Energy methods

First law of thermodynamics

Work performed on a mechanical systems by external forces

+heat that flow into the system from outside
=Increase in internal energy +increase in kinetic energy

𝛿𝛿𝑊𝑊 + 𝛿𝛿𝛿𝛿 = 𝛿𝛿𝛿𝛿 + 𝛿𝛿𝛿𝛿

To apply this law, consider a loaded member in equilibrium. Deflection

component at each point are known (𝑢𝑢, 𝑣𝑣, 𝑤𝑤). Allow infinitesimal
increment in displacement components (𝛿𝛿𝑢𝑢, 𝛿𝛿𝛿𝛿, 𝛿𝛿𝛿𝛿).

𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝑤𝑤

𝜀𝜀𝑥𝑥𝑥𝑥 = , 𝜀𝜀 = , 𝜀𝜀 =
𝜕𝜕𝜕𝜕 𝑦𝑦𝑦𝑦 𝜕𝜕𝜕𝜕 𝑧𝑧𝑧𝑧 𝜕𝜕𝑧𝑧

1 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 1 𝜕𝜕𝑤𝑤 𝜕𝜕𝜕𝜕 1 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

𝜀𝜀𝑥𝑥𝑥𝑥 = + , 𝜀𝜀𝑥𝑥𝑧𝑧 = + , 𝜀𝜀𝑦𝑦𝑧𝑧 = +
2 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 2 𝜕𝜕𝜕𝜕 𝜕𝜕𝑧𝑧 2 𝜕𝜕𝜕𝜕 𝜕𝜕𝑧𝑧

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Energy methods

So

𝜕𝜕(𝛿𝛿𝛿𝛿) 1 𝜕𝜕(𝛿𝛿𝛿𝛿) 𝜕𝜕(𝛿𝛿𝛿𝛿)

𝛿𝛿𝜀𝜀𝑥𝑥𝑥𝑥 = , 𝛿𝛿𝜀𝜀𝑥𝑥𝑥𝑥 = +
𝜕𝜕𝜕𝜕 2 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
𝜕𝜕(𝛿𝛿𝛿𝛿) 1 𝜕𝜕(𝛿𝛿𝛿𝛿) 𝜕𝜕(𝛿𝛿𝛿𝛿)
𝛿𝛿𝜀𝜀𝑦𝑦𝑦𝑦 = , 𝛿𝛿𝜀𝜀𝑥𝑥𝑥𝑥 = + (𝐴𝐴)
𝜕𝜕𝜕𝜕 2 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
𝜕𝜕(𝛿𝛿𝛿𝛿) 1 𝜕𝜕(𝛿𝛿𝛿𝛿) 𝜕𝜕(𝛿𝛿𝛿𝛿)
𝛿𝛿𝜀𝜀𝑦𝑦𝑦𝑦 = , 𝛿𝛿𝜀𝜀𝑦𝑦𝑦𝑦 = +
𝜕𝜕𝜕𝜕 2 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

Consider a volume 𝑉𝑉 with surface area 𝑆𝑆. For adiabatic conditions: no heat
flow into 𝑉𝑉 ⇒ 𝛿𝛿𝐻𝐻 = 0

For static equilibrium 𝛿𝛿𝐾𝐾 = 0

⇒ 𝛿𝛿𝑊𝑊 = 𝛿𝛿𝑈𝑈

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The work 𝛿𝛿𝛿𝛿 can be divided into work of surface force 𝛿𝛿𝑊𝑊𝑠𝑠 and work of
body force 𝛿𝛿𝑊𝑊𝐵𝐵

The stress vector acting on ds has components 𝜎𝜎𝑃𝑃𝑃𝑃 , 𝜎𝜎𝑃𝑃𝑃𝑃 , 𝜎𝜎𝑃𝑃𝑃𝑃

𝜎𝜎𝑃𝑃𝑃𝑃 = 𝑙𝑙𝜎𝜎𝑥𝑥𝑥𝑥 + 𝑚𝑚𝜎𝜎𝑥𝑥𝑥𝑥 + 𝑛𝑛𝜎𝜎𝑥𝑥𝑥𝑥

𝜎𝜎𝑃𝑃𝑦𝑦 = 𝑙𝑙𝜎𝜎𝑥𝑥𝑥𝑥 + 𝑚𝑚𝜎𝜎𝑦𝑦𝑦𝑦 + 𝑛𝑛𝜎𝜎𝑦𝑦𝑦𝑦
𝜎𝜎𝑃𝑃𝑧𝑧 = 𝑙𝑙𝜎𝜎𝑥𝑥𝑥𝑥 + 𝑚𝑚𝜎𝜎𝑦𝑦𝑦𝑦 + 𝑛𝑛𝜎𝜎𝑧𝑧𝑧𝑧

The surface force= Product of these forces with ds

∴ 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 = 𝛿𝛿𝑊𝑊𝑆𝑆 = ∫𝑆𝑆 𝜎𝜎𝑃𝑃𝑃𝑃 𝛿𝛿𝑢𝑢𝑢𝑢𝑢𝑢 + ∫𝑆𝑆 𝜎𝜎𝑃𝑃𝑃𝑃 𝛿𝛿𝑣𝑣𝑑𝑑𝑑𝑑 + ∫𝑆𝑆 𝜎𝜎𝑃𝑃𝑃𝑃 𝛿𝛿𝑤𝑤𝑑𝑑𝑑𝑑
= ∫𝑆𝑆 (𝑙𝑙𝜎𝜎𝑥𝑥𝑥𝑥 + 𝑚𝑚𝜎𝜎𝑥𝑥𝑥𝑥 + 𝑛𝑛𝜎𝜎𝑥𝑥𝑥𝑥 )𝛿𝛿𝛿𝛿𝛿𝛿𝛿𝛿 + ∫𝑆𝑆 (𝑙𝑙𝜎𝜎𝑥𝑥𝑥𝑥 + 𝑚𝑚𝜎𝜎𝑦𝑦𝑦𝑦 + 𝑛𝑛𝜎𝜎𝑦𝑦𝑦𝑦 )𝛿𝛿𝛿𝛿𝛿𝛿𝛿𝛿 +
∫𝑆𝑆 (𝑙𝑙𝜎𝜎𝑥𝑥𝑥𝑥 + 𝑚𝑚𝜎𝜎𝑦𝑦𝑦𝑦 + 𝑛𝑛𝜎𝜎𝑧𝑧𝑧𝑧 )𝛿𝛿𝛿𝛿𝛿𝛿𝛿𝛿

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For volume element, 𝑑𝑑𝑑𝑑 in volume 𝑉𝑉, body forces are given by product
of 𝑑𝑑𝑑𝑑 and body forces per unit volume (𝐵𝐵𝑥𝑥 , 𝐵𝐵𝑦𝑦 , 𝐵𝐵𝑧𝑧 )

𝛿𝛿𝑊𝑊𝐵𝐵 = ∫𝑉𝑉 (𝐵𝐵𝑥𝑥 𝛿𝛿𝑢𝑢 + 𝐵𝐵𝑦𝑦 𝛿𝛿𝑣𝑣 + 𝐵𝐵𝑧𝑧 𝛿𝛿𝑤𝑤)𝑑𝑑𝑑𝑑

𝛿𝛿𝑊𝑊 = 𝛿𝛿𝑊𝑊𝑆𝑆 + 𝛿𝛿𝑊𝑊𝐵𝐵 (𝐵𝐵)

Previously seen that summation of forces in 𝑥𝑥, 𝑦𝑦 and 𝑧𝑧 direction on a

force cube

𝜕𝜕𝜎𝜎𝑥𝑥𝑥𝑥 𝜕𝜕𝜎𝜎𝑥𝑥𝑥𝑥 𝜕𝜕𝜎𝜎𝑥𝑥𝑥𝑥

+ + + 𝐵𝐵𝑥𝑥 = 0
𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
𝜕𝜕𝜎𝜎𝑥𝑥𝑥𝑥 𝜕𝜕𝜎𝜎𝑦𝑦𝑦𝑦 𝜕𝜕𝜎𝜎𝑦𝑦𝑦𝑦
+ + + 𝐵𝐵𝑦𝑦 = 0
𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
𝜕𝜕𝜎𝜎𝑥𝑥𝑥𝑥 𝜕𝜕𝜎𝜎𝑦𝑦𝑦𝑦 𝜕𝜕𝜎𝜎𝑧𝑧𝑧𝑧
+ + + 𝐵𝐵𝑧𝑧 = 0
𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

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Also seen equation set above (A) so

𝛿𝛿𝑊𝑊 = 𝛿𝛿𝑊𝑊𝑆𝑆 + 𝛿𝛿𝑊𝑊𝐵𝐵

𝛿𝛿𝛿𝛿 = � (𝜎𝜎𝑥𝑥𝑥𝑥 𝛿𝛿𝜀𝜀𝑥𝑥𝑥𝑥 + 𝜎𝜎𝑦𝑦𝑦𝑦 𝛿𝛿𝜀𝜀𝑦𝑦𝑦𝑦 + 𝜎𝜎𝑧𝑧𝑧𝑧 𝛿𝛿𝜀𝜀𝑧𝑧𝑧𝑧 + 2𝜎𝜎𝑥𝑥𝑦𝑦 𝛿𝛿𝜀𝜀𝑥𝑥𝑦𝑦 + 2𝜎𝜎𝑥𝑥𝑧𝑧 𝛿𝛿𝜀𝜀𝑥𝑥𝑧𝑧 + 2𝜎𝜎𝑦𝑦𝑧𝑧 𝛿𝛿𝜀𝜀𝑦𝑦𝑧𝑧 )𝑑𝑑𝑑𝑑
𝑉𝑉

