This document provides the design calculations for a reinforced concrete staircase. It includes: 1. Preliminary sizing and load calculations for the inclined ramp and landing sections. Applied loads include self-weight, finishes weight, and live loads. 2. Structural analysis to calculate bending moments and shear forces. 3. Design of the longitudinal rebar, including positive moment, negative moment, and temperature rebar. Rebar sizes, amounts, and spacings are calculated. 4. Check of shear capacity to ensure the concrete can resist the shear forces without rebar. So in summary, this document performs the structural design of a reinforced concrete staircase under gravity and temperature loads. It sizes the rebar

1. ING. ERLY MARVIN ENRIQUEZ QUISPE Pág. 1
DISEÑO SÍSMICO DE ESTRUCTURAS DE
CONCRETO ARMADO
SEDE TACNA1.1 DISEÑO DE ESCALERA
P =
α =
β =
e = m
0.9
0.18 m
2.0
Ln = 2.50 m
0.25 m
1.00 m 0.150.25 m 1.50 m
CP =
1. PREDIMENSIONAMIENTO
2. METRADO DE CARGAS
a) DEL TRAMO INCLINADO:
CosØ = = 0.82
P
=
0.25
CosØ = = 0.82
P
=
0.25
0.25
( P2
+ CP2
)1/2
0.175( .2
+ .2
)1/2
CP
2
Peso Propio de la Escalera: x x =
Peso de Acabados: x =
=
Sobrecarga: x =
=
x 1.7 x =
=
b) DEL DESCANSO:
Peso Propio del descanso: x x =
Peso de Acabados: x =
=
Sobrecarga: x =
=
x 1.7 x =
=
t
CosØ
1.00
CosØ =
WU2 = 1,4WD + 1,7WL = 1.4
WU1 =
1.00 100
h =
= 0.82
2
=
0.13
0.8192
=
hm = h + = 0.16
0.175
100 Kg/m
1,4WD + 1,7WL = 1.4
2400
=
0.25
= 0.25 m
0.16 m
+
+
WU1
( P2
+ CP2
)1/2
312 Kg/m
1.00 100 100 Kg/m
0.13 1.00
0.175
1.647 Tn/m
691 Kg/mWD
591 Kg/m0.25 1.00 2400
400 Kg/m
1647 Kg/m
400 400 Kg/m
WL
691 400
1257 Kg/m
WD 412 Kg/m
1.00 400 400 Kg/m
WU2 1.257 Tn/m
WL 400 Kg/m
400412 +

2. ING. ERLY MARVIN ENRIQUEZ QUISPE Pág. 2
DISEÑO SÍSMICO DE ESTRUCTURAS DE
CONCRETO ARMADO
SEDE TACNA3. ANÁLISIS ESTRUCTURAL
Ay By
∑ MB = 0
+ x -
Ay =
∑ Fy = 0
- = 0
By =
WU1 = 1.647 Tn/m
2.700 Ay = 01.2571.075 + 1.625
2
1.075
x 1.075
WU2 = 1.257 Tn/m
L1 = 1.075 mL1 = 1.63 m
1.888 Tn
1.647 x 1.625 1.257
2
1.075 x
2.140 Tn
1.647 x 1.625 x
2.140 + By -
CHEQUEO AL CORTE:
d = = =
Vud = = - + ) =
Vc =
Vc = .1/2
x 10 .-3
=
Como: > El concreto absorve todo el corte
6.8548 Tn
2.140 1.647 0.105 1.76 Tn( 0.125
0.85 ( 0.53 x ( 210 ) x 100 x
t - 2,50 cm 13.00 cm - 2.50 cm 10.50 cm
6.85 1.76 
10.50 )
0,85(0,53(f'c)1/2
bd)
Vmáx - WU1(e/2+d)
4. DISEÑO DEL REFUERZO
a) REFUERZO LONGITUDINAL POSITIVO:
M+
= α M = 0.9 x
d = = =
- a/2)
As+
cm2
As+
cm2
 As cm2
db =
.2
As
Ab
Usar: Ø 1/2 ''
M+
10.00 cm
0,9fy(d - a/2)
1.273.1416
=
0.2 m@
Nº de varillas en
=S
b
As mín = 0.0018 b t = 0.0018 x 100 x
=
x 1000,85 x 210
=
0,85 f'c b
As+
fya = x 4200
1/2 pulg
As+ =
1.39 = 1.251 Tn - m
t - 2,50 cm 3.00 cm13.00 cm -
=
x 4200 (10.000.9
1.251 x 100000
( 0.500 =
13 = 2.34 = 3.5
x 2.54)
4
= 3.5
Ab =
2.72=
3.45
1.27
= 36.71 ≈ 20 cm=
100
2.72
cm2
Nº de varillas en b
π db2
4
=

3. ING. ERLY MARVIN ENRIQUEZ QUISPE Pág. 3
DISEÑO SÍSMICO DE ESTRUCTURAS DE
CONCRETO ARMADO
SEDE TACNA
b) REFUERZO LONGITUDINAL NEGATIVO:
M+
β
- a/2)
As-
cm2
As-
cm2
 As cm2
db =
.2
As
Ab
Usar: Ø
c) REFUERZO POR TEMPERATURA:
As cm2
db =
.2
As
Ab
Usar: Ø
1.251
2
= 1.69
a = As-
fy =
0,85 f'c b 0,85 x 210 x 100
As-
= M-
= 0.626 x 100000
0,9fy(d - a/2) 0.9 x 4200 (10.00
M-
= =
x 4200
= 2.34
0.626 Tn - m=
1.27 cm2
4 4
As mín 0.0018 x 100 x 13 = 2.34= 0.0018 b t =
1/2 pulg
Ab = π db2
= 3.1416 ( 0.500 x 2.54) =
20 cm
Nº de varillas en 1.85
Nº de varillas en b = =
2.34
= 1.85
1.27
1/2 '' @ 0.20 m
= 0.0018 b t = 0.0018 x 100 x 13 = 2.34
S =
b
=
100
= 54.14 ≈
3/8 pulg
Ab = π db2
= 3.1416 ( 0.375 x 2.54) =
30.45 ≈ 25 cm
Nº de varillas en 3.28
0.71 cm2
4 4
Nº de varillas en b = =
2.34
= 3.28
0.71
3/8 '' @ 0.25 m
S =
b
=
100
=