CEE 415 Example 11.1: Design a square tied column and a circular spiral column for concentric axial loads. Use fc = 4000 psi, fy = 60 ksi and fyt = 60 ksi Assume the loads are: PD = 150 kips and PL = 200 kips Start design with g 0.02. Pu = 1.2 PD + 1.6 PL
CEE 415 Example 7.1: We are to check deflections and bar spacing (crack width control) for an interior span of a multi-span floor beam. The floor beams frame between columns (18 x 18 in.), so the total span is 24 ft. and the clear span is 22.5 ft. Assume
CEE 415 Example 4.4 Design a doubly reinforced section for negative bending moment in the indicated girder at the face of the first interior column of the floor plan shown on the last page. We cannot use the ACI moment coefficients for this case because o
CEE 415 Example 4.3: Design reinforcement for midspan and end sections of an interior span of the one-way slab shown in the floor system used in class discussion. Assume SDL = 25 psf and LL = 60 psf (no reduction). Use fc' = 4000 psi and fy = 60 ksi. From
CEE 415 Example 3.2: Calculate Mn and -factor for the doubly reinforced section shown below. 12 in. 3 No. 7 18 in. 4 No. 8 d fc' = 4000 psi, fy = 60 ksi 1 = 0.85 As = 4(0.79 in.2) = 3.16 in.2 d h 2.5 in. = 15.5 in. As' = 3(0.60 in.2) = 1.80 in.2 d' 2.5 in
CEE 415 Example 3.1: For the singly reinforced section shown below, calculate Mn, As,min, and show that this is a tension-controlled section. 12 in. f'c = 4000 psi, fy = 60 ksi 1 = 0.85 18 in. 4 No. 8 d As = 4(0.79 in.2) = 3.16 in.2
Calculation of Mn; sta
CEE 415
Example 12.1: Slenderness evaluation of a column in a nonsway frame loaded only by axial loads. PD = 230 kips and PL = 140 kips (15% sustained); Pu = 500 kips u = 12 ft (assume k = 1.0, conservative) fc = 4000 psi, fy = 60 ksi. Aim for a square co
CEE 415 Example 11.2: Calculate points on the Pn vs. Mn interaction diagram for the column shown below. Use fc' = 4000 psi, fy = 60 ksi.
d1 = 2.5 in. d2 = 7.5 in. 12 No. 9 bars d3 = 12.5 in. d4 = 17.5 in.
Column is 20 x 20 and Ast = 12.0 in.2
1. Pure axia
CEE 415 Example 11.3 Design of column section (tied) for combined axial load and bending. Use an R-type arrangement of longitudinal steel and start with g = t = 0.025. Material properties: Given loads: fc' = 5000 psi, PD = 300 k PL = 200 k P W = 60 k fy =
Recommended design steps for singly-reinforced sections when beam dimensions are not known: Will assume material properties, loads, span lengths and moments are all known. Usually the beam self-weight is not known, but with some experience you should be a
Trial and error procedure to calculate c, then Mn for doubly reinforced sections: Assume s y Guess c
Calc. s'
=
c d cu c
Calc. fs' Calc. Cs' Calc. Cc Calc. T increase c T > Cc + Cs' Compare T with Cc + Cs'
= Ess' fy = As'(fs' 0.85fc) = 0.85 fc b 1 c = As