36 Combined Flexure and Tension.

Article 36
Combined Flexure and Tension.

A beam that is built-in at one end carries a load $P$ at the other, and is also subjected to a horizontal tensile force $Q$ applied at the same point; to ﬁnd the equation of the curve assumed by its neutral surface: Let $x, y$ be any point of the elastic curve, referred to the free end as origin, then the bending moment for this point is $Q y - P x$ . Hence, with the usual notation of the theory of ﬂexure,¹⁸

\begin{matrix} E I \frac{d^{2} y}{d x^{2}} = Q y - P x, \frac{d^{2} y}{d x^{2}} = n^{2} (y - m x), & [m = \frac{P}{Q}, n^{2} = \frac{Q}{E I} . \end{matrix}

which, on putting $y - m x = u$ , and $\frac{d^{2} y}{d x^{2}} = \frac{d^{2} u}{d x^{2}}$ , becomes

\begin{matrix} \frac{d^{2} u}{d x^{2}} = n^{2} u, \end{matrix}

whence

\begin{matrix} u = A cosh n x + B sinh n x, & [probs. 28, 30 \end{matrix}

that is,

\begin{matrix} y = m x + A cosh n x + B sinh n x . \end{matrix}

The arbitrary constants $A$ , $B$ are to be determined by the terminal conditions. At the free end $x = 0$ , $y = 0$ ; hence $A$ must be zero, and

\begin{array}{l} y & = m x + B sinh n x, \\ \frac{d y}{d x} & = m + n B cosh n x; \\ but at the ﬁxed end, x = l, and \frac{d y}{d x} = 0, hence \\ B & = - \frac{m}{n} cosh n l, \\ and accordingly \\ y & = m x - \frac{m sinh n x}{n cosh n l} . \end{array}

To obtain the deﬂection of the loaded end, ﬁnd the ordinate of the ﬁxed end by putting $x = l$ , giving

deﬂection = m (l - \frac{1}{n} tanh n l),

Prob. 104.

Compute the deﬂection of a cast-iron beam,

2 \times 2

inches section, and

6

feet span, built-in at one end and carrying a load of

100

pounds at the other end, the beam being subjected to a horizontal tension of

8000

pounds. [In this case

I = \frac{4}{3}, E = 15 \times 1 0^{6}, Q = 8000, P = 100

; hence

n = \frac{1}{50}, m = \frac{1}{80}

, deﬂection

= \frac{1}{80} (72 - 50 tanh 1.44) = \frac{1}{80} (72 - 44.69) = . 341

inches.]

Prob. 105.

If the load be uniformly distributed over the beam, say

w

per linear unit, prove that the diﬀerential equation is

E I \frac{d^{2} y}{d x^{2}} = Q y - \frac{1}{2} w x^{2}, or \frac{d^{2} y}{d x^{2}} = n^{2} (y - m x^{2}),

and that the solution is $y = A cosh n x + B sinh n x + m x^{2} + \frac{2 m}{n^{2}}$ . Show also how to determine the arbitrary constants.

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Article 36Combined Flexure and Tension.

Article 36
Combined Flexure and Tension.