,Aircraft Structures
for engineering students
Fourth Edition
Solutions Manual
T. H. G. Megson
,This page intentionally left blank
, Solutions Manual
Solutions to Chapter 1 Problems
S.1.1
The principal stresses are given directly by Eqs (1.11) and (1.12) in which
σx = 80 N/mm2, σy = 0 (or vice versa) and,τxy = 45 N/mm2. Thus, from Eq. (1.11)
80 1
σI 802 + 4 × 452
= +
2 2
i.e.
σI = 100.2 N/mm2
From Eq. (1.12)
80 1, 2
σII 80 + 4 × 452
= —
2 2
i.e.
σII = —20.2 N/mm2
The directions of the principal stresses are defined by the angle θ in Fig. 1.8(b) in
which θ is given by Eq. (1.10). Hence
2 × 45
tan 2θ 1.125
= =
80 — 0
which gives
θ = 24◦11r and θ = 114◦11r
It is clear from the derivation of Eqs (1.11) and (1.12) that the first value of θ
corresponds to σI while the second value corresponds to σII.
Finally, the maximum shear stress is obtained from either of Eqs (1.14) or (1.15).
Hence from Eq. (1.15)
100.2 — (—20.2)
τmax = = 60.2 N/mm2
2
and will act on planes at 45◦ to the principal planes.
for engineering students
Fourth Edition
Solutions Manual
T. H. G. Megson
,This page intentionally left blank
, Solutions Manual
Solutions to Chapter 1 Problems
S.1.1
The principal stresses are given directly by Eqs (1.11) and (1.12) in which
σx = 80 N/mm2, σy = 0 (or vice versa) and,τxy = 45 N/mm2. Thus, from Eq. (1.11)
80 1
σI 802 + 4 × 452
= +
2 2
i.e.
σI = 100.2 N/mm2
From Eq. (1.12)
80 1, 2
σII 80 + 4 × 452
= —
2 2
i.e.
σII = —20.2 N/mm2
The directions of the principal stresses are defined by the angle θ in Fig. 1.8(b) in
which θ is given by Eq. (1.10). Hence
2 × 45
tan 2θ 1.125
= =
80 — 0
which gives
θ = 24◦11r and θ = 114◦11r
It is clear from the derivation of Eqs (1.11) and (1.12) that the first value of θ
corresponds to σI while the second value corresponds to σII.
Finally, the maximum shear stress is obtained from either of Eqs (1.14) or (1.15).
Hence from Eq. (1.15)
100.2 — (—20.2)
τmax = = 60.2 N/mm2
2
and will act on planes at 45◦ to the principal planes.
