,
,
, Solutions to Chapter 1
Problems
S.1.1
The principal stresses are given directly by Eqs (1.11) and (1.12) in which x ˆ 80 N/mm2, y ˆ 0 (or vice
versa) and xy ˆ 45 N/mm2. Thus, from Eq. (1.11)
80 1 p•••••2•••••••••••• •••••••••2•
Iˆ ‡ 80 ‡ 4 4 5
2 2
i.e.
I ˆ 100:2 N=mm2
From Eq. (1.12)
80 1 p•••••2•••••••••••• ••••••• ••2•
II ˆ ÿ 80 ‡ 4 4 5
2 2
i.e.
II ˆ ÿ20:2 N=mm2
The directions of the principal stresses are de®ned by the angle in Fig. 1.8(b) in which is given by
Eq. (1.10). Hence
2 45
tan 2 ˆ ˆ 1:125
80 ÿ 0
which gives
ˆ 248 110 and ˆ 1148 110
It is clear from the derivation of Eqs (1.11) and (1.12) that the ®rst value of
corresponds to I while the second value corresponds to II.
Finally, the maximum shear stress is obtained from either of Eqs (1.14) or (1.15). Hence from Eq.
(1.15)
100:2 ÿ …ÿ20:2† 2
max ˆ ˆ 60:2 N=mm
2
,
, Solutions to Chapter 1
Problems
S.1.1
The principal stresses are given directly by Eqs (1.11) and (1.12) in which x ˆ 80 N/mm2, y ˆ 0 (or vice
versa) and xy ˆ 45 N/mm2. Thus, from Eq. (1.11)
80 1 p•••••2•••••••••••• •••••••••2•
Iˆ ‡ 80 ‡ 4 4 5
2 2
i.e.
I ˆ 100:2 N=mm2
From Eq. (1.12)
80 1 p•••••2•••••••••••• ••••••• ••2•
II ˆ ÿ 80 ‡ 4 4 5
2 2
i.e.
II ˆ ÿ20:2 N=mm2
The directions of the principal stresses are de®ned by the angle in Fig. 1.8(b) in which is given by
Eq. (1.10). Hence
2 45
tan 2 ˆ ˆ 1:125
80 ÿ 0
which gives
ˆ 248 110 and ˆ 1148 110
It is clear from the derivation of Eqs (1.11) and (1.12) that the ®rst value of
corresponds to I while the second value corresponds to II.
Finally, the maximum shear stress is obtained from either of Eqs (1.14) or (1.15). Hence from Eq.
(1.15)
100:2 ÿ …ÿ20:2† 2
max ˆ ˆ 60:2 N=mm
2
