12. Stairs by U Nyi Hla ngae from Myanmar.pdf

1
Department of Civil Engineering
Chapter 12
Stairs
Ref; U Nyi Hla Nge, “Reinforced Concrete Design”.

2
Introduction
❑ Stairs consist of rises and runs (or treads) and landings.
❑ The total steps and landings are called staircase.
❑ The rise is defined as the vertical distance between two steps, and the run
is the depth of the step.
❑ The landing is the horizontal part of the stair case without rises.
❑ The normal dimensions of the rises and runs in a building are related by
some empirical rules such as:
rise + run = 17 in
2 × rise + run = 25 in
rise × run = 75 in2

3
❑ In addition to dead loads, stairs must be designed for a minimum
of 100 psf except for one- and two- family dwellings for which
40 psf is specified.
Types of Stairs
❑ There are different types of stairs, which depend mainly on the type and
function of the building and on the architectural requirements.
For public building, rise ≈ 6 in
For residential building, rise = 6 in → 7.5 in
For public building, run ≈ 1 ft (12 in)
For residential buildings, run = 9 in → 12 in
run ≯8 in and ≮4 in

4
(1) Single-flight Stairs
❑ Similar to one way slab supported at both
ends
❑ The thickness of the slab is the waist. (Fig.
12.1)
❑ When the flight of stairs contains landings, it
may be more economical to provide beams at
B and C between landings. (Fig. 12.2)
❑ If such supports are not provided, the span of
the stair case will increase by the width of
the two landings and will extend between A
and D.

5
w2 lb/ft2
w2 lb/ft2
w1 lb/ft2
Landing
Run
Rise
Waist
B B
C
C
B'
Landing
Landing
Beam
Figure 12.1. Single-flight staircase

6
Load
Landing
Run
Rise
Waist
A B
C D
A B
C D
A B
C D
A B
C D
Beam
Figure 12.2. Supporting systems of one flight of a
double-flight staircase

7
❑ In residential buildings, the landing width is in the range of 4 ft
to 6 ft, and the total distance between A and D may be about 20
ft in some cases.
❑ The support conditions of A and D may be assumed hinged or
partially fixed or fixed depending on the relative stiffness of slab
and supporting element.
❑ An alternative method of supporting a single flight of stairs is to
use stringers, or edge beams, at the two sides of the stairs; the
steps are then supported between the edge beams. (Fig. 12.3)
Load
Edge beam Edge beam
Step beam
B'
Figure 12.3. Steps supported by
stringer beams

8
(2) Double-flight Stairs
❑ It is more convenient in most buildings to build the staircase in
double-flight between floors.
❑ The types commonly used are quarter-turn (Fig. 12.4) and closed – or
open-well stairs, as shown in Fig. 12.5)
❑ For the structural analysis of the stairs, each flight is treated as a
single flight and is considered supported on two or more beams, as
shown in Fig. 12.2.

9
❑ For the structural analysis of the stairs,
each flight is treated as a single flight and
is considered supported on two or more
beams, as shown in Fig. 12.2.
❑ The landing extends in the transverse
direction between two supports and is
designed as one-way slab.
❑ In the case of open-well stairs, the middle
part of the landing carries a full load,
whereas the two end parts carry half-
loading only, as shown in Fig. 12.5(d).
❑ The other half-loading is carried in the
longitudinal direction by the slab
containing stair flights. (Fig. 12.5(e))
Quarter
landing
Figure 12.4. Quarter-turn staircase

10
B
L
A A
B B
C
C
Open well
Figure 12.5. Double-flight stairs:
(a) closed-well staircase
(b) open-well staircase

11
Beam
w/2 lb/ft2
w/2 lb/ft2
w lb/ft2
w/2 lb/ft2
w/2 lb/ft2
w1 lb/ft2
End
portion
End
portion
Middle
portion
B
(Landing)
end portion
(Landing)
end portion
(Stairs)
middle portion
L
Beam
B
L
A A
B B
C
C
Open well
w/2 lb/ft2
w/2 lb/ft2
w lb/ft2
w/2 lb/ft2
w/2 lb/ft2
w1 lb/ft2
End
portion
End
portion
Middle
portion
B
(Landing)
end portion
(Landing)
end portion
(Stairs)
middle portion
L
(c) closed-well staircase
(d) Loading along C-C
(e) Loading along A-A or B-B

12
(3) Three or more flight of Stairs
❑ In some cases, where the overall dimensions of the staircase are
limited, three or four flights may be adopted. (Fig. 12.6)
❑ Each flight will be treated separately, as in the case of double-
flight staircases.