Internal energy 𝑈𝑈 = ∫𝑉𝑉 𝑈𝑈0 𝑑𝑑𝑑𝑑

𝑈𝑈0 =Internal energy/unit volume

⇒ 𝛿𝛿𝑈𝑈 = � 𝛿𝛿𝑈𝑈0 𝑑𝑑𝑑𝑑

𝑉𝑉

𝛿𝛿𝑊𝑊 = 𝛿𝛿𝛿𝛿

⇒ 𝛿𝛿𝑈𝑈0 = 𝜎𝜎𝑥𝑥𝑥𝑥 𝛿𝛿𝜀𝜀𝑥𝑥𝑥𝑥 + 𝜎𝜎𝑦𝑦𝑦𝑦 𝛿𝛿𝜀𝜀𝑦𝑦𝑦𝑦 + 𝜎𝜎𝑧𝑧𝑧𝑧 𝛿𝛿𝜀𝜀𝑧𝑧𝑧𝑧 + 2𝜎𝜎𝑥𝑥𝑥𝑥 𝛿𝛿𝜀𝜀𝑥𝑥𝑥𝑥 + 2𝜎𝜎𝑥𝑥𝑥𝑥 𝛿𝛿𝜀𝜀𝑥𝑥𝑥𝑥 + 2𝜎𝜎𝑦𝑦𝑦𝑦 𝛿𝛿𝜀𝜀𝑦𝑦𝑦𝑦

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Energy methods
𝑈𝑈0 = 𝑈𝑈0 𝜀𝜀𝑥𝑥𝑥𝑥 , 𝜀𝜀𝑦𝑦𝑦𝑦 , 𝜀𝜀𝑧𝑧𝑧𝑧 , 𝜀𝜀𝑥𝑥𝑥𝑥 , 𝜀𝜀𝑥𝑥𝑥𝑥 , 𝜀𝜀𝑦𝑦𝑦𝑦 , 𝑥𝑥, 𝑦𝑦, 𝑧𝑧, 𝑇𝑇

For displacement 𝛿𝛿𝑢𝑢, 𝛿𝛿𝑣𝑣, 𝛿𝛿𝑤𝑤 & strain components

𝛿𝛿𝜀𝜀𝑥𝑥𝑥𝑥 , 𝛿𝛿𝜀𝜀𝑦𝑦𝑦𝑦 , 𝛿𝛿𝜀𝜀𝑧𝑧𝑧𝑧 , 𝛿𝛿𝜀𝜀𝑥𝑥𝑥𝑥 , 𝛿𝛿𝜀𝜀𝑥𝑥𝑥𝑥 , 𝛿𝛿𝜀𝜀𝑦𝑦𝑦𝑦

𝜕𝜕𝑈𝑈0 𝜕𝜕𝑈𝑈0 𝜕𝜕𝑈𝑈0 𝜕𝜕𝑈𝑈0 𝜕𝜕𝑈𝑈0 𝜕𝜕𝑈𝑈0

𝛿𝛿𝑈𝑈0 = 𝛿𝛿𝜀𝜀𝑥𝑥𝑥𝑥 + 𝛿𝛿𝜀𝜀𝑦𝑦𝑦𝑦 + 𝛿𝛿𝜀𝜀𝑧𝑧𝑧𝑧 + 𝛿𝛿𝜀𝜀𝑥𝑥𝑥𝑥 + 𝛿𝛿𝜀𝜀𝑥𝑥𝑥𝑥 + 𝛿𝛿𝜀𝜀𝑦𝑦𝑦𝑦
𝜕𝜕𝜀𝜀𝑥𝑥𝑥𝑥 𝜕𝜕𝜀𝜀𝑦𝑦𝑦𝑦 𝜕𝜕𝜀𝜀𝑧𝑧𝑧𝑧 𝜕𝜕𝜀𝜀𝑥𝑥𝑥𝑥 𝜕𝜕𝜀𝜀𝑥𝑥𝑥𝑥 𝜕𝜕𝜀𝜀𝑦𝑦𝑦𝑦

Comparison with previous equation results in

𝜕𝜕𝑈𝑈0 𝜕𝜕𝑈𝑈 𝜕𝜕𝑈𝑈

𝜎𝜎𝑥𝑥𝑥𝑥 = , 𝜎𝜎𝑦𝑦𝑦𝑦 = 0 , 𝜎𝜎𝑧𝑧𝑧𝑧 = 0
𝜕𝜕𝜀𝜀𝑥𝑥𝑥𝑥 𝜕𝜕𝜀𝜀𝑦𝑦𝑦𝑦 𝜕𝜕𝜀𝜀𝑧𝑧𝑧𝑧
1 𝜕𝜕𝑈𝑈0 1 𝜕𝜕𝑈𝑈0 1 𝜕𝜕𝑈𝑈0
𝜎𝜎𝑥𝑥𝑥𝑥 = , 𝜎𝜎𝑥𝑥𝑥𝑥 = , 𝜎𝜎𝑦𝑦𝑦𝑦 =
2 𝜕𝜕𝜀𝜀𝑥𝑥𝑥𝑥 2 𝜕𝜕𝜀𝜀𝑥𝑥𝑥𝑥 2 𝜕𝜕𝜀𝜀𝑦𝑦𝑦𝑦

𝜕𝜕𝑈𝑈0
⇒ 𝜎𝜎 = ⇒ 𝑈𝑈0 = � 𝜎𝜎𝜎𝜎𝜎𝜎
𝜕𝜕𝜀𝜀
⇒ 𝑈𝑈0 =area under stress- strain diagram

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Energy methods
𝜎𝜎𝜎𝜎 = 𝑈𝑈0 + 𝐶𝐶0

𝑈𝑈0 = strain energy density

𝐶𝐶0 = complementary strain energy density

𝜕𝜕𝐶𝐶0
𝐶𝐶0 = � 𝜀𝜀𝜀𝜀𝜀𝜀 ⇒ 𝜀𝜀 =
𝜕𝜕𝜕𝜕
Since 𝜎𝜎𝜎𝜎 = 𝑈𝑈0 + 𝐶𝐶0

𝐶𝐶0 = −𝑈𝑈0 + 𝜎𝜎𝑥𝑥𝑥𝑥 𝜀𝜀𝑥𝑥𝑥𝑥 + 𝜎𝜎𝑦𝑦𝑦𝑦 𝜀𝜀𝑦𝑦𝑦𝑦 + 𝜎𝜎𝑧𝑧𝑧𝑧 𝜀𝜀𝑧𝑧𝑧𝑧 + 2𝜎𝜎𝑥𝑥𝑥𝑥 𝜀𝜀𝑥𝑥𝑥𝑥 + 2𝜎𝜎𝑥𝑥𝑥𝑥 𝜀𝜀𝑥𝑥𝑥𝑥 + 2𝜎𝜎𝑦𝑦𝑧𝑧 𝜀𝜀𝑦𝑦𝑦𝑦

Differentiate 𝜎𝜎𝑥𝑥𝑥𝑥
𝜕𝜕𝑈𝑈0 𝜕𝜕𝑈𝑈0 𝜕𝜕𝜀𝜀𝑥𝑥𝑥𝑥 𝜕𝜕𝑈𝑈0 𝜕𝜕𝜀𝜀𝑦𝑦𝑦𝑦 𝜕𝜕𝑈𝑈0 𝜕𝜕𝜀𝜀𝑧𝑧𝑧𝑧
= + +
𝜕𝜕𝜎𝜎𝑥𝑥𝑥𝑥 𝜕𝜕𝜀𝜀𝑥𝑥𝑥𝑥 𝜕𝜕𝜎𝜎𝑥𝑥𝑥𝑥 𝜕𝜕𝜀𝜀𝑦𝑦𝑦𝑦 𝜕𝜕𝜎𝜎𝑥𝑥𝑥𝑥 𝜕𝜕𝜀𝜀𝑧𝑧𝑧𝑧 𝜕𝜕𝜎𝜎𝑥𝑥𝑥𝑥
𝜕𝜕𝑈𝑈0 𝜕𝜕𝜀𝜀𝑥𝑥𝑥𝑥 𝜕𝜕𝑈𝑈0 𝜕𝜕𝜀𝜀𝑥𝑥𝑥𝑥 𝜕𝜕𝑈𝑈0 𝜕𝜕𝜀𝜀𝑦𝑦𝑦𝑦
+ + +
𝜕𝜕𝜀𝜀𝑥𝑥𝑥𝑥 𝜕𝜕𝜎𝜎𝑥𝑥𝑥𝑥 𝜕𝜕𝜀𝜀𝑥𝑥𝑥𝑥 𝜕𝜕𝜎𝜎𝑥𝑥𝑥𝑥 𝜕𝜕𝜀𝜀𝑦𝑦𝑦𝑦 𝜕𝜕𝜎𝜎𝑥𝑥𝑥𝑥
𝜕𝜕𝐶𝐶0
and 𝜀𝜀𝑥𝑥𝑥𝑥 =
𝜕𝜕𝜎𝜎𝑥𝑥𝑥𝑥
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Energy methods

