Page 47 - Aircraft Stuctures for Engineering Student

32 Basic elasticity

and
(ii)

respectively. Adding Eqs (i) and (ii) we obtain

01 + a11 = a,
Thus
ax = 80.9 - 10.9 = 70N/mm2
For an axial load P

2 p P
ax = 70N/mm = - =
A n x 502/4
whence
P = 137.4kN
Substituting for a, in either of Eqs (i) or (ii) gives

7.1. = 29.7N/mm2
From the theory of the torsion of circular section bars
Tr T x 25
T,~ = 29.7N/mm = - =
J ITX 504/32
from which
T =0.7kNm
Note that P could have been found directly in this particular case from the axial
strain. Thus, from the first of Eqs (1.47)
a, = EE, = 70000 x 1000 x lop6 = 70N/mm2

as before.

1 Timoshenko, S. and Goodier, J. N., Theory of Elasticity, 2nd edition, McGraw-Hill Book
Company, New York, 1951.
2 Wang, C. T., Applied Elasticity, McGraw-Hill Book Company, New York, 1953.

P.l.l A structural member supports loads which produce, at a particular point, a
direct tensile stress of 80 N/mm2 and a shear stress of 45 N/mm2 on the same plane.
Calculate the values and directions of the principal stresses at the point and also the
maximum shear stress, stating on which planes this will act.