________________________________________________________________________
PROBLEM (11.51) Redo Prob. 11.50, this time applying Eq. (11.21).
SOLUTION
See solution of Prob. 11.50:
(a)
62
1963.5 10Am
−
=×
20.8 56
e
L
ec
rr
==
Equation (11.21) becomes
69
275 10 1 0.8sec 28 200 10
yy
PP
AA
⎡⎤
⎛⎞
⎢⎥
×= + ⎜⎟
⎜⎟
×
⎢⎥
⎝⎠
⎣⎦
Solving by trial and error: 133
y
PA MPa
=
Then
66
133 10 (1963.5 10 ) 261
y
Pk
−
=× × = N
So, 261 3 87
all
Pk==N
(b) 20.48 56
e
L
ec
rr
==
Equation (11.21) becomes
69
275 10 1 0.48sec 28 200 10
yy
PP
AA
⎡⎤
⎛⎞
⎢⎥
×= + ⎜⎟
⎜⎟
×
⎢⎥
⎝⎠
⎣⎦
Solving by trial and error: 163
y
PA MPa
=
66
163 10 (1963.5 10 ) 320
y
Pk
−
=× × = N
and
320 3 106.7
all
Pk== N
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