1–15. A hospital located in Chicago, Illinois, has a flat roof,
where the ground snow load is 25 lb
>
ft2. Determine the design
snow load on the roof of the hospital.
SOLUTION
Ce=1.2
Ct=1.0
I=1.2
pf=0.7
CeCt
Ipg
pf=0.7(1.2)(1.0)(1.2)(25) =25.2
lb>ft2
Ans.
1–17. Wind blows on the side of the fully enclosed hospital
located on open flat terrain in Arizona. Determine the external
pressure acting on the leeward wall, if the length and width of
the building are 200 ft and the height is 30 ft.
SOLUTION
V=120
mi>h
Kzt =1.0
Kd=1.0
Ke=1.0
qh=0.00256KzKztKdKeV2
=0.00256Kz(1.0)(1.0)(1.0)(120)2
=36.864Kz
From Table 1–5, for
z=h=30
ft,
Kz=0.98
qh=36.864(0.98) =36.13
From the text,
L
B
=
200
200
=1
so
that
Cp=–0.
5
p=qGCp–qh(GCpi)
p=36.13(0.85)(–0.5) –36.13(|0.18)
p=–21.9
psf
or –8.85
psf
Ans.
1 8
Ans.
p=–9.96
psf
or
–18.6
psf
SOLUTION
V=105
mi>h
Kzt =1.0
Kd=1.0
Ke=1.0
qz=0.00256KzKztKdKeV2
=0.00256Kz(1.0)(1.0)(1.0)(105)2
=28.22Kz
From Table 1–5,
For
0…z…15
ft,
Kz=0.85
Thus,
qz=28.22(0.85) =23.99
p=qGCp–qh(GCpi)
p=23.99(0.85)(0.7) –(23.99)({0.18)
p=–9.96
psf
or
p=–18.6
psf
Ans.
1–18. The light metal storage building is on open flat terrain in
central Oklahoma. If the side wall of the building is 14 ft high,
what are the two values of the design wind pressure acting on
this wall when the wind blows on the back of the building? The
roof is essentially flat and the building is fully enclosed.
8 ft
8 ft
32 ft
sign if
qh=25.5
lb>ft2
. The sign has a width of 32 ft and a
height of 8 ft as indicated.
SOLUTION
ft>8
2 0
Ans.
pf=0.806
kN>m2
2 1
1–21. The stall has a flat roof with a slope of 40 mm
>
m.
It is located in an open field where the ground snow load is
0.84kN
>
m2. Determine the snow load that is required to design
the roof of the stall.
Ans.
pf=0.452
kN>m2
SOLUTION
Here, the slope of the
r
oof =
a40
mm
1000
mm b
*10%
=4% 65%.
Then the roof can be considered at. Since the
barn is located in an open terrain, is unheated and is an agri-
cultural building,
Ce=0.8,
Ct=1.2
and
Is=0.8
, respectively.
Here,
pg=0.84
kN>m2.
pf=0.7CeCtIspg
=0.7(0.8)(1.2)(0.8)(0.84
kN>m2)
=0.4516
kN>m2=0.452
kN>m2
Ans.
2 3
1–23. The school building has a flat roof. It is located in an
open area where the ground snow load is 0.68 kN
>
m2.
Determine the snow load that is required to design the roof.
Ans.
pf=0.457
kN>m2
SOLUTION
pf=0.7CeCtIspg
pf=0.7(0.8)(1.0)(1.20)(0.68)
=0.457
kN>m2
Ans.
2 4
*1–24. Wind blows on the side of the fully enclosed
agriculture building located on open flat terrain in Oklahoma.
Determine the external pressure acting over the windward
wall, the leeward wall, and the side walls. Also, what is the
internal pressure in the building which acts on the walls? Use
linear interpolation to determine qh. 100 ft
50 ft
wind
A
B
C
D
15 ft108
SOLUTION
qz=0.00256Kz
Kzt
Kd Ke
V2
qz=0.00256Kz
(1)(1)(1)(105)2
q15 =0.00256(0.85)(1)(1)(1)(105)2=23.9904
psf
q20 =0.00256(0.90)(1)(1)(1)(105)2=25.4016
psf
h
=15 +
1
2
(25
tan
10°)=17.204
ft
q
h
–23.9904
17.204 –15
=
25.4016 –23.9904
20 –15
qh=24.612
psf
External pressure on windward wall:
pmax =qz
GCp=23.990410.85210.82=16.3 psf
Ans.
External pressure on leeward wall:
L
B
=
50
100
=0.
5
p=qh GCp=24.612(0.85)(–0.5) =–10.5
psf
Ans.
External pressure on side walls:
p=qh GCp=24.61210.85)1–0.72=–14.6
psf
Ans.
Internal pressure:
p=–qh(GCpi)=–24.612(0.18) ={4.43
psf
Ans.
Ans.
External pressure on windward wall
pmax =16.3
psf
External pressure on leeward wall
p=–10.5
psf
External pressure on side walls
p=–14.6
psf
Internal pressure
p={4.43
psf
1–25. Wind blows on the side of the fully enclosed agriculture
building located on open flat terrain in Oklahoma. Determine
the external pressure acting on the roof. Also, what is the
internal pressure in the building which acts on the roof? Use
linear interpolation to determine qh and Cp in Fig. 1–13.
SOLUTION
qz=0.00256Kz
Kzt
Kd
Ke
V2
=0.00256Kz
(1)(1)(1)(105)2
q15 =0.00256(0.85)(1)(1)(1)(105)2=23.9904
psf
q20 =0.00256(0.90)(1)(1)(1)(105)2=25.4016
psf
h=15 +
1
2
(25
tan
10°)=17.204
ft
q
h
–23.9904
17.204 –15
=
25.4016 –23.9904
20 –15
qh=24.612
psf
External pressure on windward side of roof:
p=qh
GCp
h
L
=
17.204
50
=0.344
1
[–0.9 –(–0.7)]
(0.5 –0.25)
=
(–0.9 –C
p
)
(0.5 –0.3441)
Cp=–0.7753
p=24.612(0.85)(–0.7753) =–16.2
psf
Ans.
External pressure on leeward side of roof:
[–0.5 –(–0.3)]
(0.5 –0.25)
=
(–0.5 –C
p
)
(0.5 –0.3441)
Cp=–0.3753
p=qh
GCp
=24.612(0.85)(–0.3753) =–7.85
psf
Ans.
Internal pressure:
p=–qh(GCpi)=–24.612({0.18) ={4.43
psf
Ans.
100 ft
50 ft
wind
A
B
C
D
15 ft108