13
Up
Up
Up
Four-flight stair Three-flight stair
Figure 12.6. Three- and four-flights staircase

14
(4) Cantilever Stairs
❑ Cantilever stairs are used mostly in fire-escape stairs, and they
are supported by concrete walls or beams.
❑ The stairsteps may be of the full-flight type, projecting from one
side of the wall, the half-flight type, projecting from both sides of
supporting wall, or of the semispiral type as shown in Fig. 12.7.
❑ In this type of stairs, each step acts as a cantilever, and the main
reinforcement is placed in the tension side of the run and the bars
are anchored within the concrete wall.
❑ Temperature and shrinkage reinforcement is provided in the
transverse direction.

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❑ Another form of a cantilever stair is that using open-riser
steps supported by a central beam, as shown in Fig. 12.8.
❑ The beam has a slope similar to the flight of stairs and
receives the steps on its horizontally prepared portions.
❑ In most cases, precast concrete steps are used, with special
provisions for anchor bolts that fix the steps into the beam.

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Wall
Beam
Wall
(a) Plan
Half-flight
(b) Plan
Semispiral
A A
(c) Section A-A (d) Plan
Figure 12.7. Steps projecting from one or two sides of the supporting wall

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Finishing Anchor bolts
Main
reinforcement
Beam
Supporting beam
A A
(a)
(b)
Figure 12.8. Precast cantilever stair supported by central beam; (a) section A-A; (b) part plan

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(5) Precast flight of Stairs
❑ The speed of construction in some project requires the use of
precast flights of stairs. (Fig. 12.8)
❑ The flights may be cast separately and then fixed to cast-in-place
landings.
❑ In other cases, the flights, including the landings, are cast and then placed
in position on their supporting walls or beams.
❑ They are designed as simply supported one-way slabs with the main
reinforcement at the bottom of the stair waist.
❑ Adequate reinforcement must be provided at the joints. (Fig. 12.9)

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❑ Provision must be made for lifting and handling the precast stair
units by providing lifting holes or inserting special lifting hooks
into the concrete.
❑ Special reinforcement must be provided at critical locations to
account for tensile stresses that will occur in the stairs from the
lifting and handling processes.
Precast stair
Landing
Figure. 12.9. Joint of a precast concrete flight of stairs

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(6) Free-standing staircase
❑ The landing projects into the air without any support at its end.
❑ The stair behave in a springboard manner, causing torsional
stresses in the slab.

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Span L
Stairs Landing
Width of stairs
Open space
Width of stairs
Cantilever
Figure 12.10. Plan of a free standing staircase

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Loading Consideration in the Design of Free-standing Stairs
❑ When the live load acts on the upper flight and half the landing
width only (Fig.12.11), the upper flight will be subjected to
tensile forces in addition to bending moments, whereas the lower
flight will be subjected to compression forces, which may cause
buckling of the slab.
❑ When the live load acts on the lower flight and half the landing width
only (Fig. 12.12), the upper flight slab will be subjected to tensile forces,
whereas the lower flight will be subjected to bending moment and
compression forces.
❑ When the live load acts on both upper and lower flights, the loading of
one flight will cause the twisting of the other. The torsional stresses
developed in the stair slabs and the landing. Transverse reinforcement in
the slab and the landing must be provided in both faces of the concrete in
the shape of closed U-bars lapping at midwidth of the stairs.

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Cantilever
Loaded steps
Compression
(buckling)
A
B
C
D
(Bending)
Loaded
landing part
A
B
D
C
Loaded upper
steps
Unloaded
landing part
Unloaded
lower steps
Cantilever
Tension
(Bending)
A
B
C
D
(Bending)
Unloaded
landing part
A
B
D
C
Unloaded
upper steps
Loaded
landing part
Loaded lower
steps
Twist
Twist
Fig. 12.11. Case 1, ABC Loaded
Fig. 12.12. Case 2, DBC Loaded

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(7) Run-riser stairs
❑ Run-riser stairs are stepped underside stairs that consist of a
number of runs and risers rigidly connected without the
provision of the normal waist slab. (Fig. 12.14(a))
❑ This type of stairs has an elegant appearance and is sometime
favoured by architects.
❑ The structural analysis of run-riser stairs can be simplified by
assuming that the effect of axial forces is negligible and that the
load on each run is concentrated at the end of the run. (Fig.
12.14(b))
❑ For the analysis of a simply supported flight of stairs, consider a
simple flight of two runs, ABC, subjected to a concentrated load
P at B′.
❑ Because joint B and B′ are rigid, the moment at joint B is equal
the moment at B′.

