Design of Concrete Structures Compilation
Principles of Reinforced/Prestressed Concrete
CE 416
In Partial Fulfillment of the Requirement for
Finals Performance Innovative Task
Bachelor of Science in Civil Engineering
Submitted By Rogie Zorion J. Go
BSCE_3K
Submitted to Engr. Lorrizza Zafra
June 16,2023
Design of Singly-Reinforced Beam
Given: b= 250 mm
DL = 20 KN.m
d = 350 mm
LL = 35 KN.m
d’ = 60 mm
fc’ = 20.7 MPa
fy = 275 Mpa
Solution:
SOLVE FOR Rn
Mu = 1.2DL + 1.6LL = 1.2(20) + 1.6(35) = 80 KN.m
Rn =
Mu
Φ bd2
= 80 x 106
0.9(250)(350)2
Rn = 2.9025
SOLVE FOR p, pmax, pmin
p=
0.85𝑓𝑐 ′
𝑓𝑦
0.85(20.7)
275
2(2.9025)
(1 − √1 − 0.85(20.7))
p = 0.0116
pmax =
2𝑅𝑛
(1 − √1 − 0.85𝑓𝑐 ′ )
0.85𝑓𝑐′ 3
𝛽1
𝑓𝑦
7
0.85(20.7)
3
(0.85)
275
7
pmax = 0.0233
pmin =
1.4
𝑓𝑦
1.4
275
pmin = 0.0051
𝐴𝑠
p = 𝑏𝑑
0.0116 =
𝐴𝑠
(250 𝑥 350)
As = 1015.65 mm2
SOLVE FOR a and c
C=T
0.85𝑓𝑐 ′ 𝑎𝑏 = 𝐴𝑠𝑓𝑦
0.85(20.7)(a)(250) = (1015.65)(275)
a = 63.5
c=
𝑎
𝛽1
63.5
0.85
c = 74.7
CHECK IF TENSION-CONTROLLED
𝑓𝑠 =
600(𝑑 − 𝑐)
𝑐
𝑓𝑠 =
600(350 − 74.7)
74.7
𝑓𝑠 = 0.011 > 0.005 OK!
IF OKAY, COMPUTE NUMBER OF BARS
No. of bars =
𝐴𝑠
𝐴𝑏
1015.65
202 𝜋
4
No. of bars = 3.23 = 4 bars
Design of One-Way Slab
Given: LL = 3 kpa
L = 2.7m simple span
Main bar = 12mm dia.
fc’ = 27.6 Mpa
1m strip
Temp. bar = 10mm dia.
fy = 276 Mpa
ɤ = 24.5 kN/m3
SOLVE FOR MINIMUM SLAB THICKNESS
𝑙
ℎ𝑚𝑖𝑛 = 20 (0.4 +
𝑓𝑦
) =
700
2700
20
(0.4 +
276
) = 107.29 = 110 mm
700
SOLVE FOR EFFECTIVE DEPTH
d = 110 – 20 – ½(12) = 84 mm
SOLVE FOR Mu (consider 1m strip)
Wdl = 24.5(1)(0.12) = 2.88 kN/m
Wll = 3(1) = 3 kN/m
Wu = 1.2(2.88) + 1.6(3) = 8.26. kN/m
Mu = WL2/8 = (8.26 x 2.72)/8 = 7.53 kN.m
CHECK SLAB THICKNESS IF ACCEPTABLE FOR THE MAXIMUM MOMENT
Assume ∅ = 0.90
𝑀𝑢 = ∅
51
140
𝛽1 𝑓𝑐′𝑏𝑑 2 𝑟𝑒𝑞(1 −
7.53 𝑥 106 = 0.9
dreq= 34.60 mm
3
𝛽)
14 1
=
51
3
0.85(27.6)(1000)𝑑 2 𝑟𝑒𝑞(1 − 0.85)
140
14
dreq < d, OK!
COMPUTE FOR REQUIRED STEEL AREA
Find Rn: 𝑀𝑢 = ∅𝑅𝑛𝑏𝑑 2
7.53 𝑥 106 = 0.9𝑅𝑛(1000)(84)2
Rn = 1.186 Mpa
Solve p:
0.85𝑓𝑐 ′
𝑓𝑦
2𝑅𝑛
(1 − √1 − 0.85𝑓𝑐 ′ ) =
0.85(27.6)
2(1.186)
(1 − √1 −
)
276
0.85(27.6)
p = 0.00441
Solve As: As = pbd = (0.00441)(1000)(84) = 370.44 mm 2
Asmin = pminbd = 0.002 (1000)(110) = 220 mm2
SOLVE FOR THE REQUIRED MAIN BAR SPACING USING 12MM DIA. MAIN BARS
𝑆=
𝐴𝐵𝐴𝑅 1000
𝐴𝑆
=
𝜋 2
12 (1000)
4
370.44
= 305.31 = 310 mm
a. 3h = 3(110) = 330 mm
b. 310 mm
c. 450 mm
Use 12mm dia. Main bar spaced at 310 mm on center
COMPUTE THE REQ. TEMP. BAR SPACING
Asmin = 220 mm2
𝑆=
𝐴𝐵𝐴𝑅 1000
𝐴𝑆𝑚𝑖𝑛
=
𝜋 2
10 (1000)
4
220
= 357 mm = 360 mm
a. 360 mm
b. 5h = 5(110) = 550 mm
c. 450 mm
Use 10mm dia. Temp. bars spaced at 360 mm on center
Design of Short-Column
Given: LL = 600
Fy = 276
DL = 400
Main bar = 12 mm dia.
fc’ = 27.6
ties = 10 mm dia.
SOLVING FOR LOAD COMBINATION:
Pu = 1.2Pdl + 1.6Pll = 1.2(400) + 1.6(600) = 1440 kN
AXIAL LOAD CAPACITY OF A TIED COLUMN