SECTION 1– DESIGN FOR SIMPLE STRESSES Page 1 of 64 TENSION, COMPRESSION, SHEAR DESIGN PROBLEMS 1. The link shown, made of AISI C1045 steel, as rolled, is subjected to a tensile load of 8000 lb. Let b h 5 . 1 = . If the load is repeated but not reversed, determine the dimensions of the section with the design based on (a) ultimate strength, (b) yield strength. (c) If this link, which is 15 in. long., must not elongate more than 0.005 in., what should be the dimensions of the cross section? Problems 1 – 3. Solution: For AISI C1045 steel, as rolled (Table AT 7) ksi s u 96 = ksi s y 59 = psi E 6 10 30 × = A F s d = where lb F 8000 = bh A = but b h 5 . 1 = therefore 2 5 . 1 b A = (a) Based on ultimate strength N = factor of safety = 6 for repeated but not reversed load (Table 1.1) A F N s s u d = = 2 5 . 1 8000 6 000 , 96 b = in b 577 . 0 = say in 8 5 .
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SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 1 of 64
TENSION, COMPRESSION, SHEAR
DESIGN PROBLEMS
1. The link shown, made of AISI C1045 steel, as rolled, is subjected to a tensile load
of 8000 lb. Let bh 5.1= . If the load is repeated but not reversed, determine the
dimensions of the section with the design based on (a) ultimate strength, (b) yield
strength. (c) If this link, which is 15 in. long., must not elongate more than 0.005
in., what should be the dimensions of the cross section?
Problems 1 – 3.
Solution:
For AISI C1045 steel, as rolled (Table AT 7)
ksisu
96=
ksisy 59=
psiE 61030×=
A
Fs
d=
where
lbF 8000=
bhA =
but
bh 5.1=
therefore 25.1 bA =
(a) Based on ultimate strength
N = factor of safety = 6 for repeated but not reversed load (Table 1.1)
A
F
N
ss u
d==
25.1
8000
6
000,96
b=
inb 577.0= say in8
5.
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 2 of 64
inbh16
155.1 ==
(b) Based on yield strength
N = factor of safety = 3 for repeated but not reversed load (Table 1.1)
A
F
N
ss u
d==
25.1
8000
3
000,59
b=
inb 521.0= say in16
9.
inbh32
275.1 ==
(c) Elongation = AE
FL=δ
where,
in005.0=δ
lbF 8000=
psiE 61030×=
inL 15= 25.1 bA =
then,
AE
FL=δ
( )( )( )( )62
10305.1
158000005.0
×=
b
inb 730.0= say in4
3.
inbh8
115.1 ==
2. The same as 1 except that the material is malleable iron, ASTM A47-52, grade 35
018.
Solution:
For malleable iron, ASTM A47-52, grade 35 018(Table AT 6)
ksisu
55=
ksisy 5.36=
psiE 61025×=
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 3 of 64
A
Fs
d=
where
lbF 8000=
bhA =
but
bh 5.1=
therefore 25.1 bA =
(a) Based on ultimate strength
N = factor of safety = 6 for repeated but not reversed load (Table 1.1)
A
F
N
ss u
d==
25.1
8000
6
000,55
b=
inb 763.0= say in8
7.
inbh16
515.1 ==
(b) Based on yield strength
N = factor of safety = 3 for repeated but not reversed load (Table 1.1)
A
F
N
ss u
d==
25.1
8000
3
500,36
b=
inb 622.0= say in16
11.
inbh32
115.1 ==
(c) Elongation = AE
FL=δ
where,
in005.0=δ
lbF 8000=
psiE 61025×=
inL 15= 25.1 bA =
then,
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 4 of 64
AE
FL=δ
( )( )( )( )62
10255.1
158000005.0
×=
b
inb 8.0= say in8
7.