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(a) cross-section
Fig. 12.14. Run-riser staircase
Landing
Landing
Rise
Run
A
B
B'
C
P
Hinged or partially fixed
or fixed
P/2
P/2
S S
A
B
B' C
P/2
P/2
+
+
+
Elastic curve
(b) elastic curve of a two-run flight
(c) bending moment diagram
(hinged-support case) of a two-run
flight

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𝑀𝐵 = 𝑀𝐵′ =
𝑃𝑆
2
where, S = the width of the run
The moment in BB′ is constant and is equal to PS/2. (Fig. 12.14(c))
❑ When the rise (or riser) is absent, the stairs, ABC, act as a simply
supported beam, and the maximum bending moment occurs at midspan
with value
𝑀𝐵 =
𝑃𝐿
4
=
𝑃𝑆
2
❑ For a flight of stairs that consists of a number of runs and risers,
𝑀𝐵 = 𝑀𝐵′ = 2𝑃𝑆, 𝑀𝐶 = 𝑀𝐶′ = 3𝑃𝑆, 𝑀𝐷 = 𝑀𝐷′ = 3𝑃𝑆
𝑀𝐸 = 𝑀𝐸′ = 2𝑃𝑆

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P
P
P
P
S S S S S
A
B
B' C
C' D
D' E
E'
F
2P
2P
B
B'
2PS
C
C'
3PS
D
D'
3PS 2PS
E
E'
2PS 3PS 3PS 2PS
MB
MC MD
ME
Fig. 12.15. (a). Distribution of moments (a) bending moment due to concentrated loads

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❑ If a landing is present at one or both ends, the load on the
landing practically may be represented by concentrated loads
similar to the runs.
❑ The moment in every riser is constant and is obtained from the bending
moment diagram of a simply supported beam subjected to a uniform load.
Uniform load, w = 4P/5S
MB
MC MD
ME
Fig. 12.15. (a). Distribution of moments (a) bending moment due to uniform loads (hinged-
support case)

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❑ If the stair flight is fixed or continuous at one or both ends, the
moments can be obtained using any method of structural
analysis.
❑ From Fig. 12.16(a), the moments at the fixed ends A and C, due to a
concentrated at B′ are equal to PL/8 = PS/4.
❑ The moment at mid-span, Section B, is equal to
𝑀𝐵 =
𝑃𝐿
4
− 𝑀𝐴 =
𝑃𝑆
2
−
𝑃𝑆
4
=
𝑃𝑆
4
❑ Note that the moment in the riser BB′ is constant, and 𝑀𝐵 = 𝑀𝐵′ =
𝑃𝑆
2
❑ For a symmetrical stair flight, fixed at both ends and subjected to a
number of concentrated loads at the node of each run, the moment at the
fixed end can be calculated as follows;

30
𝑀 fixed ends =
𝑃𝑆
12
𝑛2 − 1
where, P = concentrated load at the node of the run
S = width of run
n = number of runs
When, n = 2, then
𝑀 fixed ends =
𝑃𝑆
12
4 − 1 =
𝑃𝑆
4
❑ If a landing is present at one or both ends, the load on the landing may be
represented by concentrated loads at spacing S.

31
Fig. 12.16. Fixed-end stair case: (a) loaded steps; (b) loaded beam
A
B'
C
P
P/2
P/2
S S
A
B
B' C
Elastic curve
B
L
PS/4
PS/4
PS/4
PS/4
-
+
+
+
-
-
A C
B
P
S S
L
A C
B
MA = MC
MB = PS/4
MC = PL/8 = PS/4

32
(8) Helical stairs (Open-spiral stairs)
Fig. 12.17. Helical staircase

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❑ A helical staircase is a three-dimensional structure, which
usually has a circular shape in plan.
❑ It is a distinctive type of stairs used mainly in entrance halls,
theater flyers, and special low-rise office buildings.
❑ The cost of a helical stair is much higher than that of a normal
staircase.
❑ The possible practical dimensions may be chosen as follows:
total subtended arc between 120º and 320º
stair width between 4 ft and 6 ft
stairs slab thickness between 6 in and 10 in
stair height between 10 ft and 15 ft
❑ The above information can be used as a guide to achieve a proper and
economical design of helical staircase.

