span style="border-style:solid; border-width:1px; border-color:#AAAAAA;"> → / ← </span>) represent the carry-over of moment from one end to the other end of a member.
* Step 1: As joint A is released, balancing moment of magnitude equal to the fixed end moment <math>M_{AB}^{f} = 14.700 \mathrm{\,kN \,m}</math> develops and is carried-over from joint A to joint B. * Step 2: The unbalanced moment at joint B now is the summation of the fixed end moments <math>M_{BA}^{f}</math>, <math>M_{BC}^{f}</math> and the carry-over moment from joint A. This unbalanced moment is distributed to members BA and BC in accordance with the distribution factors <math>D_{BA} = 0.2727</math> and <math>D_{BC} = 0.7273</math>. Step 2 ends with carry-over of balanced moment <math>M_{BC}=3.867 \mathrm{\,kN \,m}</math> to joint C. Joint A is a roller support which has no rotational restraint, so moment carryover from joint B to joint A is zero. * Step 3: The unbalanced moment at joint C now is the summation of the fixed end moments <math>M_{CB}^{f}</math>, <math>M_{CD}^{f}</math> and the carryover moment from joint B. As in the previous step, this unbalanced moment is distributed to each member and then carried over to joint D and back to joint B. Joint D is a fixed support and carried-over moments to this joint will not be distributed nor be carried over to joint C. * Step 4: Joint B still has balanced moment which was carried over from joint C in step 3. Joint B is released once again to induce moment distribution and to achieve equilibrium. * Steps 5 - 10: Joints are released and fixed again until every joint has unbalanced moments of size zero or neglectably small in required precision. Arithmetically summing all moments in each respective columns gives the final moment values. === Result ===
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=== Result via displacements method ===
As the Hardy Cross method provides only approximate results, with a margin of error inversely proportionate to the number of iterations, it is important{{citation needed|date=September 2012}} to have an idea of how accurate this method might be. With this in mind, here is the result obtained by using an exact method: the [[
For this, the displacements method equation assumes the following form:
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