Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

the design. Substitute the allowable force for member (1) into Equation (c) to compute the
maximum load P that would be permitted:
⎡ sin45°
⎤
P = ⎢cos45° + cos60°
⎥ F = 1.11536F
⎣ sin60°
⎦
= (1.11536)(173.28 kips)
∴ P ≤ 193.27 kips
1 1,allow
(h)
Next, use Equation (d) to compute the maximum load P that would be permitted if member
(2) controls:
⎡sin60°
⎤
P =−⎢
cos45°+ cos60°
⎥ F =−1.36603F
⎣sin45°
⎦
=−(1.36603)(93.12 kips)
∴ P ≤ −127.20 kips
2 2,allow
(i)
Why is P negative in Equation (i), and, more important, how do we interpret this negative
value? The allowable stress computed in Equation (e) made no distinction between
tension and compression stress. Accordingly, the allowable member forces computed in
Equations (f) and (g) were magnitudes only. These member forces could be tension (i.e.,
positive values) or compression (i.e., negative values) forces. In Equation (i), a maximum
load was computed as P = −127.20 kips. This result implies that the load P acts in the −x
direction, and, clearly, that is not what the problem intends. Therefore, we must conclude
that the allowable force in member (2) is actually a compression force:
P ≤−(1.36603)( − 93.12 kips) = 127.20 kips (j)
Compare the results from Equations (h) and (j) to conclude that the maximum load that
may be applied to this truss joint is
P = 127.20 kips Ans.
Member forces at maximum load P: Member (2) has been shown to control the design; in other
words, the strength of member (2) is the limiting factor or the most critical consideration. At
the maximum load P, use Equations (c) and (d) to compute the actual member forces:
and
F = 114.05 kips (T)
1
F2
=− 93.12 kips = 93.12 kips (C)
The actual normal stresses in the members are
and
F1
114.05 kips
σ 1 = = = 15.80 ksi (T)
2
A 7.22 in.
1
F2
σ = = − 93.12 kips
2
= 24.0 ksi(C)
2
A 3.88 in.
2
Note: The normal stress magnitudes in both members are less than or equal to the 24-ksi
allowable stress.
72