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Example 12.1
Design the cantilever stairs shown in Figure to carry a uniform live
load of 100 psf. Assume the rise of the steps equals 6.0 in and the
run equals 12 in. Assume weight of topping to be 20 psf. Use
normal weight concrete with fc′ = 3 ksi and fy = 60 ksi.
6'
1'
A
A
B B

35
Solution
1. Loads: assume the thickness of the slab (waist) is 4.0 in. Weight
of the assumed slab areas (A1 and A2) is
m
n
4.9"
4.9"
6"
4"
waist
m'
n'
12"
A1
A2
trapezoidal area mnn′m′=
4.9+10.9
2×12
×1×150 = 98.75 lb/ft
Total D.L = 98.75 + 20 = 118.75 lb/ft
ωu = 1.2×118.75 + 1.6×100 = 302 lb/ft = 0.302 k/ft
2. Maximum bending moment,
Mu=
ωuL2
2
=
0.302×62
2
= 5.44 k−ft/ft
Average thickness of a step, h =
10.9+4.9
2
= 7.9 in

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d = 7.9 −0.75 −
0.25
2
= 6.9 in
As=
Mu
ϕfy d− ൗ
a
2
=
5.44×12
0.9×60× 6.9− ൗ
0.4
2
= 0.18 in2
Minimum As = 0.0018bh = 0.0018×12×7.9 =0.171 in2
Use two No.3 bars per step.
Assume a = 0.4 in
Check a =
Asfy
0.85fc
′ b
=
0.18×60
0.85×3×12
= 0.35 in ≈ aassume

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3. Check flexural shear at a distance d from the face of the support
Vu= 0.302× 6−
6.9
12
=1.64 kips
ϕVc= ϕ2λ fc
′ bd = 0.75×2×1× 3000×12×6.9×
1
1000
= 6.8 kips
1
2
ϕVc=
1
2
×6.8 = 3.4 kips
Vu<
1
2
ϕVc, no shear reinforcement is required.
But it is recommended to use No.3 stirrups spaced at 4 in to hold
the main reinforcement.

38
4. The stairs must maintain equilibrium either by the weight of the
wall or by a reinforced concrete beam within the wall. In this
case, the beam will be subjected to torsional moment of 5.44 k-
ft/ft.
4"
Rise 6"
Run 12"
2-No.3 per step
No.3 @ 4"
No.3 @ 15"
4.9"
6"
6'
1'
2 No.3/step
Section A-A Section B-B

39
Example 12.2
Design the staircase shown in Figure, which carries a uniform live
load of 120 psf. Assume a rise of 7.0 in and a run of 10.75 in.
Assume weight of topping to be 7.5 psf. Use fc′ = 3 ksi and fy = 60
ksi.
5' 5'
7' 2"
17' 2"
5'
5'
2'
12'
Up
Up
B1
B2

40
Solution
1. Structural system
When no intermediate supports are used, the flight of stairs
will be supported at the ends of the upper and lower landings.
2. Loads
Minimum thickness of slab waist =
L
28
=
206
28
= 7.35 in
Use 8 in slab (waist) thickness.
Wt. of one step = trapezoidal area × 150 pcf
m
n
9.5"
9.5"
7"
8"
waist
m'
n'
10.75"
A1
A2
θ=33.07º
=
9.5+16.5
2×12
×
10.75
12
×150 = 145.6 lb/ft

41
Average wt. per foot length=145.6×
12
10.75
=162.5 Τ
lb ft Τ
length ft width
Weight of 8 in landing=
8
12
×150 = 100 Τ
lb ft Τ
length ft width
Total D.L on the stair = 162.5+7.5 = 170 lb/ft
Total D.L on landing = 100 + 7.5 = 107.5 lb/ft
ωu (on stairs) = 1.2×170 + 1.6×120 = 400 lb/ft = 0.4 k/ft
ωu (on landing) = 1.2×107.5 + 1.6×120 = 320 lb/ft = 0.32 k/ft
Because the load on the landing is carried into two directions, only half the
load will be considered in each direction.

