Access the full text.

Sign up today, get DeepDyve free for 14 days.

Aircraft Engineering and Aerospace Technology
, Volume 11 (8): 1 – Aug 1, 1939

/lp/emerald-publishing/stiffness-of-beams-and-cables-6tPqnPdf8o

- Publisher
- Emerald Publishing
- Copyright
- Copyright © Emerald Group Publishing Limited
- ISSN
- 0002-2667
- DOI
- 10.1108/eb030528
- Publisher site
- See Article on Publisher Site

314 AIRCRAFT ENGINEERIN G August, 1939 Transitio n of a Laterally Loaded Member with Tensile End Load from a "Beam " to a "Cable " By Huang Yushan, B.Sc. and M = ½ W (a - x) (11) Introduction 0 ONSIDE R a member carrying both lateral Similarly, M/M = ....(12) load and axial tension, which we .shall refer to as a " beam-tie." If the and at x = 0, member is very stiff, it acts as a beam ; but as A' = tanh µa/µa = tanli S/S (13) its stiffness is reduced, it tends to become a The curve K vs S is also plotted in Fig. 2. cable. The present practice is to assume that For large values of S, (13) reduces to those are beams and those are cables. No clear K = l/S (13A) demarcation between them has ye t been worked and for small values of S, out. This paper, by comparing the maximum K = 1 - S2/3 (13B) bending moment and the maximum deflection, Using the 5 per cent limit, then gives an account of the effect of variation of S = 20 for K = 0·05 stiffness on the change from beam to cable. A For practical purposes, an error of 5 per cent is and = 0·39 for K = 0·95 non-dimensional stiffness constant 5 is intro 2 generally allowable. So as a practical criterion, If 1 per cent accuracy is required, duced and some criteria are given for practical we may say tha t a beam-tie having " K " less application. S = 100 for K = 0·01 tha n 0·05 may be regarded as a cable, while if S = 0·17 for K = 0·99 its " K " is greater tha n 0·95 it ma y be regarded as a beam. Solving (9A) and (9B), Flat Beam-Tie Under Uniformly Distributed S = 6·2 for K = 0·05 Effect of Initial Curvature Load 1 and S = 0·35 for K = 0·95 Although the above derivations are based on For a member subjected to axial tension P If the error is limited to 1 pe r cent, the criteria a flat beam-tie, the results can be extended to and uniformly distributed lateral load w per are S = 14 for A' = 0·01 cases of beam-tie with initial curvature also. unit length, as shown in Tig. 1, the moment at and S = 0·16 for A' = 0·99 Suppose the initial curve of the uniformly an y section can be derived from the following loaded beam is equation : Flat Beam-Tie Under Concentrated Load Next consider a beam-tie under a con centrated load W at its middle. The bending where c is th e maximum camber at middle, then, moment becomes which yields The bending moment M is also the primary bending moment M , which is the bending moment of pure beam, plus the product of the The primary bending moment M is axial tension P and the deflection y, thus M=M , + Py (4) where M = ½ w (a2 — x2) (5) Fro m (4), th e deflection of a beam-tie is y = ( M-M ) / P (6) Therefore the ratio K is still the same as given in (7). Now, if this member approaches a pure beam, M approaches M and, if it approaches a pure Similarly if the initial curve of the con cable, M approaches zero. The ratio M/M , centrated loading case is assumed to be which is represented here by K, defines the y = c (x/a — 1), th e same result is found. ratio of the bending moment of the beam-tie to tha t of a pure beam. On the other hand, as Conclusion given in (6), the deflection of a pure cable is equal to — M /P. The quantity (1 — A') 0 Eased upon the above calculations, in which therefore represents the ratio of the deflection it is assumed that the deflection of the member of the beam-tie to tha t of a pure cable. is not very large, the following conclusions may Dividing (3) by (5), th e ratio K for uniformly be drawn : distributed load is given as (1) For engineering purposes, if the stiffness constant 5 of a member is higher than 10, th e member may be taken as a cable ; and on the other hand if that constant is below 0·4, it may be taken as a beam. At x = 0, where both the bending moment and (2) The maximum bending moment and the deflection are maximum : maximum deflection of a beam-tie may be estimated by means of Fig. 2, the procedure is : Let the stiffness constant S be representing µa (a) Compute the stiffness S of tha t member. (b) Find the corresponding value of K from hig. 2. The ratio K is plotted against the stiffness (c) The maximum bending moment is then constant 5 as shown in Fig. 2. equal to A' times that of a pure beam under Equation (9) can be simplified for large values th e same loading. and small values of S. It gives A", = 2/S2 for large value of S (9A) (d) The maximum deflection is equal to (1 — A') times that of a pure cable under the and = 1 — S2 for small value of S. .(9B) same loading.

Aircraft Engineering and Aerospace Technology – Emerald Publishing

**Published: ** Aug 1, 1939

Loading...

You can share this free article with as many people as you like with the url below! We hope you enjoy this feature!

Read and print from thousands of top scholarly journals.

System error. Please try again!

Already have an account? Log in

Bookmark this article. You can see your Bookmarks on your DeepDyve Library.

To save an article, **log in** first, or **sign up** for a DeepDyve account if you don’t already have one.

Copy and paste the desired citation format or use the link below to download a file formatted for EndNote

Access the full text.

Sign up today, get DeepDyve free for 14 days.

All DeepDyve websites use cookies to improve your online experience. They were placed on your computer when you launched this website. You can change your cookie settings through your browser.