Page 249 - Handbook of Civil Engineering Calculations, Second Edition

2.34 REINFORCED AND PRESTRESSED CONCRETE ENGINEERING AND DESIGN

3. Compute the minimum value of c at which the entire concrete
area is stressed to 085f c
Compute the corresponding values of P u and M u . Thus, a t 18 in. (457.2 mm);
c 18/0.85 21.8 in. (537.972 mm); f B
c E s (c d)/c 87,000(21.18 15.5)/21.18
23,300 lb/sq.in. (160,653.5 kPa); F c 30,600(18) 550,800 lb (2,449,958.4 N); F A
80,000 lb (355,840 N); F B 46,600 lb (207,276.8 N); P u 0.70(550,800 80,000
46,600) 474,200 lb (2,109,241.6 N); M u 0.70(80,000 46,600)6.5 152,000 in.·lb
(17,192.9 N·m).
4. Compute the value of c at which M u = 0; compute P 0
The bending moment vanishes when F B reaches 80,000 lb (355,840 N). From the calcula-
tion in step 3, f b 87,000(c d)/c 40,000 lb/sq.in. (275,800 kPa); therefore, c 28.7 in.
(728.98 mm); P o 0.70(550,800 160,000) 497,600 lb (2,213,324.8 N).

5. Assign other values to c, and compute P u and M u
By assigning values to c ranging from 8 to 28.7 in. (203.2 to 728.98 mm), typical calculations
are: when c 8 in. (203.2 mm), f B 40,000 lb/sq.in. ( 275,800 kPa); f A
40,000(5.5/7.5) 29,300 lb/sq.in. (202,023.5 kPa); a 6.8 in. (172.72 mm); F c
30,600(6.8) 208,100 lb (925,628.8 N); P u 0.70(208,100 58,600 80,000) 130,700
lb (581,353.6 N); M u 0.70 (208,100 5.6 138,600 6.5) 1,446,000 in.·lb
(163,369.1 N·m).
When c 10 in. (254 mm), f A 40,000 lb/sq.in. (275,800 kPa); f B 40,000
lb/sq.in. ( 275,800 kPa); a 8.5 in. (215.9 mm); F c 30,600(8.5) 260,100 lb
(1,156,924.8 N); P u 0.70(260,100) 182,100 lb (809,980 N); M u 0.70(260,100
4.75 160,000 6.5) 1,593,000 in.·lb (179,997.1 N·m).
When c 14 in. (355.6 mm), f B 87,000(14 15.5)/14 9320 lb/sq.in.
( 64,261.4 kPa); a 11.9 in. (302.26 mm); F c 30,600(11.9) 364,100 lb
(1,619,516.8 N); P u 0.70(364,100 80,000 18,600) 297,900 lb (1,325,059.2 N);
M u 0.70(364,100 3.05 98,600 6.5) 1,226,000 in.·lb (138,513.5 N·m).
6. Plot the points representing computed values of P u and M u in
the interaction diagram
Figure 21 shows these points. Pass a smooth curve through these points. Note that when
P u < P b , a reduction in M u is accompanied by a reduction in the allowable load P u .

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 N
is (a) 9.2 in. (233.68 mm); (b) 6 in. (152.4 mm).

Calculation Procedure:

1. Evaluate the eccentricity associated with balanced design
Let e denote the eccentricity of the load and e b the eccentricity associated with balanced
design. Then M u P u e. In Fig. 21, draw an arbitrary radius vector OD; then tan
ED/OE 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.