42
3. Calculate the maximum bending moment and reinforcement
5' 7' 2" 5'
17' 2"
160 lb/ft 160 lb/ft
400 lb/ft
2.23 kips
2.23 kips
Mu= 2.23×
17.2
2
− 0.16×5 ×6.1−0.4×
3.62
2
=11.71 k−ft
The moment at midspan is
d = 8 −0.75 −
0.25
2
= 7.0 in
Assume a = 0.8 in
As=
Mu
ϕfy d− ൗ
a
2
=
11.71×12
0.9×60× 7− ൗ
0.8
2
As= 0.4 in2

43
Check a =
Asfy
0.85fc
′ b
=
0.4×60
0.85×3×12
Minimum As = 0.0018bh = 0.0018×12×8 =0.18 in2 < 0.4 in2
Use No.4 bars @ 6 in. (As = 0.4 in2)
Also provide No.4 bars @ 12 in (temperature and shrinkage steel) at the top
of the slab in the region of the ends as negative steel, which must be well
anchored in the beam.
To Check ϕ, c =
a
β1
=
0.8
0.85
= 0.94 in
Net tensile strain, ∈t = ∈u
dt−c
c
=0.003×
7−0.94
0.92
= 0.0193 > 0.005;
∴ ϕ = 0.9 is satisfactory.

44
Transverse reinforcement must be provided to account for
temperature and shrinkage.
As = 0.0018×12×8 = 0.18 in2/ft
Use No.4 bars @ 12 in. (As = 0.2 in2)
4. Check minimum slab thickness
For simply−supported case, h =
L
20
=
17.2×12
20
= 10.32 in > 8 in
For the case presented here, where the slab ends are cast with supporting
beams and additional negative reinforcement is provided, minimum
thickness may be assumed to be both-end continuous case.
h =
L
28
=
17.2×12
28
= 7.4 in < 8 in

45
5. Design of landings
160 lb/ft 160 lb/ft
320 lb/ft
5' 2' 5'
12'
0.96 kips 0.96 kips
Mu= 0.96×
12
2
− 0.16×5 ×3.5 − 0.32×
12
2
= 2.8 k−ft
The moment at midspan is
Because the bars in the landing will be placed on the top of the main
reinforcement,
d = 8.0 – 0.75 – 0.5 – 0.25 = 6.375 in; Say 6.3 in

46
As=
Mu
ϕfy d− ൗ
a
2
=
2.8×12
0.9×60× 6.3− ൗ
0.4
2
= 0.102 in2
Assume a = 0.4 in
Minimum As = 0.0018bh = 0.0018×12×8 = 0.18 in2 > 0.102 in2
Use As = 0.18 in2. Use No.4 bars @ 12 in. (As = 0.20 in2)
Also provide negative steel of No.4 bars @ 12 in for the slab in the
transverse direction.
6. The transverse beam at the landing levels must be designed to carry load
from stairs (2.23 k/ft) in addition to their own weight and the weight of the
wall above.
7. Check shear
Vu= 0.96 − 0.16×
6.3
12
= 0.876 kips

47
ϕVc= ϕ2λ fc
′ bd = 0.75×2×1× 3000×12×6.3×
1
1000
= 6.211 kips
ϕVc > Vu;(OK)

48
Example 12.3
Design the run-riser stair shown in Figure for a uniform live load
of 120 psf. Use fc′ = 3 ksi and fy = 60 ksi. Assume topping weight
to be 12 psf.
5'
5'
2'
13' 4"
40" 8×10 = 80" 40"
13' 4"
9×7 = 63" = 5' 3"
A
B
C

49
Solution
1. Loads
Assume the thickness of runs and risers is 6 in.
D.L per foot width of run, 10"
6"
6"
1"
PD =
16
12
×
6
12
+
1
12
×
6
12
150+
10
12
×12 =116 lbs
D.L on the landing portion,
PD =
10
12
×
6
12
×150 +
10
12
×12 = 72.5 lbs
L.L per foot width of run,
PL =
10
12
×120 = 100 lbs

