Example 1

Calculate reactions using the Menabrea-Castigliano method

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Solution

We write the equations of static equilibrium and then proceed as in the Castigliano's method, except that the force with respect to which we take the derivative is one of the statically indeterminate reactions.

∑𝑋=0𝑅2=0∑𝑌=0𝑅1+𝑅3−5⁢𝑞⁢𝑙=0⁢(1)∗∑𝑀1=0𝑀𝑈−3⁢𝑙∗𝑅3+5⁢𝑞⁢𝑙∗52⁢𝑙=0⁢(2)∗∫𝐿11𝐸∗𝐼⁢𝑀𝑔⁡1∗𝜕𝑀𝑔⁡1𝜕𝐻+∫𝐿21𝐸∗𝐼⁢𝑀𝑔⁡2∗𝜕𝑀𝑔⁡2𝜕𝐻=0𝑀𝑔⁡1=−12∗𝑞⁢𝑥2𝑀𝑔⁡2=−12⁢𝑞⁢𝑥2+𝑅3∗(𝑥−2⁢𝑙)𝜕𝑀𝑔⁡1𝜕𝑅3=0𝜕𝑀𝑔⁡2𝜕𝑅3=𝑥−2⁢𝑙

The partial derivative of the elastic energy of the entire system with respect to the statically indeterminate quantity is equal to zero.

∫5⁢𝑙2⁢𝑙[−12⁢𝑞⁢𝑥2+𝑅3⁢(𝑥−2⁢𝑙)]∗(𝑥−2⁢𝑙)𝑑𝑥=0∫5⁢𝑙2⁢𝑙−12⁢𝑞⁢𝑥2⁢(𝑥−2⁢𝑙)𝑑𝑥+∫5⁢𝑙2⁢𝑙𝑅3⁢(𝑥−2⁢𝑙)𝑑𝑥=0−2978⁢𝑞⁢𝑙4+9∗𝑅3⁢𝑙3=0𝑅3=338⁢𝑞⁢𝑙

We add the remaining reactions

(1)∗⁢𝑅1=5⁢𝑞⁢𝑙−𝑅3=78⁢𝑞⁢𝑙(2)∗⁢𝑀𝑈=3⁢𝑅3⁢𝑙−252⁢𝑞⁢𝑙2=−18⁢𝑞⁢𝑙2