Direct integration of a beam

Each time an integration is carried out, a constant of integration needs to be obtained. These constants are determined by using either the forces at supports, or at free ends.

: For internal shear and moment, the constants can be found by analyzing the beam's free body diagram.

: For rotation and displacement, the constants are found using conditions dependent on the type of supports. For a cantilever beam, the fixed support has zero rotation and zero displacement. For a beam supported by a pin and roller, both the supports have zero displacement.

Image:Simplebeam.JPG per meter load over a 15m length.]]

Take the beam shown at right supported by a fixed pin at the left and a roller at the right. There are no applied moments, the weight is a constant 10 kN, and - due to symmetry - each support applies a 75 kN vertical force to the beam. Taking x as the distance from the pin,

: $w(x)=\; 10~\backslash textrm\{kN\}/\backslash textrm\{m\}$

Integrating,

:$V(x)=\; -\backslash int\; w(x)\backslash ,\; dx=-10x+C\_1\; (\backslash textrm\{kN\})$

where $C\_1$ represents the applied loads. For these calculations, the only load having an effect on the beam is the 75 kN load applied by the pin, applied at x=0, giving

:$V(x)=-10x+75\; (\backslash textrm\{kN\})$

Integrating the internal shear,

:$M(x)=\; \backslash int\; V(x)\backslash ,\; dx=-5x^2\; +\; 75x\; (\backslash textrm\{kN\}\; \backslash cdot\; \backslash textrm\{m\})$ where, because there is no applied moment, $C\_2\; =0$.

Assuming an EI value of 1 kN$\backslash cdot$m$\backslash cdot$m (for simplicity, real EI values for structural members such as steel are normally greater by powers of ten)

:$\backslash theta\; (x)=\; \backslash int\; \backslash frac\{M(x)\}\{EI\}\backslash ,\; dx=\; -\backslash frac\{5\}\{3\}\; x^3\; +\; \backslash frac\{75\}\{2\}\; x^2\; +\; C\_3(\backslash textrm\{m\}/\backslash textrm\{m\})$* and

:$\backslash nu\; (x)\; =\; \backslash int\; \backslash theta\; (x)\backslash ,\; dx\; =\; -\backslash frac\{5\}\{12\}\; x^4\; +\; \backslash frac\{75\}\{6\}\; x^3\; +\; C\_3\; x\; +\; C\_4\; (\backslash textrm\{m\})$

Because of the vertical supports at each end of the beam, the displacement ($\backslash nu$) at x = 0 and x = 15 m is zero. Substituting (x = 0, ν(0) = 0) and (x = 15 m, ν(15 m) = 0), we can solve for constants $C\_3$=-1406.25 and $C\_4$=0, yielding

:$\backslash theta\; (x)=\; \backslash int\; \backslash frac\{M(x)\}\{EI\}\backslash ,\; dx=\; -\backslash frac\{5\}\{3\}\; x^3\; +\; \backslash frac\{75\}\{2\}\; x^2\; -1406.25(\backslash textrm\{m\}/\backslash textrm\{m\})$ and

:$\backslash nu\; (x)\; =\; \backslash int\; \backslash theta\; (x)\backslash ,\; dx\; =\; -\backslash frac\{5\}\{12\}\; x^4\; +\; \backslash frac\{75\}\{6\}\; x^3\; -1406.25x\; (\backslash textrm\{m\})$

For the given EI value, the maximum displacement, at x=7.5 m, is approximately 440 times the length of the beam. For a more realistic situation, such as a uniform load of 1 kN and an EI value of 5,000 kN·m², the displacement would be approximately 13 cm.

• Note that for the rotation $\backslash theta$ the units are meters divided by meters (or any other units of length which reduce to unity). This is because rotation is given as a slope, the vertical displacement divided by the horizontal change.

• Bending
• Beam theory
• Euler–Bernoulli static beam equation
• Solid Mechanics
• Virtual Work

• Hibbeler, R.C., Mechanics Materials, sixth edition; Pearson Prentice Hall, 2005. {{ISBN|0-13-191345-X}}.

• [http://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/double-integration-method-beam-deflections Beam Deflection by Double Integration Method]

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