50
Factored load, Pu = 1.2PD + 1.6PL
Pu1 = 1.2(116) + 1.6(100) = 299 lbs = 0.299 kips (on stairs)
Pu2 = 1.2(72.5) + 1.6(100) = 247 lbs = 0.247 kips (on landing)
RA= RC =
9Pu1
2
+
3Pu2
2
=
9×0.299
2
+
3×0.247
2
= 1.716 kips
40" 8×10 = 80" 40"
13' 4"
9×7 = 63" = 5' 3"
A
B
C
Pu2
2
Pu2
2
Pu2
2
Pu1
Pu1
Pu1
Pu1
RA = 1.716 kips
RC = 1.716 kips
1 2 3 4
5
6
7
8
1
2
3
4
5
6
7

51
2. Calculate the bending moment at midspan
Mu at B = 1.716×80 −
0.247
2
50+60+70 − 0.299 10+20+30+40
Mu at B = 85.15 kip−in
If the ends are assumed to be partially fixed, the midspan moment
will be reduced, and some negative moment will exist at the ends.
2. Calculate the reinforcement required at midspan section
h = 6 in, d = 6 – 1 = 5.0 in
Assume a = 0.7 in
As=
Mu
ϕfy d− ൗ
a
2
=
85.15
0.9×60× 5.0 − ൗ
0.7
2
= 0.34 in2
Check a =
Asfy
0.85fc
′ b
=
0.34×60
0.85×3×12

52
To Check ϕ, c =
a
β1
=
0.7
0.85
= 0.82 in
Net tensile strain, ∈t = ∈u
dt−c
c
= 0.003×
5−0.82
0.82
= 0.01538 > 0.005;
∴ ϕ = 0.9 is satisfactory.
Use No.4 bars @ 6 in. (As = 0.39 in2) horizontally and vertically in
closed stirrups form.
For the distribution steel, As(min) = 0.0018×12×6 = 0.18 in2 per step.
Use No.3 bars spaced @ 6 in (As = 0.22 in2).
For each step corner, use three No.3 bars (As = 0.33 in2).

53
4. The moment and reinforcement required at other sections can
also be determined. For example, at point 4,
Mu = 1.716×40 −
0.247
2
10+20+30 = 61.23 k−in
Assume a = 0.5 in
As=
Mu
ϕfy d− ൗ
a
2
=
61.23
0.9×60× 5.0 − ൗ
0.5
2
= 0.24 in2
Check a =
Asfy
0.85fc
′ b
=
0.24×60
0.85×3×12
Use No.4 bars @ 8 in (As = 0.29 in2) for the landing and No.4 bars @ 6 in
(As = 0.390 in2 for the steps.
For the distribution steel, As(min) = 0.0018×12×6 = 0.13 in2.
Use No.3 bars spaced @ 8 in (As = 0.166 in2).

54
5. Check reinforcement required in the transverse direction of
landing, d = 4.5 in
Load per square foot on the landing is
247
10×2
×12 = 148 psf
Mu=
0.148×122
8
×12 = 32 kip−in
Assume a = 0.3 in
As=
Mu
ϕfy d− ൗ
a
2
=
32
0.9×60× 4.5 − ൗ
0.3
2
= 0.136 in2
Check a =
Asfy
0.85fc
′ b
=
0.136×60
0.85×3×12
As(min) = 0.0018×12×6 = 0.13 in2 /ft.

55
Use No.3 @ 8 in (As = 0.166 in2).
Also provide No.3 @ 8 in as negative steel in the transverse
direction.
6. If aa uniform load is assumed to be acting on the flight of stairs instead of
concentrated loads, comparable results will be obtained. For example,
ultimate node load was calculated to be 299 lb acting over 10 in run.
Therefore, load per foot is
299
10×2
×12 = 358.8 lb/ft over the stair portion and
247
10×2
×12 = 148 lb/ft over the landing portion

56
40" 8×10 = 80" 40"
148 lb/ft
358.8 lb/ft
148 lb/ft
R = 1689.3 lb R = 1689.3 lb
Mu= 1689.3×80−148×
40
12
×60−358.8×
40
12
×80 ×
1
1000
= 81.6 k−in
Therefore, approximate maximum moment at midspan is
(close to 85.15 k-in obtained in Step 2)
Moment at other sections can be easily be found and the design performed
in Step 4.

12. Stairs by U Nyi Hla ngae from Myanmar.pdf

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