EQUIPMENT DESIGN
1. A steel steel cable cable of unif unifor orm m crosscross-se sect ctio ion n of 3 cm 2 and length of 200 meters is suspended vertically. Calculate the total elongation elongation of the steel cable cable if steel weighs 1.5 kg/m. kg/m. E=2.1x10 6 kg/ cm2. SL
=
δ
E S
d δ =
E
δ
∫ 0
d δ
=
But : S =
dL ;
=
wL
L
AE
∫ 0
W A
=
wL A
dL
wL2
δ
2 AE
(1.5)() ( 200) δ = ( 2) ( 3) ( 2.1 x10 ) 2
6
= 0.004762 m
2. A steel bar 5” wide, wide, 1” thick, thick, & 12” long is is stressed stressed by a tensile tensile load of of 80,000 pounds pounds along along the Y2 axis. The modulus of elasticity elasticity is 30,000 pounds/ in . Calculate the the dimensions dimensions of the the stressed bar and the equivalent lateral lateral stress. Take Poisson’s Constant Constant k, ε x ε y
0.3
=
Length = y =12”
Width= x = 5” P=80,000 lb
Axial Stress
P=80,000 lb
Thickness=Z= 1”
=S y =
Axial Strain =εy
Total Elongation Final Lenght , L y
=
P A
=
Sy E
80,000 lb
(5)(1) in 2
=
16,000 30 x 106
=16,000 lb
in
2
=5.33 x 10 −4 (tension )
=δy =εy L y = 5.33x10 −4 (12 inches) =0.0064 inches =12 +0.0064 =12.0064 inches
f
=εx =εz =(k ) (εy ) =(0.3) 5.33x10 −4 =1.599 x 10 −4 (5) =7.995 x 10 −4 Re duction in Width =δx =εx L x =1.599 x 10 −4 =5 −7.995 x 10 −4 =4.9992005 inch Final Width =L Lateral Strain
x
f
Re duction in Thickness Final Thickness Lateral Stress
inch (compression )
=L z
f
=δz =εz L z = 1.599 x 10 −4 (1) =1.599 x 10 −4 − 4 =1 −1.599 x 10 =0.9998401inch
inch (compressio n )
=Sx =Sz =εx E = 1.599 x 10−4 30 x 106 =4,797 lb / in 2
1
3. A tapered round post, 5 meters high, 25 cm diameter at the top and 40 cm at the base, is fixed at its base. If a horizontal force of 1000 N were applied to the top of the post; a) What would be the maximum stress in the post? b) How far up from the base of the base would this maximum stress be? 25 cm 1000 Newtons
y
y x
=
x=
5 meters
P
Similar ∆ 500
x
7.5 7.5 y
o
r
= 0.015 y 500 r = x + 7.5 =0.015 y +12.5
∑ Mo =0
m c 5 . 7
(1000 ) ( y ) −( P ) ( r ) = 0 P =
1000 y
=
r
1000 y 0.015 y
m c 5 . 2 1
40 cm
+12.5
1000 y S = dS dt
P A
0.015 y +12.5
=
( 0.015 y +12.5)
π
=0 =
2
=
1000 y
( 0.015 y +12.5) 3
π
(1000 ) ( 0.015 y +12.5) 3 − 3 y ( 0.015 y +12.5) 2 ( 0.015) (π ) ( 0.015 y +12.5) 6
( 0.015 y +12.5) = 0.045 y y
= 416.67 cm
a. S max
=
10000( 416.67 )
[( 0.015)( 416.67 ) +12.5]
π
b. Height from Base
2
= 20.12 N / cm2
=500 − 416.67 = 83.33 cm
2
4. A steel tube made of No. SAE 9255 is subjected to a tensile load of 8,000 pounds. If the thickness of the tube is 10 percent of its outside diameter find the outside diameter of the tube if factor of safety is 4 based on the ultimate strength of 135,000 psi. D Factor of Safety = N
Sw = A
=
135,000 4
P Sw
=
=
Su Sw
=3,750 lb / in 2
8,000 33,750
= 0.237 in
2
π D − d ; but d = 0.8D A= 2
d
2
4
0.237
=
π D ( . D ) − 0 8 4 2
2
D = 0.916 inch
5. A horizontally installed cylindrical tank is to be used as sulfuric acid storage tank. Tank dimension being; diameter=4 m, length=5 m. What is the stored weight of sulfuric acid inside the tank when the gauge glass level reads 1.75 m? Assume the density of sulfuric acid=1.85 kg/liter.
