c = 18/0.85 = 21.8 in (537.972 mm);/* = €<£& - d)lc = 87,000(21.18 - 15.5)721.18 =
23,300 lb/in^2 (160,653.5 kPa); F 0 = 30,600(18) = 550,800 Ib (2,449,958.4 N); FA = 80,000
Ib (355,840 N); FB = 46,600 Ib (207,276.8 N); Pn = 0.70(550,800 + 80,000 + 46,600) =
474,200 Ib (2,109,241.6 N); Mu = 0.70(80,000 - 46,600)6.5 = 152,000 in-lb (17,192.9
N-m).
- Compute the value of c at which Mu = O; compute P 0
The bending moment vanishes when FB reaches 80,000 Ib (355,840 N). From the calcula-
tion in step 3,/ 6 = 87,000(c - d)/c = 40,000 lb/in^2 (275,800 kPa); therefore, c = 28.7 in
(728.98 mm); P 0 = 0.70(550,800 + 160,000) = 497,600 Ib (2,213,324.8 N). - Assign other values to c, and compute Pu and Mu
By assigning values to c ranging from 8 to 28.7 in (203. 2 to 728.98 mm), typical calcula-
tions are: when c = 8 in (203.2 mm), fB = -40,000 lb/in^2 (-275,800 kPa); fA =
40,000(5.5/7.5) = 29,300 lb/in^2 (202,023.5 kPa); a = 6.8 in (172.72 mm); F 0 = 30,600(6.8)
= 208,100 Ib (925,628.8 N); P 14 = 0.70(208,100 + 58,600 - 80,000) = 130,700 Ib
(581,353.6 N); Mn = 0.70 (208,100 x 5.6 + 138,600 x 6.5) = 1,446,000 in-lb (163,369.1
N-m).
When c = 10 in (254 mm), fA = 40,000 lb/in^2 (275,800 kPa); fB = -40,000 lb/in^2
(-275,800 kPa); a = 8.5 in (215.9 mm); F 0 = 30,600(8.5) = 260,100 Ib (1,156,924.8 N); Pu
= 0.70(260,100) = 182,100 Ib (809,980 N); Mn = 0.70(260,100 x 4.75 + 160,000 x 6.5) =
1,593,000 in-lb (179,997.1 N-m).
When c = 14 in (355.6 mm)JB = 87,000(14 - 15.5)/14 = -9320 lb/in^2 (-64,261.4kPa);
a = 11.9 in (302.26 mm); F 0 = 30,600(11.9) = 364,100 Ib (1,619,516.8 N); Pn =
0.70(364,100 + 80,000 - 18,600) = 297,900 Ib (1,325,059.2 N); Mn = 0.70(364,100 x
3.05 + 98,600 x 6.5) = 1,226,000 in-lb (138,513.5 N-m). - Plot the points representing computed values of P 11 and Mu in
the interaction diagram
Figure 21 shows these points. Pass a smooth curve through these points. Note that when
Pn < Pb, a reduction in Mn is accompanied by a reduction in the allowable load Pn.
AXIAL-LOAD CAPACITY OF
RECTANGULAR MEMBER
The member analyzed in the previous calculation procedure is to carry an eccentric longi-
tudinal load. Determine the allowable ultimate load if the eccentricity as measured from TV
is (a) 9.2 in (233.68 mm); (b) 6 in (152.4 mm).Calculation Procedure:- Evaluate the eccentricity associated with balanced design
Let e denote the eccentricity of the load and eb the eccentricity associated with balanced
design. Then Mn = Pne. In Fig. 21, draw an arbitrary radius vector OD; then tan 0 =
EDIOE = eccentricity corresponding to point D.
Proceeding along the interaction diagram from A to C, we see that the value of c in-
creases and the value of e decreases. Thus, c and e vary in the reverse manner. To evalu-
ate the allowable loads, it is necessary to identify the portion of the interaction diagram to
which each eccentricity applies.
From the computations of the previous calculation procedure, eb = M^JPb =