Structural Concrete - Hassoun

16.6 Design of Frame Hinges 577
iii. Equate Eqs. I and II and solve to get a = 6.313 in. and P n = 113.5 K. Check f s ′ = 87 (c −
d ′ )∕c ≤ f y ∶ c = a∕0.85 = 7.43 in. and f s ′ = 87(7.43 − 2.5)∕7.43 = 58 ksi, which is close
to the 60 ksi assumed in the calculations. Choose no. 3 ties spaced at 16 in.
iv. Check φ: d t = 17.5 in.
( )
dt − c
ε t = 0.003 = 0.00407
c
( ) 250
φ = 0.65 +(ε t − 0.002) = 0.823
3
φP n = 0.823(113.5) =93.3K > 80 K
The section is adequate.
6. Check the adequacy of the column section at midheight, 7.5 ft from A: M u = 336.8/2 = 168.4 K ⋅ ft.
P u = 80 + 2.5(half the column weight) =82.5K
Use A s = three no. 9 bars and A ′ s = three no. 7 bars. In an approach similar to step 5, φP n = 122
K > 82.5 K (no. 10 bars can be terminated, and they have to be extended a development length
below the midheight of the column).
7. Design the hinge at A: M u = 0, H = 22.5 K, P u = 85 K.
a. Choose a Mesnager hinge. Using Eqs. 16.3a and Eqs. 16.3b, R 1 = 72 K and R 2 = 27 K. (Refer
to Fig. 16.19 with θ = 30 ∘ )
A s1 = R 1 72
=
0.55f y 0.55 × 60 = 2.2in.2
3 no. 9
3 no. 6
No. 3 hinge ties
3 no. 8
2 no. 7
3 no. 8
4 no. 3 ties
2 no. 7
Figure 16.19
Example 16.2: Hinge details.