r-h
r
θ r
h
b
Filled Area = Area of the segment = Area of the sector-Area of the triangle Area of the sector = π r 2 θ
360
Area of the triangle = b( r − h ) 2
But:
sin
θ 2
=
b
r
2
3
And θ = cos −1 r − h 2
r
Therefore: Filled Area A f
θ θ = πr 2 − ( r − h ) r sin 2 360 θ r − h = π r 2 − r sin cos−1 r 360
Volume of liquid:
Vf
r − h πr 2 2 cos −1 r − h −1 r − h −1 r − h r − r sin L cos −1 r h L cos r sin cos r h =πr 2 ( − ) = − ( − ) 180 360 r r r π( 2 ) 2 2 − 1.75 2 − 1.75 5 2 1 75 cos −1 . = − (2 ) sin cos −1 ( − ) 180 2 2 =26.43 m 3
mass of sulfuric acid
=(26.43 )(1.85 ) = 48.9 mt
6. A steel wire accurately graduated in centimeter was used to measure the depth of a deep-well. The cross-sectional area of the wire was 0.195 cm 2 and a 115 kg weight was attached to the wire to lower it into the well. According to the graduated scale, the measurement read 600 meters. The steel wire weighs 0.10 kg/m. What is the depth of the well? Assume E=2.1x10 6kg/cm2. Let: d = depth δ T = total elongation δ1 = elongation due to weight of wire δ 2 = elongation due to mass of 115 kg weight
d = 600 + δT
δT =δ1 +δ2 =
wL2 2AE
+
WL AE
2 0.1)(600 ) ( (115 )(600 ) = + 6 ( 2 )(0.195 )(2.1x10 ) (0.195 )(2.1x10 6 )
=0.04396 + 0.1685 =0.21246 m d = depth =600 + 0.21246 = 600.21246
meters
7. A storage tank is to store a compressed air whose pressure is to be maintained at 190 psig at a maximum temperature of 150 oF. The tank has an inside diameter of 10 ft. The material to be used is S-1 steel. It is desired to use a triple riveted butt joint. Determine the following: a) Plate thickness 4
b) c) d) e)
Strap thickness Rivet hole diameter Pitch Joint efficiency
= = ( ) ( ) − = = (= ) ( ) = = ( − = − =
Given P
8
R t
:
190
psi
12
48 inches
2
Pr
Se From pp Sa
60
Table
11 ,000
Pr
te
/
in
190
S
in
lb
selecting
Assume
:
plate
t
23
:
c.. d . e.
rivet
/
int
32,
t
1
2,
0.834
pp
thickness, hole p
Eff
0.
1 " ,
thickness
Pitch ,
Jo
thickn
32
Table6
b. strap
0
23
te
Plate
2
48
te
From
1 ,
11 ,000
Assume
a.
3
62
1 inc
o
3
diameter , 4
3
/
4
83.4%
8. A low carbon steel sheet 3/8” thick, lined with aluminum 1/8” thick is used to form a low pressure receiver 10 inches inside diameter. A force of 5000 lb is evenly distributed in the circumferential shell. Find the stress in both the steel and aluminum. Assume E steel=30x106 Ealuminum=10.3 x10 6 psi. 5,000 lb Aluminum 1/8” thick
Carbon steel 3/8” thick
5
/
d
inc
I.D. =10”
δA = δS PL PL = AE A AE S But : P = PA + PS 5,000 − PS P = S AE A AE S 2 2 A A = πr = π (5 + 3 / 8 + 1 / 8) 2 − (5 + 3 / 8) 2 = 4.2725 in
I.D. =10”
= πr 2 = π(5 + 3 / 8) 2 − (5) 2 = 12.223 in 2 5,000 − Ps PS = (4.2725) 10.3x10 6 (12.223)30 x10 6 PS = 4 ,464.23 lb PA =5,000 − 4,464.23 =535.77 lb AS
SS
=
SA
=
4,464.23 12.223 535.77 4.2725
=365.24 psi
=125.46 psi
9. A trader finds that he needs a storage tank for alcohol. He wants to economize. He locates a steels scrap dealer with one piece of scrap 1/8” thick rectangular steel plate measuring 5’x25’. a) What will be the diameter of the largest cylindrical tank with flat top and hemi spherical bottom that can make? b) What will be the capacity of the tank in liters? c) How many kilos of alcohol can be stored in the tank assuming that maximum alcohol temperature will be 95oF (neglect expansion of steel assume density of alcohol=0.80gm/cc).