Taking derivation other components

𝜕𝜕𝐶𝐶0 𝜕𝜕𝐶𝐶0 𝜕𝜕𝐶𝐶0

𝜀𝜀𝑥𝑥𝑥𝑥 = , 𝜀𝜀 = , 𝜀𝜀 =
𝜕𝜕𝜎𝜎𝑥𝑥𝑥𝑥 𝑦𝑦𝑦𝑦 𝜕𝜕𝜎𝜎𝑦𝑦𝑦𝑦 𝑧𝑧𝑧𝑧 𝜕𝜕𝜎𝜎𝑧𝑧𝑧𝑧
𝜕𝜕𝐶𝐶0 𝜕𝜕𝐶𝐶0 𝜕𝜕𝐶𝐶0
𝜀𝜀𝑥𝑥𝑦𝑦 = , 𝜀𝜀 = , 𝜀𝜀 =
𝜕𝜕𝜎𝜎𝑥𝑥𝑦𝑦 𝑥𝑥𝑧𝑧 𝜕𝜕𝜎𝜎𝑥𝑥𝑧𝑧 𝑦𝑦𝑦𝑦 𝜕𝜕𝜎𝜎𝑦𝑦𝑦𝑦

Hooke’s law: Anisotropic material

𝜕𝜕𝑈𝑈0
= 𝜎𝜎𝑥𝑥𝑥𝑥 = 𝑐𝑐11 𝜀𝜀𝑥𝑥𝑥𝑥 + 𝑐𝑐12 𝜀𝜀𝑦𝑦𝑦𝑦 + 𝑐𝑐13 𝜀𝜀𝑧𝑧𝑧𝑧 + 𝑐𝑐14 𝛾𝛾𝑥𝑥𝑥𝑥 + 𝑐𝑐15 𝛾𝛾𝑥𝑥𝑥𝑥 + 𝑐𝑐16 𝛾𝛾𝑦𝑦𝑦𝑦
𝜕𝜕𝜀𝜀𝑥𝑥𝑥𝑥
⋮ ⋮
𝜕𝜕𝑈𝑈0
= 𝜎𝜎𝑦𝑦𝑦𝑦 = 𝑐𝑐61 𝜀𝜀𝑥𝑥𝑥𝑥 + 𝑐𝑐62 𝜀𝜀𝑦𝑦𝑦𝑦 + 𝑐𝑐63 𝜀𝜀𝑧𝑧𝑧𝑧 + 𝑐𝑐64 𝛾𝛾𝑥𝑥𝑥𝑥 + 𝑐𝑐65 𝛾𝛾𝑥𝑥𝑥𝑥 + 𝑐𝑐66 𝛾𝛾𝑦𝑦𝑦𝑦
𝜕𝜕𝛾𝛾𝑦𝑦𝑦𝑦

8
Energy methods

Hence
𝜕𝜕 2 𝑈𝑈0 𝜕𝜕 2 𝑈𝑈0
= 𝐶𝐶12 = 𝐶𝐶21 , = 𝐶𝐶13 = 𝐶𝐶31
𝜕𝜕𝜀𝜀𝑥𝑥𝑥𝑥 𝜕𝜕𝜀𝜀𝑦𝑦𝑦𝑦 𝜕𝜕𝜀𝜀𝑥𝑥𝑥𝑥 𝜕𝜕𝜀𝜀𝑧𝑧𝑧𝑧
𝜕𝜕 2 𝑈𝑈0 𝜕𝜕 2 𝑈𝑈0
= 𝐶𝐶56 = 𝐶𝐶65 , = 𝐶𝐶46 = 𝐶𝐶64
𝜕𝜕𝛾𝛾𝑥𝑥𝑥𝑥 𝜕𝜕𝛾𝛾𝑦𝑦𝑧𝑧 𝜕𝜕𝛾𝛾𝑦𝑦𝑧𝑧 𝜕𝜕𝛾𝛾𝑥𝑥𝑦𝑦

Strain energy density for anisotropic material

1 2
1 1
𝑈𝑈0 = 𝑐𝑐11 𝜀𝜀𝑥𝑥𝑥𝑥 + 𝑐𝑐12 𝜀𝜀𝑥𝑥𝑥𝑥 𝜀𝜀𝑦𝑦𝑦𝑦 + ⋯ ⋯ + 𝑐𝑐16 𝜀𝜀𝑥𝑥𝑥𝑥 𝛾𝛾𝑦𝑦𝑦𝑦
2 2 2
1 1 2
1
+ 𝑐𝑐21 𝜀𝜀𝑦𝑦𝑦𝑦 𝜀𝜀𝑥𝑥𝑥𝑥 + 𝑐𝑐22 𝜀𝜀𝑦𝑦𝑦𝑦 + ⋯ ⋯ + 𝑐𝑐26 𝜀𝜀𝑦𝑦𝑦𝑦 𝛾𝛾𝑦𝑦𝑦𝑦
2 2 2
⋮ ⋮
1 1 1
+ 𝑐𝑐61 𝛾𝛾𝑦𝑦𝑦𝑦 𝜀𝜀𝑥𝑥𝑥𝑥 + 𝑐𝑐62 𝛾𝛾𝑦𝑦𝑦𝑦 𝜀𝜀𝑦𝑦𝑦𝑦 + ⋯ ⋯ + 𝑐𝑐66 𝛾𝛾𝑦𝑦𝑦𝑦 2
2 2 2

9
Energy methods

Strain energy density for isotropic material

1 2
1 1
𝑈𝑈0 = 𝑐𝑐11 𝜀𝜀1 + 𝑐𝑐12 𝜀𝜀1 𝜀𝜀2 + 𝑐𝑐13 𝜀𝜀1 𝜀𝜀3
2 2 2
1 1 2
1
+ 𝑐𝑐12 𝜀𝜀2 𝜀𝜀1 + 𝑐𝑐22 𝜀𝜀2 + 𝑐𝑐23 𝜀𝜀2 𝜀𝜀3
2 2 2
1 1 1
+ 𝑐𝑐13 𝜀𝜀3 𝜀𝜀1 + 𝑐𝑐23 𝜀𝜀3 𝜀𝜀2 + 𝑐𝑐33 𝜀𝜀3 2
2 2 2

Since naming of principal axes is arbit

𝑐𝑐11 = 𝑐𝑐22 = 𝑐𝑐33 = 𝑐𝑐1 , 𝑐𝑐12 = 𝑐𝑐13 = 𝑐𝑐23 = 𝑐𝑐2

1
𝑈𝑈0 = 𝜆𝜆(𝜀𝜀1 + 𝜀𝜀2 + 𝜀𝜀3 )2 +𝐺𝐺 𝜀𝜀1 2 + 𝜀𝜀2 2 + 𝜀𝜀3 2
2
𝑐𝑐1 −𝑐𝑐2
𝜆𝜆 = 𝑐𝑐2 , 𝐺𝐺 = (Lame constants)
2

10
Energy methods
Equation can be written in terms of invariants
1
𝑈𝑈0 = 𝜆𝜆 + 𝐺𝐺 𝐼𝐼1 2 − 2𝐺𝐺𝐼𝐼2
2

𝐼𝐼1 = 𝜀𝜀𝑥𝑥𝑥𝑥 + 𝜀𝜀𝑦𝑦𝑦𝑦 + 𝜀𝜀𝑧𝑧𝑧𝑧

𝐼𝐼2 = 𝜀𝜀𝑥𝑥𝑥𝑥 𝜀𝜀𝑦𝑦𝑦𝑦 + 𝜀𝜀𝑥𝑥𝑥𝑥 𝜀𝜀𝑧𝑧𝑧𝑧 + 𝜀𝜀𝑦𝑦𝑦𝑦 𝜀𝜀𝑧𝑧𝑧𝑧 − 𝜀𝜀𝑥𝑥𝑦𝑦 2 − 𝜀𝜀𝑥𝑥𝑧𝑧 2 − 𝜀𝜀𝑦𝑦𝑧𝑧 2

1
𝑈𝑈0 = 𝜆𝜆(𝜀𝜀𝑥𝑥𝑥𝑥 + 𝜀𝜀𝑦𝑦𝑦𝑦 + 𝜀𝜀𝑧𝑧𝑧𝑧 )2
2
+𝐺𝐺 𝜀𝜀𝑥𝑥𝑥𝑥 2 + 𝜀𝜀𝑦𝑦𝑦𝑦 2 + 𝜀𝜀𝑧𝑧𝑧𝑧 2 + 2𝜀𝜀𝑥𝑥𝑥𝑥 2 + 2𝜀𝜀𝑥𝑥𝑥𝑥 2 + 2𝜀𝜀𝑦𝑦𝑦𝑦 2

𝜎𝜎𝑥𝑥𝑥𝑥 = 𝜆𝜆𝑒𝑒 + 2𝐺𝐺𝜀𝜀𝑥𝑥𝑥𝑥 , 𝜎𝜎𝑥𝑥𝑦𝑦 = 2𝐺𝐺𝜀𝜀𝑥𝑥𝑦𝑦

⇒ 𝜎𝜎𝑦𝑦𝑦𝑦 = 𝜆𝜆𝜆𝜆 + 2𝐺𝐺𝜀𝜀𝑦𝑦𝑦𝑦 , 𝜎𝜎𝑥𝑥𝑥𝑥 = 2𝐺𝐺𝜀𝜀𝑥𝑥𝑧𝑧
𝜎𝜎𝑧𝑧𝑧𝑧 = 𝜆𝜆𝜆𝜆 + 2𝐺𝐺𝜀𝜀𝑧𝑧𝑧𝑧 , 𝜎𝜎𝑦𝑦𝑦𝑦 = 2𝐺𝐺𝜀𝜀𝑦𝑦𝑦𝑦

