Designing a Rectangular Beam
In order to understand the steps of how to solve this numerical problem,
check out the document on the Design Procedure of Rectangular beams.
Q1: Design a beam whose width is 9 inches and bear a bending
moment of 650,000 lbin. Take fc’ = 2,500 Psi and fy = 60,000 Psi
Given Data:
b = 9 in
MU = 650000 lbin
fc’ = 2500 Psi
fy = 60000 Psi
Solution:
As we already have bending moment, so
dmin = √MU/0.205fc’b = √650000/(0.205*2500*9) = 11.87 in ≈
12 in
Overall depth = dmin + c.c = 12 + 1.5 = 13.5 in
Direct Method:
W = (0.85)fc’/fy = (0.85)(2500/60000) = 0.0354
R = MU/bd2 = 650000/(9)(12)2 = 501.54 lb/in2
ρ = w[1-√1 - 2.614R/fc’] = 0.0354[1 - √1 - 2.614(501.54/2500)]
ρ = 0.010987178
AS = ρbd = (0.010987178)(9)(12) = 1.18 in2
Trial Method:
ASmin = 200/fy (b*d) = (200/60000)(9*12) = 0.36 in2
ASmax = 0.375(0.85)2(2500/60000)(9*12) = 1.21 in2
Trial 01
Assume a = d/3 = 12/3 = 4 in
AS = MU/0.9fy(d - a/2) = 650000/0.9(60000)(12 - 4/2) =1.203 in2
Trial 02
a = ASfy/0.85fc’b = (1.203)(60000)/0.85(2500)(9) = 3.77 in
AS = MU/0.9fy(d - a/2) = 650000/0.9(60000)(12 - 3.77/2) = 1.19 in2
Trial 03
a = ASfy/0.85fc’b = (1.19)(60000)/0.85(2500)(9) = 3.73 in
, AS = MU/0.9fy(d - a/2) = 650000/0.9(60000)(12 - 3.73/2) = 1.18 in2
As trial 03 is less than ASmax and greater than ASmin and is nearly equal
to the Trial 02, So, AS = 1.18 in2 (OK)
________________________________________________
Q2: Design a beam whose width is 10 inches and bear a bending
moment of 650,000 lbin. Take fc’ = 3000 Psi and fy = 60,000 Psi
Given Data:
b = 10 in
MU = 650000 lbin
fc’ = 3000 Psi
fy = 60000 Psi
Solution:
As we already have bending moment, so
dmin = √MU/0.205fc’b = √650000/(0.205*3000*10) = 10.28 in ≈
11 in
Overall depth = dmin + c.c = 11 + 1.5 = 12.5 in
Direct Method:
W = (0.85)fc’/fy = (0.85)(3000/60000) = 0.0425
R = MU/bd2 = 650000/(10)(11)2 = 537.19 lb/in2
ρ = w[1-√1 - 2.614R/fc’] = 0.0425[1 - √1 - 2.614(537.19/3000)]
ρ = 0.0115
AS = ρbd = (0.0115)(10)(11) = 1.26 in2
Trial Method:
ASmin = 200/fy (b*d) = (200/60000)(10*11) = 0.3667 in2
ASmax = 0.375(0.85)2(2500/60000)(10*11) = 1.49 in2
Trial 01
Assume a = d/3 = 11/3 = 3.667 in
AS = MU/0.9fy(d - a/2) = 650000/0.9(60000)(11 - 3.667/2) =1.3131 in2
Trial 02
a = ASfy/0.85fc’b = (1.3131)(60000)/0.85(3000)(10) = 3.089 in
AS = MU/0.9fy(d - a/2) = 650000/0.9(60000)(11 - 3.089/2) = 1.27 in2
In order to understand the steps of how to solve this numerical problem,
check out the document on the Design Procedure of Rectangular beams.
Q1: Design a beam whose width is 9 inches and bear a bending
moment of 650,000 lbin. Take fc’ = 2,500 Psi and fy = 60,000 Psi
Given Data:
b = 9 in
MU = 650000 lbin
fc’ = 2500 Psi
fy = 60000 Psi
Solution:
As we already have bending moment, so
dmin = √MU/0.205fc’b = √650000/(0.205*2500*9) = 11.87 in ≈
12 in
Overall depth = dmin + c.c = 12 + 1.5 = 13.5 in
Direct Method:
W = (0.85)fc’/fy = (0.85)(2500/60000) = 0.0354
R = MU/bd2 = 650000/(9)(12)2 = 501.54 lb/in2
ρ = w[1-√1 - 2.614R/fc’] = 0.0354[1 - √1 - 2.614(501.54/2500)]
ρ = 0.010987178
AS = ρbd = (0.010987178)(9)(12) = 1.18 in2
Trial Method:
ASmin = 200/fy (b*d) = (200/60000)(9*12) = 0.36 in2
ASmax = 0.375(0.85)2(2500/60000)(9*12) = 1.21 in2
Trial 01
Assume a = d/3 = 12/3 = 4 in
AS = MU/0.9fy(d - a/2) = 650000/0.9(60000)(12 - 4/2) =1.203 in2
Trial 02
a = ASfy/0.85fc’b = (1.203)(60000)/0.85(2500)(9) = 3.77 in
AS = MU/0.9fy(d - a/2) = 650000/0.9(60000)(12 - 3.77/2) = 1.19 in2
Trial 03
a = ASfy/0.85fc’b = (1.19)(60000)/0.85(2500)(9) = 3.73 in
, AS = MU/0.9fy(d - a/2) = 650000/0.9(60000)(12 - 3.73/2) = 1.18 in2
As trial 03 is less than ASmax and greater than ASmin and is nearly equal
to the Trial 02, So, AS = 1.18 in2 (OK)
________________________________________________
Q2: Design a beam whose width is 10 inches and bear a bending
moment of 650,000 lbin. Take fc’ = 3000 Psi and fy = 60,000 Psi
Given Data:
b = 10 in
MU = 650000 lbin
fc’ = 3000 Psi
fy = 60000 Psi
Solution:
As we already have bending moment, so
dmin = √MU/0.205fc’b = √650000/(0.205*3000*10) = 10.28 in ≈
11 in
Overall depth = dmin + c.c = 11 + 1.5 = 12.5 in
Direct Method:
W = (0.85)fc’/fy = (0.85)(3000/60000) = 0.0425
R = MU/bd2 = 650000/(10)(11)2 = 537.19 lb/in2
ρ = w[1-√1 - 2.614R/fc’] = 0.0425[1 - √1 - 2.614(537.19/3000)]
ρ = 0.0115
AS = ρbd = (0.0115)(10)(11) = 1.26 in2
Trial Method:
ASmin = 200/fy (b*d) = (200/60000)(10*11) = 0.3667 in2
ASmax = 0.375(0.85)2(2500/60000)(10*11) = 1.49 in2
Trial 01
Assume a = d/3 = 11/3 = 3.667 in
AS = MU/0.9fy(d - a/2) = 650000/0.9(60000)(11 - 3.667/2) =1.3131 in2
Trial 02
a = ASfy/0.85fc’b = (1.3131)(60000)/0.85(3000)(10) = 3.089 in
AS = MU/0.9fy(d - a/2) = 650000/0.9(60000)(11 - 3.089/2) = 1.27 in2
