Derivation: Pinned-Pinned Column with Load Applied at Arbitrary Location

Last updated November 18, 2022
By Ian Story

Position the load a distance b from the bottom and a from the top

The top and bottom of the beam will have different deflection equations.

Reactions

(1) $\begin{equation*}\Sigma M_0 = -P_c_rv_B(b) + RL = 0\end{equation*}$

(2) $\begin{equation*}R = \frac{P_c_rv_B(b)}{L}\end{equation*}$

Bottom

(3) $\begin{equation*}\Sigma M_x = M(x) + P_c_rv_B(x)- Rx = 0\end{equation*}$

(4) $\begin{equation*}M(x) = -P_c_rv_B(x) + Rx = -P_c_rv_B(x) + \frac{P_c_rv_B(b)x}{L}\end{equation*}$

(5) $\begin{equation*}M(x) + P_c_rv_B(x) - \frac{P_c_rv_B(b)x}{L}\end{equation*}$

Use moment-curvature relationship: $EIv''(x) = M(x)$

(6) $\begin{equation*}EIv_B''(x) + P_c_rv_B(x) - \frac{P_c_rv_B(b)x}{L} = 0\end{equation*}$

(7) $\begin{equation*}v_B''(x) + k^2v_B(x) - \frac{k^2v_B(b)x}{L} = 0\end{equation*}$

where:

(8) $\begin{align*}k = \sqrt{\frac{P_c_r}{EI}} \\\end{align*}$

Use general solution for column buckling differential equation. Set the final integration constant to zero due to boundary condition ( $v_B(0) = 0$ )

(9) $\begin{equation*}v_B(x) = Asin(kx) + Bcos(kx) + Cx\end{equation*}$

where:

(10) $\begin{align*}C = \frac{v_B(b)}{L} \\\end{align*}$

Check solution works:

(11) $\begin{equation*}v_B'(x) = Akcos(kx) - Bksin(kx) + C\end{equation*}$

(12) $\begin{equation*}v_B''(x) = -Ak^2sin(kx) - Bk^2cos(kx)\end{equation*}$

(13) $\begin{equation*}[-Ak^2sin(kx) - Bk^2cos(kx)] + k^2[Asin(kx) + Bcos(kx) + \frac{v_B(b)x}{L}] - \frac{k^2v_B(b)x}{L} = 0\end{equation*}$

All the terms cancel out and we get 0 = 0, so this solution works

Top

(14) $\begin{equation*}\Sigma M_x = M(x) + Rx = 0\end{equation*}$

(15) $\begin{equation*}M(x) = -Rx = -\frac{Pv_B(b)x}{L}\end{equation*}$

Use moment-curvature relationship: $v''(x) = EIM(x)$

(16) $\begin{equation*}EIv_T''(x) = \frac{-Pv_B(b)x}{L}\end{equation*}$

(17) $\begin{equation*}v_T''(x) = \frac{-k^2v_B(b)x}{L}\end{equation*}$

where:

(18) $\begin{align*}k = \sqrt{\frac{P_c_r}{EI}} \\\end{align*}$

Integrate twice, setting the final integration constant to zero due to boundary condition $(v_T(0) = 0$

(19) $\begin{equation*}v_T(x) = \frac{-k^2v_B(b)x^3}{6L}+Dx\end{equation*}$

Boundary Conditions

The moment for the bottom section at x = 0 is zero (pin connection), which gives us the following boundary condition:

(20) $\begin{align*}v_B''(0) = 0 \\\end{align*}$

The two sections meet at the point of load, which gives us three additional boundary conditions:

(21) $\begin{align*}v_B(b) = v_T(a) \\v_B'(b) = -v_T'(a) \\v_B''(b) = v_T''(a) \\\end{align*}$

Solve for First Boundary Condition

(22) $\begin{equation*}v_B''(0) = -Ak^2sin(0) - Bk^2cos(0) = 0 \Longrightarrow B = 0\end{equation*}$

Solve for Second Boundary Condition

(23) $\begin{equation*}v_B(b) = Asin(kb) + Cb = \frac{-k^2v_B(b)a^3}{6L}+Da = v_T(a)\end{equation*}$

(24) $\begin{equation*}Asin(kb) + Cb = \frac{-k^2(Asin(kb) + Cb)a^3}{6L}+Da\end{equation*}$

(25) $\begin{equation*}Da = (Asin(kb) + Cb)(1 + \frac{-k^2a^3}{6L})\end{equation*}$

(26) $\begin{equation*}D = E(Asin(kb) + Cb)\end{equation*}$

where:

(27) $\begin{align*}E = \frac{1}{a} + \frac{k^2a^2}{6L} \\\end{align*}$

Solve for Third Boundary Condition

(28) $\begin{equation*}v_B'(b) = Akcos(kb) + C = F(Asin(kb) + Cb)-E(Asin(kb) + Cb) = -v_T'(a)\end{equation*}$

where:

(29) $\begin{align*}F = \frac{k^2a^2}{2L} \\\end{align*}$

(30) $\begin{equation*}Akcos(kb) - FAsin(kb) + EAsin(kb) = FCb - ECb - C\end{equation*}$

(31) $\begin{equation*}A[kcos(kb) - Fsin(kb) + Esin(kb)] = C[Fb - Eb - 1]\end{equation*}$

(32) $\begin{equation*}\frac{A}{C} = \frac{Fb - Eb - 1}{kcos(kb) - Fsin(kb) + Esin(kb)}\end{equation*}$

