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Graphical Method of Determining the Deflexion of Laterally Loaded Struts

Graphical Method of Determining the Deflexion of Laterally Loaded Struts Graphical Method of Determining the Deflexion of Laterally Loaded Struts By N. A. Townsend, A.F.R.Ae.S., M.I.Ae.S. TABLE I Bending Slope EI Moment = X ∑M.∆x. "M" Δx M.Δx. in. lb./in. in. straight line and the curve and dividing by E.I. HE solution of the differential equations for 0 0 This method is only applicable to the cases the varied cases of laterally loaded struts 1100 2 2200 where a polar diagram can be constructed. How­ present arithmetical difficulties particularly 2 2200 1900 2 3800 ever R. & M. 1233 covers many of the cases met as the number of the integration constants con­ 4 6000 tained in a problem increase. The time taken to with in routine stress analysis of airframes. 2600 2 5200 obtain an acceptable solution can be extremely To illustrate the method a fully worked example 3250 2 6500 long if the first answer arrived at does not seem to is presented and it is apparent that the evaluation 8 17700 3900 2 7800 be of the right order, thereby indicating an error of the eight integration constants in the mathe­ 10 25500 matical analysis presents a greater source of error in the evaluation of the constants. 4500 2 9000 12 34500 than does the graphical method. A graphical method besides giving the shape 5100 2 10200 of the deflexion curve and value at any position 14 Steel tube 1½in. dia. 17 S.W.G. 5550 2 11100 along the strut, is more satisfactory as any error Material spec. T.50. 16 55800 in shape is immediately noticeable. The bend­ 5850 2 11700 I =·0663 in.4 18 67500 ing moment diagram, an integral part of the E=28xl0 6lb./sq.in . 5970 2 11940 method, is also available so that the final solution 20 79440 μ=·0614. 5950 2 11900 of the problem presents the bending moment and 22 91340 FIG. 1 represents the loads on the strut and deflection at any point along the strut. 5730 2 11460 FIG. 2 the "polar diagram" for this loading. For 24 102800 The graphical method consists of drawing the 5420 2 10840 clearness the majority of the constructional lines polar diagram for the particular case of laterally 26 113640 used in forming the diagram have been omitted. 5080 2 10160 loaded strut and constructing the bending moment Values of the bending moment at every two-inch diagram from it. The graphical integration of this 4620 2 9240 of span are also shown in FIG. 2. 133040 curve represents the slope of the deflected strut 4100 2 8200 in terms of E & I. The graphical integration of (Concluded overleaf) 32 141240 3550 2 7100 the plotted slope curve gives the total deflexion 34 148340 of the strut—also in terms of E & I. The require­ 2930 2 5860 ment that there is no deflexion at either end of 2100 2 4200 the strut is represented on the plotted total 158400 1200 2 2400 deflexion curve by joining the ends of the curve with a straight line. The actual value of the deflexion at any point along the strut is found by reading off the differences of the ordinates of the August 1946 269 http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Aircraft Engineering and Aerospace Technology Emerald Publishing

Graphical Method of Determining the Deflexion of Laterally Loaded Struts

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Publisher
Emerald Publishing
Copyright
Copyright © Emerald Group Publishing Limited
ISSN
0002-2667
DOI
10.1108/eb031404
Publisher site
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Abstract

Graphical Method of Determining the Deflexion of Laterally Loaded Struts By N. A. Townsend, A.F.R.Ae.S., M.I.Ae.S. TABLE I Bending Slope EI Moment = X ∑M.∆x. "M" Δx M.Δx. in. lb./in. in. straight line and the curve and dividing by E.I. HE solution of the differential equations for 0 0 This method is only applicable to the cases the varied cases of laterally loaded struts 1100 2 2200 where a polar diagram can be constructed. How­ present arithmetical difficulties particularly 2 2200 1900 2 3800 ever R. & M. 1233 covers many of the cases met as the number of the integration constants con­ 4 6000 tained in a problem increase. The time taken to with in routine stress analysis of airframes. 2600 2 5200 obtain an acceptable solution can be extremely To illustrate the method a fully worked example 3250 2 6500 long if the first answer arrived at does not seem to is presented and it is apparent that the evaluation 8 17700 3900 2 7800 be of the right order, thereby indicating an error of the eight integration constants in the mathe­ 10 25500 matical analysis presents a greater source of error in the evaluation of the constants. 4500 2 9000 12 34500 than does the graphical method. A graphical method besides giving the shape 5100 2 10200 of the deflexion curve and value at any position 14 Steel tube 1½in. dia. 17 S.W.G. 5550 2 11100 along the strut, is more satisfactory as any error Material spec. T.50. 16 55800 in shape is immediately noticeable. The bend­ 5850 2 11700 I =·0663 in.4 18 67500 ing moment diagram, an integral part of the E=28xl0 6lb./sq.in . 5970 2 11940 method, is also available so that the final solution 20 79440 μ=·0614. 5950 2 11900 of the problem presents the bending moment and 22 91340 FIG. 1 represents the loads on the strut and deflection at any point along the strut. 5730 2 11460 FIG. 2 the "polar diagram" for this loading. For 24 102800 The graphical method consists of drawing the 5420 2 10840 clearness the majority of the constructional lines polar diagram for the particular case of laterally 26 113640 used in forming the diagram have been omitted. 5080 2 10160 loaded strut and constructing the bending moment Values of the bending moment at every two-inch diagram from it. The graphical integration of this 4620 2 9240 of span are also shown in FIG. 2. 133040 curve represents the slope of the deflected strut 4100 2 8200 in terms of E & I. The graphical integration of (Concluded overleaf) 32 141240 3550 2 7100 the plotted slope curve gives the total deflexion 34 148340 of the strut—also in terms of E & I. The require­ 2930 2 5860 ment that there is no deflexion at either end of 2100 2 4200 the strut is represented on the plotted total 158400 1200 2 2400 deflexion curve by joining the ends of the curve with a straight line. The actual value of the deflexion at any point along the strut is found by reading off the differences of the ordinates of the August 1946 269

Journal

Aircraft Engineering and Aerospace Technology – Emerald Publishing

Published: Aug 1, 1946

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