Where 𝑒𝑒 = 𝜀𝜀𝑥𝑥𝑥𝑥 + 𝜀𝜀𝑦𝑦𝑦𝑦 + 𝜀𝜀𝑧𝑧𝑧𝑧 = 𝐼𝐼1

11
Energy methods
1 1 1 + 𝑣𝑣
𝜀𝜀𝑥𝑥𝑥𝑥 = 𝜎𝜎𝑥𝑥𝑥𝑥 − 𝑣𝑣 𝜎𝜎𝑦𝑦𝑦𝑦 + 𝜎𝜎𝑧𝑧𝑧𝑧 , 𝜀𝜀𝑥𝑥𝑥𝑥 = 𝜎𝜎 = 𝜎𝜎𝑥𝑥𝑥𝑥
𝐸𝐸 2𝐺𝐺 𝑥𝑥𝑥𝑥 𝐸𝐸
1 1 1 + 𝑣𝑣
𝜀𝜀𝑦𝑦𝑦𝑦 = 𝜎𝜎𝑦𝑦𝑦𝑦 − 𝑣𝑣 𝜎𝜎𝑥𝑥𝑥𝑥 + 𝜎𝜎𝑧𝑧𝑧𝑧 , 𝜀𝜀𝑥𝑥𝑥𝑥 = 𝜎𝜎𝑥𝑥𝑧𝑧 = 𝜎𝜎𝑥𝑥𝑥𝑥
𝐸𝐸 2𝐺𝐺 𝐸𝐸
1 1 1 + 𝑣𝑣
𝜀𝜀𝑧𝑧𝑧𝑧 = 𝜎𝜎𝑧𝑧𝑧𝑧 − 𝑣𝑣 𝜎𝜎𝑥𝑥𝑥𝑥 + 𝜎𝜎𝑦𝑦𝑦𝑦 , 𝜀𝜀𝑦𝑦𝑦𝑦 = 𝜎𝜎 = 𝜎𝜎𝑦𝑦𝑦𝑦
𝐸𝐸 2𝐺𝐺 𝑦𝑦𝑦𝑦 𝐸𝐸

𝐺𝐺(3𝜆𝜆 + 2𝐺𝐺)
⇒ 𝐸𝐸 =
𝜆𝜆 + 𝐺𝐺
𝜆𝜆
𝑣𝑣 =
2(𝜆𝜆 + 𝐺𝐺)
𝑣𝑣𝑣𝑣 3𝑣𝑣𝐺𝐺
𝜆𝜆 = =
(1 + 𝑣𝑣)(1 − 2𝑣𝑣) (1 + 𝑣𝑣)
𝐸𝐸
𝐺𝐺 =
2(1 + 𝑣𝑣)
𝐸𝐸
𝐾𝐾 =
3(1 − 2𝑣𝑣)

12
Castigliano’s theorem on deflections
Castigliano’s second theorem

If an elastic system is supported so that rigid-body displacements of the

system are prevented, and if certain concentrated forces of magnitudes
𝐹𝐹1, 𝐹𝐹2, . . , 𝐹𝐹𝑝𝑝 act on the system, in addition to distributed loads and thermal
strains, the displacement component 𝑞𝑞𝑖𝑖 of the point of application of the
force 𝐹𝐹𝑖𝑖, is determined by the equation

𝜕𝜕𝐶𝐶
𝑞𝑞𝑖𝑖 = , 𝑖𝑖 = 1,2, … , 𝑝𝑝
𝜕𝜕𝐹𝐹𝑖𝑖

Castigliano’s theorem is restricted to small displacements of the structure.

The complementary energy 𝐶𝐶 of a structure composed of m members may
be expressed by the relation
𝑚𝑚

𝐶𝐶 = � 𝐶𝐶𝑖𝑖
𝑖𝑖=1

13
Castigliano’s theorem on deflections
Example
a beam that is supported on rigid supports and subjected to external
concentrated forces

Solution
The rotation 𝜃𝜃 resulting m the deformations is given by
1 𝜕𝜕𝜕𝜕 1 𝜕𝜕𝜕𝜕
𝜃𝜃 = +
𝑏𝑏 𝜕𝜕𝐹𝐹1 𝑏𝑏 𝜕𝜕𝐹𝐹2
14
Castigliano’s theorem on deflections
Considering the magnitudes of 𝐹𝐹1 and 𝐹𝐹1 to be functions of 𝑆𝑆

𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝐹𝐹1 𝜕𝜕𝜕𝜕 𝜕𝜕𝐹𝐹2

= +
𝜕𝜕𝜕𝜕 𝜕𝜕𝐹𝐹1 𝜕𝜕𝜕𝜕 𝜕𝜕𝐹𝐹2 𝜕𝜕𝜕𝜕

In particular, 𝑆𝑆 equals to 𝐹𝐹1 and 𝐹𝐹2

𝜕𝜕𝐹𝐹1� 𝜕𝜕𝐹𝐹2�
⇒ 𝜕𝜕𝜕𝜕 = 𝜕𝜕𝜕𝜕 = 1

We can obtain

𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 1 𝜕𝜕𝜕𝜕

⇒ = + ⇒ 𝜃𝜃 =
𝜕𝜕𝐹𝐹 𝜕𝜕𝐹𝐹1 𝜕𝜕𝐹𝐹2 𝑏𝑏 𝜕𝜕𝐹𝐹

Since the equal and opposite forces 𝐹𝐹1 , 𝐹𝐹2 constitute a couple of magnitude

𝜕𝜕𝜕𝜕
𝑀𝑀 = 𝑏𝑏𝑏𝑏 ⇒ 𝜃𝜃 =
𝜕𝜕𝑀𝑀𝑖𝑖
15
Castigliano’s theorem on deflections
If we limit our consideration to linear elastic material behavior and small
displacement, then the load-deflection relation is linear⇒

𝜕𝜕𝜕𝜕
𝑞𝑞𝑖𝑖 = , 𝑖𝑖 = 1,2, … , 𝑝𝑝
𝜕𝜕𝐹𝐹𝑖𝑖
𝜕𝜕𝜕𝜕
𝜃𝜃𝑖𝑖 = , 𝑖𝑖 = 1,2, … , 𝑠𝑠
𝜕𝜕𝑀𝑀𝑖𝑖

the strain energy 𝑈𝑈

𝑈𝑈 = 𝑈𝑈 𝐹𝐹1 , 𝐹𝐹2 , … , 𝐹𝐹𝑃𝑃 , 𝑀𝑀1 , 𝑀𝑀2 , … , 𝑀𝑀𝑠𝑠 = � 𝑈𝑈0 𝑑𝑑𝑑𝑑

𝑈𝑈0 is the strain-energy density and for linear elastic, isotropic,

homogeneous materials

1 2 2 2
𝑣𝑣
𝑈𝑈0 = 𝜎𝜎 + 𝜎𝜎𝑦𝑦𝑦𝑦 + 𝜎𝜎𝑧𝑧𝑧𝑧 − 𝜎𝜎 𝜎𝜎 + 𝜎𝜎𝑦𝑦𝑦𝑦 𝜎𝜎𝑧𝑧𝑧𝑧 + 𝜎𝜎𝑧𝑧𝑧𝑧 𝜎𝜎𝑥𝑥𝑥𝑥
2𝐸𝐸 𝑥𝑥𝑥𝑥 𝐸𝐸 𝑥𝑥𝑥𝑥 𝑦𝑦𝑦𝑦
1 2 2 2
+ 𝜎𝜎𝑦𝑦𝑦𝑦 + 𝜎𝜎𝑧𝑧𝑧𝑧 + 𝜎𝜎𝑥𝑥𝑥𝑥
2𝐺𝐺
16
Castigliano’s theorem on deflections
Axial loading member

The total strain energy 𝑈𝑈𝑁𝑁

1
𝑈𝑈𝑁𝑁 = � 𝑑𝑑𝑈𝑈𝑁𝑁 = � 𝑁𝑁𝑁𝑁𝑒𝑒𝑧𝑧
2

17
Castigliano’s theorem on deflections

Noting that

𝜀𝜀 = 𝜎𝜎𝑧𝑧𝑧𝑧 /𝐸𝐸
� 𝑧𝑧𝑧𝑧
𝜎𝜎𝑧𝑧𝑧𝑧 = 𝑁𝑁/𝐸𝐸

𝑁𝑁 2
⇒ 𝑈𝑈𝑁𝑁 = � 𝑑𝑑𝑧𝑧
2𝐸𝐸𝐸𝐸

For abrupt changes in material properties, we can account approximately

for the changes

𝑁𝑁12
𝐿𝐿1 𝐿𝐿2
𝑁𝑁22 𝐿𝐿3
𝑁𝑁32
𝑈𝑈𝑁𝑁 = � 𝑑𝑑𝑑𝑑 + � 𝑑𝑑𝑑𝑑 + � 𝑑𝑑𝑑𝑑
0 2𝐸𝐸 𝐴𝐴
1 1 𝐿𝐿1 2𝐸𝐸 𝐴𝐴
2 2 𝐿𝐿2 2𝐸𝐸 𝐴𝐴
3 3