inbh16
515.1 ==
3. The same as 1 except that the material is gray iron, ASTM 30.
Solution:
For ASTM 30 (Table AT 6)
ksisu
30= , no ys
psiE 6105.14 ×=
Note: since there is no ys for brittle materials. Solve only for (a) and (c)
A
Fs
d=
where
lbF 8000=
bhA =
but
bh 5.1=
therefore 25.1 bA =
(a) Based on ultimate strength
N = factor of safety = 7 ~ 8 say 7.5 (Table 1.1)
A
F
N
ss u
d==
25.1
8000
5.7
000,30
b=
inb 1547.1= say in16
31 .
inbh32
2515.1 ==
(c) Elongation = AE
FL=δ
where,
in005.0=δ
lbF 8000=
psiE 6105.14 ×=
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 5 of 64
inL 15= 25.1 bA =
then,
AE
FL=δ
( )( )( )( )62
105.145.1
158000005.0
×=
b
inb 050.1= say in16
11 .
inbh32
1915.1 ==
4. A piston rod, made of AISI 3140 steel, OQT 1000 F (Fig. AF 2), is subjected to a
repeated, reversed load. The rod is for a 20-in. air compressor, where the
maximum pressure is 125 psig. Compute the diameter of the rod using a design
factor based on (a) ultimate strength, (b) yield strength.
Solution:
From Fig. AF 2 for AISI 3140, OQT 1000 F
ksisu
5.152=
ksisy 5.132=
( ) ( ) kipslbforceF 27.39270,39125204
2====
π
From Table 1.1, page 20
8=u
N
4=yN
(a) Based on ultimate strength
u
u
s
FNA =
( )( )5.152
27.398
4
2 =dπ
ind 62.1= say in8
51
(b) Based on yield strength
y
y
s
FNA =
( )( )5.132
27.394
4
2 =dπ
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 6 of 64
ind 23.1= say in4
11
5. A hollow, short compression member, of normalized cast steel (ASTM A27-58,
65 ksi), is to support a load of 1500 kips with a factor of safety of 8 based on the
ultimate strength. Determine the outside and inside diameters if io
DD 2= .
Solution:
ksisu
65=
8=u
N
kipsF 1500=
( ) ( )4
34
44
22222 iiiio
DDDDDA
πππ=−=−=
( )( )65
15008
4
32
===u
ui
s
FNDA
π
inDi 85.8= say in8
78
inDD io4
317
8
7822 =
==
6. A short compression member with io DD 2= is to support a dead load of 25 tons.
The material is to be 4130 steel, WQT 1100 F. Calculate the outside and inside
diameters on the basis of (a) yield strength, (b) ultimate strength.
Solution:
From Table AT 7 for 4130, WQT 1100 F
ksisu 127=
ksisy 114=
From Table 1.1 page 20, for dead load
4~3=uN , say 4
2~5.1=yN , say 2
Area, ( ) ( )4
34
44
22222 iiiio
DDDDDA
πππ=−=−=
kipstonsF 5025 ==
(a) Based on yield strength
( )( )114
502
4
32
===y
yi
s
FNDA
π
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 7 of 64
inDi 61.0= say in8
5
inDD io4
11
8
522 =
==
(b) Based on ultimate strength
( )( )127
504
4
32
===u
ui
s
FNDA
π
inDi 82.0= say in8
7
inDD io4
31
8
722 =
==
7. A round, steel tension member, 55 in. long, is subjected to a maximum load of
7000 lb. (a) What should be its diameter if the total elongation is not to exceed
0.030 in? (b) Choose a steel that would be suitable on the basis of yield strength if
the load is gradually applied and repeated (not reversed).
