By T. Heddle

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A heavy cable running above ground is supported on vertical poles. The route changes direction by 30° at one pole, where the side strain is taken symmetrically by a stay at 70° to the horizontal, attached to the top of the pole. Each part of the cable leaves the pole at 10° to the horizontal and has a tension of 250 N . Calculate the tension in the stay, assuming no horizontal force due to the pole itself. 8. W O R K E D EXAMPLE A square plate of metal of uniform thickness has a side of length 30 cm.

WTxTS = wRXSR. 4) N o w if wi = weight per square centimetre of the plate, then 2 Wt = w i X 2 0 / 2 = 200wi 2 and wR = w i X ( 3 0 - 2 0 2 / 2 ) = 700wi. But TS = AS-AT = 30 cos 4 5 ° - ( f ) 20 cos 45° = 1 1 . 8 cm. Then, substituting in eqn. 8 = lOOwiXSR. 4 cm. 4 cm from the centre of the square on the line of symmetry. An alternative method is to apply the principle of moments t o the case of a knife edge under the corner A, running parallel t o DB (Fig. 13). If a force F equal t o the weight of the complete square ws is applied upward under the centre of the square, the 31 COMBINATIONS O F FORCES plate will be supported.

However, if there COMBINATIONS O F FORCES 43 were longitudinal components of external force, a resultant tension or thrust would be produced. If the force at the free end had been upward, the directions of Fs and Mb, shown in Fig. 24, would have been reversed. h. part. Thus in Fig. 24, Fs is positive but Mb is negative. This convention makes the shearing force positive where the bending moment diagram has a positive upward gradient; in fact Fsequals the gradient of Mb as seen in Fig. 23. The values of Fs and Mb at Ρ and other parts of the beam may be found if the reactions at the supports A and C in Fig.