6
Area of steel plate = Total area of cylindical tank with flat top hemispherical bottom
= ( 5 ft )( 25 ft ) =125 ft =π r + 2π r + 2π rH 125 −3π r H = 2
AS
2
2
2
2τ r
V olume of tank = V =π r 2 H +
V
= (π r
2
dV dr
)125 −3π r 2π r
2
4 π r 2 3
1
3
2 125 3 2 5 + 3 π r = 2 r − 2 π r + 3 π r = 62.5r − 6 π r 3
3
3
3
= 0 = 62.5 − 2.5π r
2
r = 2.82 ft 125 −3π ( 2.82 )
H
=
V
= ( 62 .5)( 2.82) −
2
2π ( 2.82 ) 5 6
= 2.82 ft
( 2.82 ) =117 .54 ft = 3,326 liters
π
3
3
10. A spherical carbon steel storage tank for ammonia has an inside diameter of 30ft. All joints are welded with backing strip. If the tank is to be used at a working pressure of 50 psig and at a temperature of 80 oF, estimate the required wall thickness. Assume no corrosion allowance is necessary and joint efficiency = 80%. Express your answer to the nearest 1/8 th of an inch. For S-1 carbon steel at 80 oF, pp 62, Table 3-1 Hesse and Rushton S = 11, 000 psi
t =
PR 2Se − 0.2P
Diameter=
( 50 ) ( 30/2 ) ( 12 ) ft t= = = 0.584 " =305/8 " 2Se − 0.2P 2 (11,000 ) ( 0.80 ) − 0.2 ( 11,000 ) PR
11. A vertical cylinder 8 feet high and 5 ft diameter is to sit on 4 equally spaced concrete posts. The outer edge of the concrete posts are flushed with the outer perimeter of the tank. When full, the tank and its content weighs 15,000lb.The compressive strength of the concrete mix is 1000 psi. A factor of safety of 4 is required by the building code. 7
a) Calculate the diameter of the posts. b) Calculate the minimum horizontal force to topple the tank . F
S a
=
Sc factor of safety
=P
Sa
A
; A
=
P Sa
= 1000 =250 psi
8 ft
4
=15,000 =60 in
15,000 lb
2
250
L
A post
d
= 60 =15 in = π d 2
4
2
4
= ( 4 )(15 ) = 4.37 inches
D =5 ft
π
S
D2
= ( 5) = 2S 2
2
S = 3.536 ft
Σ M = 0 o
( F ) ( h ) − (W ) ( L ) = 0
F =
W L h
3.536 2 =
(15,000 )
=
8
3,315
lb
12. A chemical engineer was commissioned to design a vertical cylindrical tank with flat bottom and a conical roof. The tank must be able to hold a maximum 4,500 m 3 of water for firefighting purposes. Ease of climbing the tank and bearing capacity allows a maximum height of 16.5 meters from the bottom of the tank up to the rim of the tank cylinder. Normal working practice dictates that the
8
normal working capacity of the tank is 90%of the total tank volume. The tank roof has a 10%incline. Suitable steel plates available for construction the tank are in 4’x 8’sheets. a) What are the complete tank dimensions? b) What is the total area of steel plate needed? c) What is the number of sheets needed ?