Solution:
(a) AE
FL=δ or
E
FLA
δ=
where,
lbF 7000=
inL 55=
in030.0=δ
psiE6
1030×=
( )( )( )( )6
2
1030030.0
557000
4 ×== dA
π
ind 74.0= say in4
3
(b) For gradually applied and repeated (not reversed) load
3=yN
( )( )
( )psi
A
FNs
y
y 534,47
75.04
70003
2
===π
ksisy 48≈
say C1015 normalized condition ( ksisy 48= )
8. A centrifuge has a small bucket, weighing 0.332 lb. with contents, suspended on a
manganese bronze pin (B138-A, ½ hard) at the end of a horizontal arm. If the pin
is in double shear under the action of the centrifugal force, determine the diameter
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 8 of 64
needed for 10,000 rpm of the arm. The center of gravity of the bucket is 12 in.
from the axis of rotation.
Solution:
From Table AT 3, for B138-A, ½ hard
ksisus 48=
rg
WF 2ω=
where
lbW 332.0= 22.32 fpsg =
( )sec1047
60
000,102
60
2rad
n===
ππω
inr 12=
( ) ( ) kipslbrg
WF 3.11300,1111047
2.32
332.0 22 ==== ω
From Table 1.1, page 20
4~3=N , say 4
u
u
s
FNA =
( )( )48
3.114
42
2 =
d
π for double shear
ind 774.0= say in32
25
CHECK PROBLEMS
9. The link shown is made of AISIC1020 annealed steel, with inb4
3= and
inh2
11= . (a) What force will cause breakage? (b) For a design factor of 4 based
on the ultimate strength, what is the maximum allowable load? (c) If 5.2=N
based on the yield strength, what is the allowable load?
Problem 9.
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 9 of 64
Solution:
For AISI C1020 annealed steel, from Table AT 7
ksisu 57=
ksisy 42=
(a) AsF u=
2125.1
2
11
4
3inbhA =
==
( )( ) kipsF 64125.157 ==
(b) u
u
N
AsF =
4=uN
2125.1
2
11
4
3inbhA =
==
( )( )kipsF 16
4
125.157==
(c) y
y
N
AsF =
5.2=yN
2125.1
2
11
4
3inbhA =
==
( )( )kipsF 9.18
2
125.142==
10. A ¾-in.bolt, made of cold-finished B1113, has an effective stress area of 0.334 sq.
in. and an effective grip length of 5 in. The bolt is to be loaded by tightening until
the tensile stress is 80 % of the yield strength, as determined by measuring the
total elongation. What should be the total elongation?
Solution:
E
sL=δ
from Table AT 7 for cold-finished B1113
ksisy 72=
then, ( ) ksiss y 6.57728.080.0 ===
ksipsiE 000,301030 6 =×=
( )( )in
E
sL0096.0
000,30
56.57===δ
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 10 of 64
11. A 4-lb. weight is attached by a 3/8-in. bolt to a rotating arm 14-in. from the center
of rotation. The axis of the bolts is normal to the plane in which the centrifugal
force acts and the bolt is in double shear. At what speed will the bolt shear in two
if it is made of AISI B1113, cold finish?
Solution:
From Table AT 7, psiksisus 000,6262 ==
( ) 2
2
2209.08
3
4
12 inA =
= π
Asrg
WF us== 2ω
( ) ( )( )2209.0000,62142.32
4 2 =ω
sec74.88 rad=ω
74.8860
2==
nπω
rpmn 847=
12. How many ¾-in. holes could be punched in one stroke in annealed steel plate of
AISI C1040, 3/16-in. thick, by a force of 60 tons?
Solution:
For AISI C1040, from Figure AF 1
ksisu 80=
( ) ksiksiss uus 608075.075.0 ===
24418.0
16
3
4
3intdA =
== ππ
kipstonsF 12060 ==
n = number of holes
( )( )5
604415.0
120===
usAs
Fn holes
13. What is the length of a bearing for a 4-in. shaft if the load on the bearing is 6400
lb. and the allowable bearing pressure is 200 psi of the projected area?
Solution:
WpDL =
where
psip 200=
inD 4=
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 11 of 64
lbW 6400=
( )( ) 64004200 =L
inL 8=
BENDING STRESSES
DESIGN PROBLEMS
14. A lever keyed to a shaft is inL 15= long and has a rectangular cross section of
th 3= . A 2000-lb load is gradually applied and reversed at the end as shown; the
material is AISI C1020, as rolled. Design for both ultimate and yield strengths. (a)
What should be the dimensions of a section at ina 13= ? (b) at inb 4= ? (c) What
should be the size where the load is applied?