18
Castigliano’s theorem on deflections
Strain Energy for axially loaded Springs
The external force 𝑄𝑄 is applied slowly so that it remains in equilibrium.
The potential energy 𝑈𝑈
𝛿𝛿
𝑈𝑈 = ∫ 𝑑𝑑𝑑𝑑 = ∫0 𝑄𝑄𝑄𝑄𝑄𝑄

the complementary strain energy 𝐶𝐶 of the spring

𝑃𝑃
𝐶𝐶 = ∫ 𝑑𝑑𝐶𝐶 = ∫0 𝑥𝑥𝑑𝑑𝑃𝑃

19
Castigliano’s theorem on deflections
For linear spring, 𝐶𝐶 = 𝑈𝑈 ⇒
𝛿𝛿
𝐶𝐶 = 𝑈𝑈 = ∫0 𝑄𝑄𝑄𝑄𝑄𝑄

𝐹𝐹 = 𝑘𝑘𝑘𝑘 ⇒
𝛿𝛿 1
𝐶𝐶 = 𝑈𝑈 = ∫0 𝐹𝐹𝑑𝑑𝑑𝑑 = 𝑘𝑘𝛿𝛿 2
2

Strain Energies 𝑼𝑼𝑴𝑴 and 𝑼𝑼𝑺𝑺 for Beams

20
Castigliano’s theorem on deflections
Consider a beam of uniform cross section. The loads 𝑃𝑃, 𝑄𝑄, and 𝑅𝑅 lie in the
(𝑦𝑦, 𝑧𝑧) plane. The flexure formula

𝑀𝑀𝑥𝑥 𝑦𝑦
𝜎𝜎𝑧𝑧𝑧𝑧 =
𝐼𝐼𝑥𝑥

For elastic material behavior, the angle 𝑑𝑑𝑑𝑑 varies linearly with 𝑀𝑀𝑥𝑥 . So the
strain energy 𝑈𝑈𝑀𝑀 for the beam caused by 𝑀𝑀𝑥𝑥 becomes

1
𝑈𝑈𝑀𝑀 = � 𝑑𝑑 𝑈𝑈𝑀𝑀 = � 𝑀𝑀𝑥𝑥 𝑑𝑑𝑑𝑑
2

𝜎𝜎𝑧𝑧𝑧𝑧
Noting that 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑒𝑒𝑧𝑧 /𝑦𝑦 and 𝑑𝑑𝑒𝑒𝑧𝑧 = 𝜀𝜀𝑧𝑧𝑧𝑧 𝑑𝑑𝑑𝑑. Assuming that 𝜀𝜀𝑧𝑧𝑧𝑧 = and
𝐸𝐸
𝑀𝑀𝑥𝑥 𝑦𝑦
𝜎𝜎𝑧𝑧𝑧𝑧 = , we may write 𝑈𝑈𝑀𝑀 in the form
𝐼𝐼𝑥𝑥

𝑀𝑀𝑥𝑥 2
𝑈𝑈𝑀𝑀 = � 𝑑𝑑𝑑𝑑
2𝐸𝐸𝐼𝐼𝑥𝑥
21
Castigliano’s theorem on deflections

We define an average value of 𝜎𝜎𝑧𝑧𝑧𝑧 as 𝜏𝜏 = 𝑉𝑉𝑦𝑦 /𝐴𝐴. Then, the shear strain energy for the
beam 𝑈𝑈𝑠𝑠 resulting from shear 𝑉𝑉𝑦𝑦 is

′ 𝑘𝑘
𝑈𝑈𝑠𝑠 = � 𝑘𝑘 𝑑𝑑 𝑈𝑈𝑠𝑠 = � 𝑉𝑉𝑦𝑦 𝑑𝑑𝑒𝑒𝑦𝑦
2

Noting that 𝑑𝑑𝑒𝑒𝑧𝑧 = 𝛾𝛾𝛾𝛾𝛾𝛾 and 𝛾𝛾 = 𝜏𝜏/𝐺𝐺, 𝜏𝜏 = 𝑉𝑉𝑦𝑦 /𝐴𝐴, we may write 𝑈𝑈𝑠𝑠 in the form

𝑘𝑘𝑉𝑉𝑦𝑦 2
𝑈𝑈𝑠𝑠 = � 𝑑𝑑𝑑𝑑
2𝐺𝐺𝐺𝐺

In the engineering theory of beams, the stress component 𝜎𝜎𝑧𝑧𝑧𝑧 is assumed to be

uniform over thickness b.

𝑉𝑉𝑦𝑦 𝑄𝑄
𝜎𝜎𝑧𝑧𝑧𝑧 =
𝐼𝐼𝑥𝑥 𝑏𝑏

22
Castigliano’s theorem on deflections
Fortunately, in practical problems, the shear strain energy 𝑈𝑈𝑠𝑠 is often small
compared to 𝑈𝑈𝑀𝑀 . Hence, for practical problems, the need for exact values of
Us is not critical.

When exact values of 𝑘𝑘 are not generally available, as an expedient

approximation, the correction coefficient 𝑘𝑘 may be obtained as the ratios of
the shear stress at the neutral surface of the beam.

For rectangular cross section:

𝑏𝑏𝑏 ℎ
𝑉𝑉𝑦𝑦 𝑄𝑄 𝐴𝐴 𝑄𝑄𝑄𝑄𝑄 2 4 ℎ
𝑘𝑘 = � = = = 1.50
𝐼𝐼𝑥𝑥 𝑏𝑏 𝑉𝑉𝑦𝑦 𝐼𝐼𝑥𝑥 𝑏𝑏 1 3
12 𝑏𝑏ℎ

This value is larger and hence more conservative than the more exact value
1.20. Nevertheless, since the more precise value is known, we recommend
that 𝑘𝑘 = 1.20 be used for rectangular cross sections.

23
Castigliano’s theorem on deflections
Strain Energies 𝑼𝑼𝑻𝑻 for Torsion

For elastic material behavior, the angle 𝑑𝑑𝑑𝑑 varies linearly with 𝑇𝑇. So the the
total torsional strain energy 𝑈𝑈𝑇𝑇 for the torsional member becomes
1
𝑈𝑈𝑇𝑇 = � 𝑑𝑑 𝑈𝑈𝑇𝑇 = � 𝑇𝑇𝑑𝑑𝑑𝑑
2

𝜏𝜏 𝑇𝑇𝑇𝑇
Noting that 𝑑𝑑𝑑𝑑 = 𝛾𝛾 𝑑𝑑𝑑𝑑 and 𝛾𝛾 = , τ = , we may write 𝑈𝑈𝑇𝑇 in the form
𝐺𝐺 𝐽𝐽
𝑇𝑇 2
𝑈𝑈𝑇𝑇 = � 𝑑𝑑𝑑𝑑
2𝐺𝐺𝐺𝐺
24
Castigliano’s theorem on deflections
Example

Consider weights 𝑊𝑊1 and 𝑊𝑊2 supported by the

linear springs shown. The spring constants are 𝑘𝑘1
and 𝑘𝑘2 . Assume that the weights are applied slowly
so that the system is always in equilibrium as the
springs are stretched from their initially stretched
lengths

Solution

Under the displacements 𝑞𝑞1 and 𝑞𝑞2 , spring 1 undergoes

an elongation 𝑞𝑞1 and spring2 undergoes an elongation
𝑞𝑞2 − 𝑞𝑞1 . Hence, the potential energy stored in the
springs is

1 2
1 2
𝑈𝑈 = 𝑘𝑘1 𝑞𝑞1 + 𝑘𝑘1 𝑞𝑞2 − 𝑞𝑞1
2 2

25
Castigliano’s theorem on deflections
The internal tension forces in springs 1 and 2 are

𝐹𝐹1 = 𝑘𝑘1 𝑞𝑞1 , 𝐹𝐹1 = 𝑘𝑘1 (𝑞𝑞2 − 𝑞𝑞1 )

By equilibrium,
𝐹𝐹1 = 𝑊𝑊1 + 𝑊𝑊2 , 𝐹𝐹2 = 𝑊𝑊2

The potential energy in terms of 𝑊𝑊1 , 𝑊𝑊2 , 𝑘𝑘1 and 𝑘𝑘2 is

1 𝑊𝑊1 + 𝑊𝑊2 2
1 𝑊𝑊2 2
𝑈𝑈 = +
2 𝑘𝑘1 2 𝑘𝑘2

We obtain

𝜕𝜕𝜕𝜕 𝑊𝑊1 + 𝑊𝑊2

𝑞𝑞1 = =
𝜕𝜕𝑊𝑊1 𝑘𝑘1

𝜕𝜕𝜕𝜕 𝑊𝑊1 + 𝑊𝑊2 𝑊𝑊2

𝑞𝑞2 = = +
𝜕𝜕𝑊𝑊2 𝑘𝑘1 𝑘𝑘2
26
Statically Determinate Structures

In the analysis of many engineering structures,

the equations of static equilibrium are both
necessary and sufficient to solve for unknown
reactions and for internal actions in the
members of the structure.

Structures for which the equations of static

equilibrium are sufficient to determine the
unknown tensions, shears, etc., uniquely
are said to be statically determinate
structures.