Solve for Fourth Boundary Condition

(33) $\begin{equation*}v_B''(b) = -Aksin(kb) = -G(Asin(kb) + Cb) = v_T''(a)\end{equation*}$

where:

(34) $\begin{align*}G = \frac{k^2a}{L} \\\end{align*}$

(35) $\begin{equation*}-Ak^2sin(kb) + GAsin(kb) = -GCb\end{equation*}$

(36) $\begin{equation*}\frac{A}{C} = \frac{Gb}{k^2sin(kb) - Gsin(kb)}\end{equation*}$

Equate and Collect Terms

(37) $\begin{equation*}\frac{Fb - Eb - 1}{kcos(kb) - Fsin(kb) + Esin(kb)} = \frac{Gb}{k^2sin(kb) - Gsin(kb)}\end{equation*}$

(38) $\begin{equation*}(Fb - Eb - 1)(k^2sin(kb) - Gsin(kb)) = Gb(kcos(kb) - Fsin(kb) + Esin(kb))\end{equation*}$

Divide both sides by sin(kb)

(39) $\begin{equation*}(Fb - Eb - 1)(k^2 - G) = Gb(kcot(kb) - F + E)\end{equation*}$

(40) $\begin{equation*}Fk^2b - FGb - Ek^2b + EGb - k^2 + G = Gkbcot(kb) - FGb + EGb\end{equation*}$

(41) $\begin{equation*}Fk^2b - FGb - Ek^2b + EGb - k^2 + G - Gkbcot(kb) + FGb - EGb = 0\end{equation*}$

Expand and Simplify

(42) $\begin{equation*}(\frac{k^2a^2}{2L})k^2b - (\frac{k^2a^2}{2L})(\frac{k^2a}{L})b - (\frac{1}{a} + \frac{k^2a^2}{6L})k^2b + (\frac{1}{a} + \frac{k^2a^2}{6L})(\frac{k^2a}{L})b - k^2 + (\frac{k^2a}{L}) - (\frac{k^2a}{L})kbcot(kb) + (\frac{k^2a^2}{2L})(\frac{k^2a}{L})b - (\frac{1}{a} + \frac{k^2a^2}{6L})(\frac{k^2a}{L})b = 0\end{equation*}$

(43) $\begin{equation*}\frac{k^4a^2b}{2L} - \frac{k^4a^3b}{2L^2} - \frac{k^2b}{a} - \frac{k^4a^2b}{6L} + \frac{k^2b}{L} + \frac{k^4a^3b}{6L^2} - k^2 + \frac{k^2a}{L} - \frac{k^3ab}{L}cot(kb) + \frac{k^4a^3b}{2L^2} - \frac{k^2b}{L} - \frac{k^4a^3b}{6L^2} = 0\end{equation*}$

Cancel terms (note: $a + b = L$ , so $\frac{k^2b}{L} + \frac{k^2a}{L} = k^2$ )

(44) $\begin{equation*}\frac{k^4a^2b}{3L} - \frac{k^3ab}{L}cot(kb) - \frac{k^2b}{a} - \frac{k^2b}{L} = 0\end{equation*}$

Multiply by $L$

(45) $\begin{equation*}\frac{k^4a^2b}{3} - k^3abcot(kb) - k^2b(\frac{L}{a} + 1) = 0\end{equation*}$

Multiply by $b^3$

(46) $\begin{equation*}\frac{k^4a^2b^4}{3} - k^3ab^4cot(kb) - k^2b^4(\frac{L}{a} + 1) = 0\end{equation*}$

(47) $\begin{equation*}\frac{a^2}{3}X^4 - abcot(X)X^3 - b^2(\frac{L}{a} + 1)X^2 = 0\end{equation*}$

where:

(48) $\begin{align*}X = kb \\\end{align*}$

Graph to Find Values

This is a transcendental equation, so there is no algebraic solution. We will need to use a graphing utility or numeric solver to identify roots. Solving the above equation for a range of values gives the following table.

Effective length factors are calculated by converting to the standard column buckling format:

(49) $\begin{equation*}kb = b\frac{L}{L}\sqrt{\frac{P_c_r}{EI}}\end{equation*}$

(50) $\begin{equation*}\frac{\frac{kb}{\pi(b/L)}\pi}{L} = \sqrt{\frac{P_c_r}{EI}}\end{equation*}$

(51) $\begin{equation*}\frac{\pi}{KL} = \sqrt{\frac{P_c_r}{EI}}\end{equation*}$

(52) $\begin{equation*}\frac{\pi^2}{(KL)^2} = \frac{P_c_r}{EI}\end{equation*}$

(53) $\begin{equation*}P_c_r = \frac{\pi^2EI}{(KL)^2}\end{equation*}$

where:

(54) $\begin{align*}K = \frac{\pi(b/L)}{kb} \\\end{align*}$

b/L	kb	Effective Length Factor (K)
1.00	3.141	1.00
0.95	3.134	0.95
0.90	3.110	0.91
0.85	3.070	0.87
0.80	3.013	0.83
0.75	2.936	0.80
0.70	2.835	0.78
0.65	2.708	0.75
0.60	2.551	0.74
0.55	2.366	0.73
0.50	2.160	0.73
0.45	1.945	0.73
0.40	1.731	0.73
0.35	1.524	0.72
0.30	1.327	0.71
0.25	1.141	0.69
0.20	0.962	0.65
0.15	0.787	0.60
0.10	0.608	0.52
0.05	0.408	0.38
0.00	0.000	0.00

Derivation: Pinned-Pinned Column with Load Applied at Arbitrary Location

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