Problem 14.
Solution:
For AISI C1020, as rolled, Table AT 7
ksisu 65=
ksisy 49=
Design factors for gradually applied and reversed load
8=uN
4=yN
12
3thI = , moment of inertial
but th 3=
36
4hI =
Moment Diagram (Load Upward)
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 12 of 64
Based on ultimate strength
u
u
N
ss =
(a) I
Fac
I
Mcs ==
2
hc =
kipslbsF 22000 ==
( )( )
==
36
2132
8
654h
h
s
inh 86.3=
inh
t 29.13
86.3
3===
say
ininh2
145.4 ==
inint2
115.1 ==
(b) I
Fbc
I
Mcs ==
2
hc =
kipslbsF 22000 ==
( )( )
==
36
242
8
654h
h
s
inh 61.2=
inh
t 87.03
61.2
3===
say
inh 3=
int 1=
(c)
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 13 of 64
413
35.4
4
3
−
−=
− h
inh 33.2=
413
15.1
4
1
−
−=
− t
int 78.0=
say
inh 625.2= or inh8
52=
15. A simple beam 54 in. long with a load of 4 kips at the center is made of cast steel,
SAE 080. The cross section is rectangular (let bh 3≈ ). (a) Determine the
dimensions for 3=N based on the yield strength. (b) Compute the maximum
deflection for these dimensions. (c) What size may be used if the maximum
deflection is not to exceed 0.03 in.?
Solution:
For cast steel, SAE 080 (Table AT 6)
ksisy 40=
psiE 61030×=
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 14 of 64
From Table AT 2
Max. ( )( )
inkipsFL
M −=== 544
544
4
12
3bhI =
but bh 3=
36
4hI =
(a) I
Mc
N
ss
y
y==
2
hc =
( )
=
36
254
3
404h
h
inh 18.4=
inh
b 39.13
18.4
3===
say inh2
14= , inin
hb
2
115.1
3
5.4
3====
(b) ( )( )
( ) ( )( )in
EI
FL0384.0
12
5.45.1103048
544000
48 3
6
33
=
×
==δ
(c)
=
3648
4
3
hE
FLδ
( )( ) ( )( )( )46
3
103048
3654400003.0
h×=
inh 79.4=
inh
b 60.13
79.4
3===
say ininh4
1525.5 == , inin
hb
4
3175.1
3
25.5
3====
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 15 of 64
16. The same as 15, except that the beam is to have a circular cross section.
Solution:
(a) I
Mc
N
ss
y
y==
64
4dI
π=
2
dc =
34
32
64
2
d
M
d
dM
sππ
=
=
( )3
5432
3
40
dπ=
ind 46.3=
say ind2
13=
(b) EI
FL
48
3
=δ
64
4dI
π=
( )( )( )
( )( )( )in
dE
FL0594.0
5.3103048
54400064
48
6446
3
4
3
=×
==ππ
δ
(c) ( )4
3
48
64
dE
FL
πδ =
( )( )( )( ) 46
3
103048
5440006403.0
dπ×=
ind 15.4=
say ind4
14=
17. A simple beam, 48 in. long, with a static load of 6000 lb. at the center, is made of
C1020 structural steel. (a) Basing your calculations on the ultimate strength,
determine the dimensions of the rectangular cross section for bh 2= . (b)
Determine the dimensions based on yield strength. (c) Determine the dimensions
using the principle of “limit design.”