27
Statically Determinate Structures

For statically determinate structures, the strain energy 𝑈𝑈 for a structure is

equal to the sum of the strain energies of its members. The loading of the jth
member of the structure is

𝑈𝑈𝑗𝑗 = 𝑈𝑈𝑁𝑁𝑁𝑁 + 𝑈𝑈𝑀𝑀𝑀𝑀 + 𝑈𝑈𝑆𝑆𝑆𝑆 + 𝑈𝑈𝑇𝑇𝑇𝑇

With the total strain energy U of the structure known, the deflection 𝑞𝑞𝑖𝑖 of
the structure at the location of a concentrated force 𝐹𝐹𝑖𝑖

𝜕𝜕𝜕𝜕
𝑞𝑞𝑖𝑖 =
𝜕𝜕𝐹𝐹𝑖𝑖

𝑁𝑁𝑗𝑗 𝜕𝜕𝑁𝑁𝑗𝑗 𝑘𝑘𝑗𝑗 𝑉𝑉𝑗𝑗 𝜕𝜕𝑉𝑉𝑗𝑗 𝑀𝑀𝑗𝑗 𝜕𝜕𝑀𝑀𝑗𝑗 𝑇𝑇𝑗𝑗 𝜕𝜕𝑇𝑇𝑗𝑗

= ∑𝑚𝑚
𝑗𝑗=1 ∫ 𝐸𝐸 𝑑𝑑𝑑𝑑 + ∫ 𝑑𝑑𝑑𝑑 + ∫ 𝑑𝑑𝑑𝑑 + ∫ 𝑑𝑑𝑑𝑑
𝑗𝑗 𝐴𝐴𝑗𝑗 𝜕𝜕𝐹𝐹𝑖𝑖 𝐺𝐺𝑗𝑗 𝐴𝐴𝑗𝑗 𝜕𝜕𝐹𝐹𝑖𝑖 𝐸𝐸𝑗𝑗 𝐼𝐼𝑗𝑗 𝜕𝜕𝐹𝐹𝑖𝑖 𝐺𝐺𝑗𝑗 𝐽𝐽𝑗𝑗 𝜕𝜕𝐹𝐹𝑖𝑖

28
Statically Determinate Structures
the angle (slope) change 𝜃𝜃𝑖𝑖 of the structure at the location of a concentrated
moment 𝑀𝑀𝑖𝑖

𝜕𝜕𝜕𝜕
𝑞𝑞𝑖𝑖 =
𝜕𝜕𝑀𝑀𝑖𝑖

𝑁𝑁𝑗𝑗 𝜕𝜕𝑁𝑁𝑗𝑗 𝑘𝑘𝑗𝑗 𝑉𝑉𝑗𝑗 𝜕𝜕𝑉𝑉𝑗𝑗 𝑀𝑀𝑗𝑗 𝜕𝜕𝑀𝑀𝑗𝑗 𝑇𝑇𝑗𝑗 𝜕𝜕𝑇𝑇𝑗𝑗

= ∑𝑚𝑚
𝑗𝑗=1 ∫ 𝐸𝐸 𝑑𝑑𝑑𝑑 + ∫ 𝐺𝐺 𝐴𝐴 𝜕𝜕𝑀𝑀 𝑑𝑑𝑑𝑑 +∫ 𝑑𝑑𝑑𝑑 + ∫ 𝐺𝐺 𝐽𝐽 𝜕𝜕𝑀𝑀 𝑑𝑑𝑑𝑑
𝑗𝑗 𝐴𝐴𝑗𝑗 𝜕𝜕𝑀𝑀𝑖𝑖 𝑗𝑗 𝑗𝑗 𝑖𝑖 𝐸𝐸𝑗𝑗 𝐼𝐼𝑗𝑗 𝜕𝜕𝑀𝑀𝑖𝑖 𝑗𝑗 𝑗𝑗 𝑖𝑖

Example
Determine the deflection under load 𝑃𝑃 of the cantilever beam. Assume
that the beam length 𝐿𝐿 is more than five times the beam depth ℎ.

29
Statically Determinate Structures
Since 𝐿𝐿 > 5ℎ, the strain energy 𝑈𝑈𝑆𝑆 is small and will be neglected, so the
total strain energy is

𝐿𝐿
𝑀𝑀𝑥𝑥2
𝑈𝑈 = 𝑈𝑈𝑀𝑀 = � 𝑑𝑑𝑑𝑑
0 2𝐸𝐸𝐼𝐼 𝑥𝑥

By Castigliano’s theorem, the deflection 𝑞𝑞𝑝𝑝 is

𝐿𝐿
𝜕𝜕𝜕𝜕 𝑀𝑀𝑥𝑥 𝜕𝜕𝑀𝑀𝑥𝑥
𝑞𝑞𝑝𝑝 = =� 𝑑𝑑𝑑𝑑
𝜕𝜕𝜕𝜕 0 𝐸𝐸𝐼𝐼𝑥𝑥 𝜕𝜕𝜕𝜕

𝜕𝜕𝑀𝑀𝑥𝑥
Since 𝑀𝑀𝑥𝑥 = 𝑃𝑃𝑃𝑃 ⇒ = 𝑧𝑧
𝜕𝜕𝜕𝜕

𝐿𝐿
𝑃𝑃𝑧𝑧 2 𝑃𝑃𝐿𝐿3
𝑞𝑞𝑝𝑝 = � 𝑑𝑑𝑑𝑑 =
0 𝐸𝐸𝐼𝐼𝑥𝑥 3𝐸𝐸𝐼𝐼𝑥𝑥

This result agrees with elementary beam theory.

30
Statically Indeterminate Structures
In general, statically indeterminate structures contain one or more
redundant members or supports. For statically indeterminate structures,
the equations of static equilibrium are not sufficient to determine these
internal actions.

The structures in (a) do not

contain redundant reactions
but do contain redundant
members. Hence, the truss
is statically indeterminate.

A statically determinate may be made statically indeterminate by the

addition of a member. Alternatively, a statically indeterminate structure
may be rendered statically determinate if certain members, internal
actions, or supports are removed.
31
Statically Indeterminate Structures
In particular, we shown that for internal redundant force or external
redundant reaction(𝐹𝐹1 , 𝐹𝐹2 , … )

𝜕𝜕𝜕𝜕
=0
𝜕𝜕𝐹𝐹𝑖𝑖

For every internal redundant moment or external redundant moment (𝑀𝑀1 ,

𝑀𝑀2 , … )
𝜕𝜕𝜕𝜕
=0
𝜕𝜕𝑀𝑀𝑖𝑖

let 𝑁𝑁𝐵𝐵𝐵𝐵 be the redundant internal action for the pin-joined truss in (a). It is
not obvious that
𝜕𝜕𝜕𝜕
=0
𝜕𝜕𝑁𝑁𝐵𝐵𝐵𝐵
′ ′′
To prove it, it is necessary to distinguish between tensions 𝑁𝑁𝐵𝐵𝐵𝐵 and 𝑁𝑁𝐵𝐵𝐵𝐵 .

32
Statically Indeterminate Structures

′
The displacement of point H in the direction of 𝑁𝑁𝐵𝐵𝐵𝐵 is given by

𝜕𝜕𝜕𝜕
𝑞𝑞𝑁𝑁𝐵𝐵𝐵𝐵
′ = ′
𝜕𝜕𝑁𝑁𝐵𝐵𝐵𝐵

′′
And the direction of 𝑁𝑁𝐵𝐵𝐵𝐵 is given by

𝜕𝜕𝜕𝜕
𝑞𝑞𝑁𝑁𝐵𝐵𝐵𝐵
′′ = ′′
𝜕𝜕𝑁𝑁𝐵𝐵𝐵𝐵

They are collinear and have equal magnitudes but opposite senses. So we
have

𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
′ + ′′ = 0
𝜕𝜕𝑁𝑁𝐵𝐵𝐵𝐵 𝜕𝜕𝑁𝑁𝐵𝐵𝐵𝐵

33
Statically Indeterminate Structures
′ ′′
Since 𝑁𝑁𝐵𝐵𝐵𝐵 = 𝑁𝑁𝐵𝐵𝐵𝐵 = 𝑁𝑁𝐵𝐵𝐵𝐵

𝜕𝜕𝜕𝜕
⇒ =0
𝜕𝜕𝑁𝑁𝐵𝐵𝐵𝐵

Similarly

𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

= 0, = 0, =0
𝜕𝜕𝑁𝑁 𝜕𝜕𝑉𝑉 𝜕𝜕𝑀𝑀

If the displacement magnitudes 𝑞𝑞1 or 𝑞𝑞2 of the end of the beam are known,
we may compute the reaction 𝑅𝑅 for the loaded beam by the relations

𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕
𝑞𝑞1 = − 𝑜𝑜𝑜𝑜 𝑞𝑞2 =
𝜕𝜕𝑅𝑅 𝜕𝜕𝑅𝑅

34
Statically Indeterminate Structures
Example: uniformly Loaded Propped Cantilever Beam
Consider the reaction 𝑅𝑅 at the right end of the beam shown below as the
redundant. The right-hand support is conceived to be removed, and the
external force 𝑅𝑅 is applied.
(a) Determine the reaction 𝑅𝑅, enforcing the condition that 𝑞𝑞𝑞𝑞 = 0.
(b) Determine the mid-span displacement of the beam.

