Problem 3.12. Timoshenko beam with built-in ends

A Timoshenko beam built-in at both ends is subjected to uniformly distributed load. Deflection function is given:

$v (x) = \frac{p}{24 E I} (x^{4} - 2 x^{3} l + x^{2} l^{2}) + \frac{p}{S} (\frac{- x^{2}}{2} + \frac{l x}{2})$
Based on the above deflection give the rotation and the average shear strain functions. Draw the moment and shear force diagrams.

Solve Problem

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Give the rotation funcion, Χ_y and the shear strain, γ_y.

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$χ_{y} = \frac{p}{E I} (\frac{x^{3}}{6} - \frac{x^{2} l}{4} + \frac{x l^{2}}{12})$

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$γ_{y} = \frac{p}{S} (- x + \frac{l}{2})$

Derive moment and shear force distribution. Draw their diagrams.

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Steps

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Step 1. Separate bending and shear displacements from the given function.

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The deflection function consists of two parts, the bending and the shear displacements are:

See Eq.(3-63)

$v = v_{B} + v_{S} where v_{B} (x) = \frac{p}{24 E I} (x^{4} - 2 x^{3} l + x^{2} l^{2}) v_{S} (x) = \frac{p}{S} (\frac{- x^{2}}{2} + \frac{l x}{2})$

Step 2. From bending displacement determine the function of the rotation of the cross section and its derivative.

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$\frac{d χ_{y}}{d x} = \frac{d^{2} v_{B}}{d x^{2}} \to χ_{y} = \frac{d v_{B}}{d x} + D = \frac{d}{d x} (\frac{p}{24 E I} (x^{4} - 2 x^{3} l + x^{2} l^{2})) + D = \frac{p}{24 E I} (4 x^{3} - 6 x^{2} l + 2 x l^{2}) since χ_{y} (0) = 0 \to D = 0$

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$κ_{z} = - \frac{d χ_{y}}{d x} = - \frac{d}{d x} (\frac{p}{24 E I} (4 x^{3} - 6 x^{2} l + 2 x l^{2})) = - \frac{p}{24 E I} (12 x^{2} - 12 x l + 2 l^{2})$

Step 3. From shear displacement give the average shear strain function.

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$γ_{y} = \frac{d v_{S}}{d x} = \frac{d}{d x} (\frac{p}{S} (\frac{- x^{2}}{2} + \frac{l x}{2})) = \frac{p}{S} (- x + \frac{l}{2})$

Step 4. Determine bending moment distribution from $κ_{z}$ .

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Eq.(3-49)

$M (x) = E I_{z} κ_{z} = \frac{- p}{24} (12 x^{2} - 12 x l + 2 l^{2}) M (0) = M (L) = - \frac{p l^{2}}{12} M (\frac{L}{2}) = \frac{p l^{2}}{24}$

Step 5. Determine shear force distribution from the shear strain.

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Eq.(3-51)

$V = S γ_{y} = p (- x + \frac{l}{2}) V (0) = \frac{p l}{2} V (L) = - \frac{p l}{2}$

Results

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The given deflection function consists of two parts, the bending and the shear displacements:

See Eq.(3-63)

v = v_{B} + v_{S} where v_{B} (x) = \frac{p}{24 E I} (x^{4} - 2 x^{3} l + x^{2} l^{2}) v_{S} (x) = \frac{p}{S} (\frac{- x^{2}}{2} + \frac{l x}{2})

From bending displacement the function of the rotation of the cross section and its derivative can be determined:

Eq.(3-63)

$κ_{z} = - \frac{d χ_{y}}{d x} = - \frac{d}{d x} (\frac{p}{24 E I} (4 x^{3} - 6 x^{2} l + 2 x l^{2})) = - \frac{p}{24 E I} (12 x^{2} - 12 x l + 2 l^{2})$

The average shear strain function is derived from the shear displacement:

$γ_{y} = \frac{d v_{S}}{d x} = \frac{d}{d x} (\frac{p}{S} (\frac{- x^{2}}{2} + \frac{l x}{2})) = \frac{p}{S} (- x + \frac{l}{2})$

We can determine the bending moment distribution from $κ_{z}$ .

Eq.(3-49)

$M (x) = E I_{z} κ_{z} = \frac{- p}{24} (12 x^{2} - 12 x l + 2 l^{2}) M (0) = M (L) = - \frac{p l^{2}}{12} M (\frac{L}{2}) = \frac{p l^{2}}{24}$

The shear force distribution is obtained from the shear strain.

Eq.(3-51)

$V = S γ_{y} = p (- x + \frac{l}{2}) V (0) = \frac{p l}{2} V (L) = - \frac{p l}{2}$