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 16 of 64
Solution:
From Table AT 7 and Table 1.1
ksisu 65=
ksisy 48=
4~3=uN , say 4
2~5.1=yN , say 2
( )( )kipsin
FLM −=== 72
4
486
4
I
Mcs =
2
hc =
12
3bhI =
but 2
hb =
24
4hI =
34
12
24
2
h
M
h
hM
s =
=
(a) Based on ultimate strength
3
12
h
M
N
ss
u
u ==
( )3
7212
4
65
h=
inh 76.3=
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 17 of 64
inh
b 88.12
76.3
2===
say ininh4
3375.3 == , inin
hb
8
71875.1
2
75.3
2====
(b) Based on yield strength
3
12
h
M
N
ss
y
y==
( )3
7212
2
48
h=
inh 30.3=
inh
b 65.12
30.3
2===
say ininh2
135.3 == , inin
hb
4
3175.1
2
5.3
2====
(c) Limit design (Eq. 1.6)
4
2bhsM y=
( )4
24872
2hh
=
inh 29.2=
inh
b 145.12
29.2
2===
say ininh2
125.2 == , inin
hb
4
1125.1
2
5.2
2====
18. The bar shown is subjected to two vertical loads, 1F and 2F , of 3000 lb. each, that
are inL 10= apart and 3 in. ( a , d ) from the ends of the bar. The design factor is 4
based on the ultimate strength; bh 3= . Determine the dimensions h and b if the
bar is made of (a) gray cast iron, SAE 111; (b) malleable cast iron, ASTM A47-
52, grade 35 018; (c) AISI C1040, as rolled (Fig. AF 1). Sketch the shear and
moment diagrams approximately to scale.
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 18 of 64
Problems18, 19.
Solution:
lbRRFF 30002121 ====
Moment Diagram
( )( ) inkipsinlbsaRM −=−=== 99000330001
N = factor of safety = 4 based on us
12
3bhI =
2
hc =
3612
34
3
hh
h
I =
=
(a) For gray cast iron, SAE 111
ksisu 30= , Table AT 6
34
18
36
2
h
M
h
hM
I
Mc
N
ss u =
===
( )3
918
4
30
hs ==
inh 78.2=
inh
b 93.03
78.2
3===
say inh 5.3= , inb 1=
(b) For malleable cast iron, ASTM A47-52, grade 35 018
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 19 of 64
ksisu 55= , Table AT 6
34
18
36
2
h
M
h
hM
I
Mc
N
ss u =
===
( )3
918
4
55
hs ==
inh 28.2=
inh
b 76.03
28.2
3===
say inh4
12= , inb
4
3=
(c) For AISI C1040, as rolled
ksisu 90= , Fig. AF 1
34
18
36
2
h
M
h
hM
I
Mc
N
ss u =
===
( )3
918
4
90
hs ==
inh 93.1=
inh
b 64.03
93.1
3===
say inh8
71= , inb
8
5=
19. The same as 18, except that 1F acts up ( 2F acts down).
Solution:
[ ]∑ = 0AM lbRR 187521 ==
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 20 of 64
Shear Diagram
Moment Diagram
=M maximum moment = 5625 lb-in = 5.625 kips-in
(a) For gray cast iron
3
18
h
M
N
ss u ==
( )3
625.518
4
30
h=
inh 38.2=
inh
b 79.03
38.2
3===
say inh4
12= , inb
4
3=
(b) For malleable cast iron
3
18
h
M
N
ss u ==
( )3
625.518
4
55
h=
inh 95.1=
inh
b 65.03
95.1
3===
say inh8
71= , inb
8
5=
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 21 of 64
(c) For AISI C1040, as rolled
3
18
h
M
N
ss u ==
( )3
625.518
4
90
h=
inh 65.1=
inh
b 55.03
65.1
3===
say inh2
11= , inb
2
1=
20. The bar shown, supported at A and B , is subjected to a static load F of 2500 lb.
at 0=θ . Let ind 3= , inL 10= and bh 3= . Determine the dimensions of the
section if the bar is made of (a) gray iron, SAE 110; (b) malleable cast iron,
ASTM A47-52, grade 32 510; (c) AISI C1035 steel, as rolled. (d) For economic
reasons, the pins at A, B, and C are to be the same size. What should be their
diameter if the material is AISI C1035, as rolled, and the mounting is such that
each is in double shear? Use the basic dimensions from (c) as needed. (e) What
sectional dimensions would be used for the C1035 steel if the principle of “limit
design” governs in (c)?