35
Statically Indeterminate Structures
Solution
A free-body diagram of the beam is shown in (b). The bending moment in
the beam at section x is

1
𝑀𝑀 = 𝑅𝑅𝑅𝑅 − 𝑤𝑤𝑥𝑥 2
2

The moment at section x caused by the load is

𝑚𝑚 = 𝑥𝑥

Neglecting the effect of shear, we have

𝐿𝐿
𝑀𝑀𝑀𝑀 1 𝐿𝐿 1
𝑞𝑞𝑅𝑅 = � 𝑑𝑑𝑑𝑑 = � (𝑅𝑅𝑥𝑥 − 𝑤𝑤𝑥𝑥 3 )𝑑𝑑𝑑𝑑
2
0 𝐸𝐸𝐸𝐸 𝐸𝐸𝐸𝐸 0 2
Or
𝑅𝑅𝐿𝐿3 𝑤𝑤𝐿𝐿4
𝑞𝑞𝑅𝑅 = −
3𝐸𝐸𝐸𝐸 8𝐸𝐸𝐸𝐸
36
Statically Indeterminate Structures
Since R is the reaction of the fixed support, 𝑞𝑞𝑅𝑅 = 0. Therefore
3
𝑅𝑅 = 𝑤𝑤𝑤𝑤
8

(b) To determine the mid-span deflection of the beam, apply a (downward)

𝐿𝐿
unit load at 𝑥𝑥 = 𝐿𝐿/2. For 0 ≤ 𝑥𝑥 ≤ ,
2
3 1
𝑀𝑀 = 𝑤𝑤𝑤𝑤𝑤𝑤 − 𝑤𝑤𝑥𝑥 2
8 2
𝑚𝑚 = 0
𝐿𝐿
For ≤ 𝑥𝑥 ≤ 𝐿𝐿,
2
3 1
𝑀𝑀 = 𝑤𝑤𝑤𝑤𝑤𝑤 − 𝑤𝑤𝑥𝑥 2
8 2
𝐿𝐿
𝑚𝑚 = − 𝑥𝑥
2
Hence

1 𝐿𝐿 3 1 𝐿𝐿 𝑤𝑤𝐿𝐿4
𝑞𝑞𝐿𝐿 = � 𝑤𝑤𝑤𝑤𝑤𝑤 − 𝑤𝑤𝑥𝑥 2 − 𝑥𝑥 𝑑𝑑𝑑𝑑 =
2 𝐸𝐸𝐸𝐸 𝐿𝐿 8 2 2 192𝐸𝐸𝐸𝐸
2
37
Application of energy method
Energy methods are used widely to obtain solutions to elasticity problems
and determine deflections of structures and machines

Principle of stationary potential energy

Consider a system with finite number of degrees of freedom 𝑥𝑥1 , 𝑥𝑥2 , … , 𝑥𝑥𝑛𝑛

𝛿𝛿𝛿𝛿 = 𝑄𝑄1 𝛿𝛿𝑥𝑥1 + 𝑄𝑄2 𝛿𝛿𝑥𝑥2 … + 𝑄𝑄𝑖𝑖 𝛿𝛿𝑥𝑥𝑖𝑖 … + 𝑄𝑄𝑛𝑛 𝛿𝛿𝑥𝑥𝑛𝑛

where (𝑄𝑄1 , 𝑄𝑄2 , … , 𝑄𝑄𝑖𝑖 , … , 𝑄𝑄𝑛𝑛 ) are components of the generalized load.

⇒ 𝑄𝑄𝑖𝑖 is a force if 𝛿𝛿𝑥𝑥𝑖𝑖 is a translation

and

𝑄𝑄𝑖𝑖 is a moment if 𝛿𝛿𝑥𝑥𝑖𝑖 is a rotation

38
Application of energy method
For a deformable body, the virtual work 𝛿𝛿𝛿𝛿 can be separated into the sum

𝛿𝛿𝛿𝛿 = 𝛿𝛿𝑊𝑊𝑒𝑒 + 𝛿𝛿𝑊𝑊𝑖𝑖

where 𝑊𝑊𝑒𝑒 is the virtual work of external forces and 𝑊𝑊𝑖𝑖 is the virtual work of
internal forces

⇒ 𝛿𝛿𝑊𝑊𝑒𝑒 = 𝑃𝑃1 𝛿𝛿𝑥𝑥1 + 𝑃𝑃2 𝛿𝛿𝑥𝑥2 … + 𝑃𝑃𝑖𝑖 𝛿𝛿𝑥𝑥𝑖𝑖 … + 𝑃𝑃𝑛𝑛 𝛿𝛿𝑥𝑥𝑛𝑛

Where (𝑃𝑃1 , 𝑃𝑃2 , … , 𝑃𝑃𝑖𝑖 , … , 𝑃𝑃𝑛𝑛 ) are called the components of generalized
external load.

Imagine that the virtual displacement takes the system around any closed
path. In our applications, we consider only systems that undergo elastic
behavior.

⇒ 𝛿𝛿𝑊𝑊𝑖𝑖 = −𝛿𝛿𝛿𝛿

where 𝛿𝛿𝛿𝛿 is the elastic strain energy.

39
Application of energy method
For a conservative system

𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

𝛿𝛿𝑈𝑈 = 𝛿𝛿𝑥𝑥 + 𝛿𝛿𝑥𝑥 + … + 𝛿𝛿𝑥𝑥
𝜕𝜕𝑥𝑥1 1 𝜕𝜕𝑥𝑥2 2 𝜕𝜕𝑥𝑥𝑛𝑛 𝑛𝑛

⇒ 𝑄𝑄1 𝛿𝛿𝑥𝑥1 + 𝑄𝑄2 𝛿𝛿𝑥𝑥2 … + 𝑄𝑄𝑖𝑖 𝛿𝛿𝑥𝑥𝑖𝑖 … + 𝑄𝑄𝑛𝑛 𝛿𝛿𝑥𝑥𝑛𝑛

𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

= 𝑃𝑃1 𝛿𝛿𝑥𝑥1 + 𝑃𝑃2 𝛿𝛿𝑥𝑥2 … + 𝑃𝑃𝑖𝑖 𝛿𝛿𝑥𝑥𝑖𝑖 … + 𝑃𝑃𝑛𝑛 𝛿𝛿𝑥𝑥𝑛𝑛 − 𝛿𝛿𝑥𝑥1 − 𝛿𝛿𝑥𝑥2 − … − 𝛿𝛿𝑥𝑥𝑛𝑛
𝜕𝜕𝑥𝑥1 𝜕𝜕𝑥𝑥2 𝜕𝜕𝑥𝑥𝑛𝑛

Or
𝜕𝜕𝜕𝜕
𝑄𝑄𝑖𝑖 = 𝑃𝑃𝑖𝑖 −
𝜕𝜕𝑥𝑥𝑖𝑖

If the component 𝑄𝑄𝑖𝑖 vanish, then the system is in equilibrium.

𝜕𝜕𝜕𝜕
𝑃𝑃𝑖𝑖 =
𝜕𝜕𝑥𝑥𝑖𝑖
40
Application of energy method
Example
Two bars 𝐴𝐴𝐴𝐴 and 𝐶𝐶𝐶𝐶 of lengths 𝐿𝐿1 and 𝐿𝐿2 , respectively are attached to a rigid
foundation at points 𝐴𝐴 and 𝐶𝐶. The cross area of bar 𝐴𝐴𝐴𝐴 is 𝐴𝐴1 and that of bar 𝐶𝐶𝐶𝐶 is 𝐴𝐴2 .
The corresponding modulus of elasticity are 𝐸𝐸1 and 𝐸𝐸2 . Under the action of
horizontal and vertical forces P and Q, pin 𝐵𝐵 undergoes finite horizontal and vertical
displacement with components u and v. The bars 𝐴𝐴𝐴𝐴 and 𝐶𝐶𝐶𝐶 remain linearly elastic.

(a) Derive formulas for P and Q in terms of u and v

𝐸𝐸 𝐴𝐴 𝐸𝐸 𝐴𝐴
(b) Let 1𝐿𝐿 1 = 𝐾𝐾1 =2 N/mm, 2𝐿𝐿 2 = 𝐾𝐾2 =3 N/mm
1 2
𝑏𝑏1 = ℎ = 400𝑚𝑚𝑚𝑚 and 𝑏𝑏2 = 300𝑚𝑚𝑚𝑚.
For 𝑢𝑢 = 30𝑚𝑚𝑚𝑚 and v = 30𝑚𝑚𝑚𝑚
Determine the values of P and Q.
(c) Consider the equilibrium of
The pin 𝐵𝐵 in the displacement
position 𝐵𝐵 ∗ and verify the results
(d) For small displacement components 𝑢𝑢 and
𝑣𝑣, linearize the formulas for 𝑃𝑃 and 𝑄𝑄

41
Application of energy method
(a) For this problem the generalized external forces are 𝑃𝑃1 = 𝑃𝑃 and 𝑃𝑃1 = 𝑄𝑄 and the
generalized coordinates are 𝑥𝑥1 = 𝑢𝑢 and 𝑥𝑥2 = 𝑣𝑣.