Gross Volume =
4,500 0.9
R r
= 5,000
θ
Setting the height , H =16.5 meers Tan θ = 0.1
V = π r 2 H =5,000 = πr 2 (16.5) r = 9.82 meters
θ = tna −1 0.1 = 5.71o
A = πr 2 +2πrH +πR 2
Cos θ =
= π(9.2) +2π(9.82)(16.5) +π(1.005) 2
A =1,627 m
=17,515 ft
2
Number of plates =
2
(9.82)
2
R =
2
17,515
(4) (8)
= 547.34 say
548
r
R r
Cos θ
r
=
=1.005 r
Cos 5.71o
plates
13. Determine the number of bags of cement and the quantities of other aggregates (m 3) required for a pyramide structure with a base of 10 meter square and a height of 12 meters. The concrete mix used is class AA (12:0.5:1.0) and the following data can be used. Specific gravity: cement=3.1, sand=2.65, gravel=2.5, Voids: cement=0.51, sand=0.38, gravel=0.33 Use 20 liters water per bag cement. Note: 1 bag cement= 40 kg Aggregate
Quantity/batch
True Volume,
True mass 3
Cement Sand Gravel Water
12 bags 0.50m3 1.00m3 20lit/bag cement
[(12)(40)]/[(1000)(3.1)] (0.50)(1-0.38) (1.00)(1-0.33) (20)(12)/1000
total V structure
W structure
m 0.1548 0.31 0.67 0.24
mt 0.48 0.8215 1.675 0.24
(0.1548)(3.1) (0.31)(2.65) (0.67)(2.5)
1.3748
=
Bh 3
(10 ) (12 )
=
3
= 400
m3
3.2165 = ( 400 ) = 935 .85 1.3748
Aggregate Cement Sand Gravel Water
3.2165
2
mt
Quantity, total (12 /1.3748)(400 3,492 bags (0.5/1.378)(400) 145.5 m 3 (1/1.378)(400) 291 m 3 (0.24/1.378)(400) 71.2 m 3
14. A continuous evaporator is operated with a given feed material under conditions in with the concentration of the product remains constant. The feed rate at the start of a cycle after tubes have been cleaned has been found to be 10,000lbs/h. After 48 hours of continuous operation, test have show that the feed rate decreases to 5000lbs/h. The reduction in capacity is due to true scale 9
formation. If the down time per cycle for emptying, evaporation be operated between cleaning in order to obtain the maximum of product per 30 days?
= ( F )( t )( N )
P
−
dF dt
F
= K t
−dF = K ∫ dt 0
∫
F o
− (F − F ) = Kt F = F − Kt o
o
Solving for K : using 1st condition; K = N =
F 0
− F
=
10,000
− 5,000
t 48 ( 30 days )( 24hr/day ) (t + 6)hr/cycle
=
= 104.17
720 t + 6
cycles
P = ( F )( t )( N )
720 = 720 F t − Kt P = (F − Kt)( t ) t + 6 t + 6 ( t + 6 )( F − 2Kt ) − ( F t − Kt ) dP = 0 = 720 dt ( t + 6 ) Kt + 12Kt − 6F = 0 2
o
o
2
o
o
2
2
o
t = t
−b ±
b2 2a
= 18.74
− 4ac
− ( 12 )( 104.17 ) ± ( 12x104.12 ) + 4( 104.17 )( 6 )(10,000 ) 2
=
2( 104.17 )
hr
15. Analyze the stresses involved in riveted lap joint shown. The diameter d is ½” ; the plate width w is 2”; the plate thickness t is ¼” and the marginal distance m is ¾”. Asssume the ultimate stresses for the plate and rivet materials: Tensile stress, ST = 60,000 psi Bearing, compressive stress, SB;C= 90,000 psi Shearing stress, SS = 45,000 psi
t= ¼” t= ¼” ¾” ½” ¾”
¾”
m
m
d
w =2”
a.
Failure of rivet by single shear :
A s
=
2
π d
4 10
F=SSπd2/4 = (45,000)π(1/2)2/4 = 8,836 lb d b.
Failure by tearing of plate: (tensile stress) F=ST(w-d) t = 60,000(2-0.5)(0.25) = 22,500 lb
w =2”
t
d
w =2”
¾” c.
Failure by shearing out of end-plate F= SS(t)(m)(2) = (45,000)(1/4)(3/4)(2) = 16,875 lb
t= ¼”
x 2 (both sides) ¾”
d. Failure by crushing of rivet or plate (bearing stress) d = ½” t= ¼” F=SC;B (d)(t)= (90,000)(1/2)(1/4)= 11,250 lb
Joint Efficiencies:
Solid Plate=100% F= (60,000)(1/4)(2) = 30,000 lb a.
Shearing of Rivet = 8,836/30,000= 29.5%
b. Tearing of Plate = 22,500/30,000 = 75% c.
Shearing out of End Plate = 16,875/30,000 = 56.3%
d. Crushing of Rivet or Plate = 11,250/30,000 = 37.5% The joint that may fail firs t= (a) the shearing of rivet
11
