# V. Twist in a Tapered Tube

To find the twist at the free end of a hollow tapered shaft of uniform thickness.

## Reference

Hand calculation using the following reference:

Mechanics of Materials, F.P.Beer & E.R.Johnston, 1981, McGraw-Hill Review Problem 3.120, Page 149.

## Problem

The beam in the following figure which has the following geometric, load, and section properties: L = 2 m, outside diameter at fixed end = 80 mm, outside diameter at free end = 40 mm, uniform wall thickness of 10 mm, , IZ = 5,000 cm4 , E = 200 KN/mm2 , T = 2.0 KNm, and Poisson's ratio = 0.3.

## Theoretical Solution

According to the reference, the twist at the free end is

$Φ A = ( T L 4 π G t ) ( C A + C B C A 2 C B 2 )$
where
 CA = centerline radius at free end = 40 mm/2 - 10 mm/2 = 15 mm CB = center line radius at fixed end = 80 mm/2 - 10 mm/2 = 35 mm G = $E 2 ( 1 + ν ) = 200 k N / m m 2 2 ( 1 + 0.3 ) = 76.9$

## Comparison

Table 1. Comparison of results
Twist at free end (radians) 0.0751 0.0751 none

The file C:\Users\Public\Public Documents\STAAD.Pro CONNECT Edition\Samples\ Verification Models\01 Beams\Twist in a Tapered Tube.STDis typically installed with the program.

STAAD SPACE TORSION ON CONICAL SHAFT
START JOB INFORMATION
ENGINEER DATE 18-Sep-18
END JOB INFORMATION
*
* REFERENCE : MECHANICS OF MATERIALS, F.P.BEER &amp; E.R.JOHNSTON
* 1981, MCGRAW-HILL
*
* REVIEW PROBLEM 3.120, PAGE 149. TWIST AT FREE END SHOULD BE
*
UNIT METER KN
JOINT COORDINATES
1 0 0 0; 2 0 2 0;
MEMBER INCIDENCES
1 1 2;
UNIT MMS KN
MEMBER PROPERTY AMERICAN
1 PRIS ROUND STA 80 END 40 THI 10
UNIT METER KN
DEFINE MATERIAL START
ISOTROPIC MATERIAL1
E 2e+08
POISSON 0.3
END DEFINE MATERIAL
UNIT MMS KN
CONSTANTS
MATERIAL MATERIAL1 ALL
SUPPORTS
1 FIXED
UNIT METER KN
2 MY 2
PERFORM ANALYSIS
PRINT JOINT DISPLACEMENTS
FINISH


   JOINT DISPLACEMENT (CM   RADIANS)    STRUCTURE TYPE = SPACE