For the geometry, the elongations 𝑒𝑒1 and 𝑒𝑒2 of bars 1 and 2 can be obtained in
terms of 𝑢𝑢 and 𝑣𝑣

𝐿𝐿1 + 𝑒𝑒1 2 = 𝑏𝑏1 + 𝑢𝑢 2

+ ℎ + 𝑣𝑣 2 , 𝐿𝐿1 2 = 𝑏𝑏1 2 + ℎ2
𝐿𝐿2 + 𝑒𝑒2 2 = 𝑏𝑏2 + 𝑢𝑢 2
+ ℎ + 𝑣𝑣 2 , 𝐿𝐿2 2 = 𝑏𝑏2 2 + ℎ2 (a)
Solving for (𝑒𝑒1 , 𝑒𝑒2 ), we obtain

𝑒𝑒1 = 𝑏𝑏1 + 𝑢𝑢 2 + ℎ + 𝑣𝑣 2 − 𝐿𝐿1

𝑒𝑒2 = 𝑏𝑏2 + 𝑢𝑢 2 + ℎ + 𝑣𝑣 2 − 𝐿𝐿2 (b)

Since each bar remains linearly elastic, the strain energies 𝑈𝑈1 and 𝑈𝑈2 of bars 𝐴𝐴𝐴𝐴
and 𝐶𝐶𝐶𝐶 are
1 𝐸𝐸1 𝐴𝐴1 2
𝑈𝑈1 = 𝑁𝑁1 𝑒𝑒1 = 𝑒𝑒
2 2𝐿𝐿1 1
1 𝐸𝐸2 𝐴𝐴2 2
𝑈𝑈2 = 𝑁𝑁2 𝑒𝑒2 = 𝑒𝑒 (c)
2 2𝐿𝐿2 2
42
Application of energy method
Where 𝑁𝑁1and 𝑁𝑁1 are the tension forces in the two bars. The elongations of the two
bars are given by the relation 𝑒𝑒𝑖𝑖 = 𝑁𝑁𝑖𝑖 𝐿𝐿𝑖𝑖 /𝐸𝐸𝑖𝑖 𝐴𝐴𝑖𝑖 .

The total strain energy 𝑈𝑈 for the structure is equal to the sum 𝑈𝑈1 + 𝑈𝑈2 of the strain
energies of the two bars: therefore by Eq.(c)

𝐸𝐸1 𝐴𝐴1 2 𝐸𝐸2 𝐴𝐴2 2

𝑈𝑈 = 𝑒𝑒 + 𝑒𝑒 (d)
2𝐿𝐿1 1 2𝐿𝐿2 2

The magnitudes of 𝑃𝑃 and 𝑄𝑄 are obtained by differentiation of Eq.(d) with respect to 𝑢𝑢

and 𝑣𝑣 respectively. Thus,

𝜕𝜕𝜕𝜕 𝐸𝐸1 𝐴𝐴1 𝑒𝑒1 𝜕𝜕𝑒𝑒1 𝐸𝐸2 𝐴𝐴2 𝑒𝑒2 𝜕𝜕𝑒𝑒2

𝑃𝑃 = = +
𝜕𝜕𝜕𝜕 𝐿𝐿1 𝜕𝜕𝜕𝜕 𝐿𝐿2 𝜕𝜕𝜕𝜕

𝜕𝜕𝜕𝜕 𝐸𝐸1 𝐴𝐴1 𝑒𝑒1 𝜕𝜕𝑒𝑒1 𝐸𝐸2 𝐴𝐴2 𝑒𝑒2 𝜕𝜕𝑒𝑒2

𝑄𝑄 = = + (e)
𝜕𝜕𝜕𝜕 𝐿𝐿1 𝜕𝜕𝜕𝜕 𝐿𝐿2 𝜕𝜕𝜕𝜕

43
Application of energy method
The partial derivatives of 𝑒𝑒1 and 𝑒𝑒2 with respect to 𝑢𝑢 and 𝑣𝑣 are obtained from Eq.(b).
Taking the derivatives and substituting in Eq.(c). We find

𝐸𝐸1 𝐴𝐴1 (𝑏𝑏1 +𝑢𝑢) 𝑏𝑏1 + 𝑢𝑢 2 + ℎ + 𝑣𝑣 2 − 𝐿𝐿1

𝑃𝑃 =
𝐿𝐿1 𝑏𝑏1 + 𝑢𝑢 2 + ℎ + 𝑣𝑣 2
𝐸𝐸2 𝐴𝐴2 (𝑏𝑏2 −𝑢𝑢) 𝑏𝑏2 − 𝑢𝑢 2 + ℎ + 𝑣𝑣 2 − 𝐿𝐿2
−
𝐿𝐿2 𝑏𝑏2 − 𝑢𝑢 2 + ℎ + 𝑣𝑣 2

(f)
𝐸𝐸1 𝐴𝐴1 (ℎ + 𝑢𝑢) 𝑏𝑏1 + 𝑢𝑢 2 + ℎ + 𝑣𝑣 2 − 𝐿𝐿1
𝑄𝑄 =
𝐿𝐿1 𝑏𝑏1 + 𝑢𝑢 2 + ℎ + 𝑣𝑣 2
𝐸𝐸2 𝐴𝐴2 (ℎ + 𝑢𝑢) 𝑏𝑏2 − 𝑢𝑢 2 + ℎ + 𝑣𝑣 2 − 𝐿𝐿2
−
𝐿𝐿2 𝑏𝑏2 − 𝑢𝑢 2 + ℎ + 𝑣𝑣 2

(b) Substitution of the values 𝐾𝐾𝐾, 𝐾𝐾𝐾, 𝑏𝑏𝑏, 𝑏𝑏𝑏, ℎ, 𝐿𝐿𝐿, 𝐿𝐿𝐿, 𝑢𝑢, and 𝑣𝑣 into (𝑓𝑓)

𝑃𝑃 = 43.8𝑁𝑁 𝑎𝑎𝑎𝑎𝑎𝑎 𝑄𝑄 = 112.4𝑁𝑁 (𝑔𝑔)

44
Application of energy method
(c) The values of 𝑃𝑃 and 𝑄𝑄 may e verified by determining the tension forces𝑁𝑁1
and 𝑁𝑁2 in the two bars, determining directions of the axes of the two bars
for the deformed configuration and applying equations of equilibrium to a
free-body diagram of pin 𝐵𝐵∗ . Elongations 𝑒𝑒1 = 49.54𝑚𝑚𝑚𝑚 and 𝑒𝑒2 = 16.24𝑚𝑚𝑚𝑚
are given by Eq.(b). The tension forces 𝑁𝑁1 and 𝑁𝑁2 are

𝑁𝑁1 = 𝑒𝑒1 𝐾𝐾1 = 99.08𝑁𝑁

𝑁𝑁2 = 𝑒𝑒2 𝐾𝐾2 = 48.72𝑁𝑁

Angles 𝜃𝜃 ∗ and 𝜙𝜙 ∗ for the directions of the axes of the two bars for the
deformed configurations are found to be 0.7739 and 0.5504 rad,
respectively. The free-body diagram of pin 𝐵𝐵∗ is shown in Figure. The
equations of equilibrium are

∑ 𝐹𝐹𝑥𝑥 = 𝑃𝑃 − 𝑁𝑁1 𝑠𝑠𝑠𝑠𝑠𝑠(𝜃𝜃 ∗ ) + 𝑁𝑁2 𝑠𝑠𝑠𝑠𝑠𝑠(𝜙𝜙 ∗ ); hence, 𝑃𝑃 = 43.8𝑁𝑁

∑ 𝐹𝐹𝑥𝑥 = 𝑃𝑃 − 𝑁𝑁1 𝑠𝑠𝑠𝑠𝑠𝑠(𝜃𝜃 ∗ ) + 𝑁𝑁2 𝑠𝑠𝑠𝑠𝑠𝑠(𝜙𝜙 ∗ ); hence, 𝑄𝑄 = 112.4𝑁𝑁

These values of 𝑃𝑃 and 𝑄𝑄 agree with those of Eq.(g).

45
Application of energy method
(d) If displacements 𝑢𝑢 and 𝑣𝑣 are very small compared to 𝑏𝑏1 and 𝑏𝑏2 , and, hence with
respect to 𝐿𝐿1 and 𝐿𝐿2 , simple approximate expressions for 𝑃𝑃 and 𝑄𝑄 may be obtained.
For example, we find by the binomial expansion to linear terms 𝑢𝑢 and 𝑣𝑣 that

2 2
𝑏𝑏1 𝑢𝑢 ℎ𝑣𝑣
𝑏𝑏1 + 𝑢𝑢 + ℎ + 𝑣𝑣 = 𝐿𝐿1 + +
𝐿𝐿1 𝐿𝐿1
2 2
𝑏𝑏2 𝑢𝑢 ℎ𝑣𝑣
𝑏𝑏2 − 𝑢𝑢 + ℎ + 𝑣𝑣 = 𝐿𝐿2 − +
𝐿𝐿2 𝐿𝐿2

With these approximations, Eq.(f) yield the linear relations

𝐸𝐸1 𝐴𝐴1 𝑏𝑏1 𝐸𝐸2 𝐴𝐴2 𝑏𝑏2

𝑃𝑃 = (𝑏𝑏1 𝑢𝑢 + ℎ𝑣𝑣) + (𝑏𝑏2 𝑢𝑢 − ℎ𝑣𝑣)
𝐿𝐿1 3 𝐿𝐿2 3
𝐸𝐸1 𝐴𝐴1 ℎ 𝐸𝐸2 𝐴𝐴2 ℎ
𝑄𝑄 = (𝑏𝑏1 𝑢𝑢 + ℎ𝑣𝑣) + (−𝑏𝑏2 𝑢𝑢 + ℎ𝑣𝑣)
𝐿𝐿1 3 𝐿𝐿2

If these equations are solved for the displacements 𝑢𝑢 and 𝑣𝑣, the resulting
relations are identical to those derived by means of Castigliano’s theorem on
deflections for linearly elastic materials

46