Answered: An aluminum pipe (modulus of elasticity…

Steel Design (Activate Learning with these NEW titles from Engineering!)

6th Edition

ISBN:9781337094740

Author:Segui, William T.

Publisher:Segui, William T.

Chapter1: Introduction

Section: Chapter Questions

Problem 1.5.6P: The data in Table 1.5.3 were obtained from a tensile test of a metal specimen with a rectangular...

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An aluminum pipe (modulus of elasticity 10.1x10^6 psi) is fixed at one end and is subjected to a tensile load of 127.5 kips at the free end. The length and cross-sectional area of the pipe is given as 36.1 in and 9.11 in^2. Assume that the aluminum pipe has clockwise torques (100 lb-in) acting at the midpoint and free end of the pipe. (a) Derive the element stiffness matrix

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Step 1

Element stiffness matrix for a cantilever beam subjected to both axial load and moment will be the combination of the axial stiffness matrix( $K_{a x i a l}$ ) and the $(K_{b e n d i n g})$ .

$K_{a x i a l}$ = $[\begin{matrix} \frac{A E}{L} & 0 \end{matrix}]$

$K_{b e n d i n g} = [\begin{matrix} \frac{2 E I}{L} & \frac{L}{E I} \end{matrix}]$

Where A is the cross-sectional area of the beam = $9.11 i n^{2}$

E is the modulus of elasticity, E = $10.1 \times 10^{6} p s i$

L is the length of the beam = 36.1 in

I is the moment of inertia of the beam cross section.

If Area, $A = \frac{π}{4} d^{2} = 9.11$

$d = \sqrt{(\frac{9.11 \times 4}{π})} = 3.4 i n$

So moment of inertia, $I = \frac{π}{64} \times d^{4}$

$I = \frac{π}{64} \times 3 . 4^{4} = 6.56 i n^{4}$

Axial load, P = 127.5 kips

Torque at mid point and free end is same and its value is, M = 100 lb-in

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