Problems 20, 21.
Solution:
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 22 of 64
[ ]∑ = 0AM ( )2500133 =B
R
lbRB
833,10=
[ ]∑ = 0BM ( )2500103 =A
R
lbRA
8333=
Shear Diagram
Moment Diagram
( )( ) inkipsinlbM −=−== 25000,25102500
bh 3=
12
3bhI =
36
4hI =
2
hc =
34
18
36
2
h
M
h
hM
I
Mcs =
==
(a) For gray cast iron, SAE 110
ksisu
20= , Table AT 6
6~5=N , say 6 for cast iron, dead load
3
18
h
M
N
ss u ==
( )3
2518
6
20
h=
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 23 of 64
inh 13.5=
inh
b 71.13
==
say inh4
15= , inb
4
31=
(b) For malleable cast iron, ASTM A47-32 grade 32510
ksisu
52= , ksisy 34=
4~3=N , say 4 for ductile, dead load
3
18
h
M
N
ss u ==
( )3
2518
4
52
h=
inh 26.3=
inh
b 09.13
==
say inh4
33= , inb
4
11=
(c) For AISI C1035, as rolled
ksisu
85= , ksisy 55=
4=N , based on ultimate strength
3
18
h
M
N
ss u ==
( )3
2518
4
85
h=
inh 77.2=
inh
b 92.03
==
say inh 3= , inb 1=
(d) For AISI C1035, as rolled
ksissu
64=
4=N , kipsRB
833.10=
A
R
N
ss Bsu
s==
22
242 DDA
ππ=
=
2
2
833.10
4
64
D
ss π
==
inD 657.0=
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 24 of 64
say inD16
11=
(e) Limit Design
4
2bhsM y=
For AISI C1035 steel, ksisy 55=
3
hb =
( )4
35525
2hh
M
==
inh 76.1=
inh
b 59.03
==
say ininh8
71875.1 == , inb
8
5=
21. The same as 20, except that o30=θ . Pin B takes all the horizontal thrust.
Solution:
θcosFF
V=
[ ]∑ = 0AM VB
FR 133 =
( ) 30cos2500133 =B
R
lbRB
9382=
[ ]∑ = 0BM VA
FR 103 =
( ) 30cos2500103 =A
R
lbRA
7217=
Shear Diagram
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 25 of 64
Moment Diagram
( )( ) inkipsinlbM −=−== 65.21650,21102165
3
18
h
Ms =
(a) For gray cast iron, SAE 110
ksisu
20= , Table AT 6
6~5=N , say 6 for cast iron, dead load
3
18
h
M
N
ss u ==
( )3
65.2118
6
20
h=
inh 89.4=
inh
b 63.13
==
say inh4
15= , inb
4
31=
(b) For malleable cast iron, ASTM A47-32 grade 32510
ksisu
52= , ksisy 34=
4~3=N , say 4 for ductile, dead load
3
18
h
M
N
ss u ==
( )3
65.2118
4
52
h=
inh 11.3=
inh
b 04.13
==
say inh 3= , inb 1=
(c) For AISI C1035, as rolled
ksisu
85= , ksisy 55=
4=N , based on ultimate strength
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 26 of 64
3
18
h
M
N
ss u ==
( )3
65.2118
4
85
h=
inh 64.2=
inh
b 88.03
==
say inh8
52= , inb
8
7=
(d) For AISI C1035, as rolled
ksissu
64=
4=N , lbRBV
9382=
lbFFRHBH
125030sin2500sin ==== θ
( ) ( )22222 12509382 +=+=BHBVB
RRR
lbRB
9465=
A
R
N
ss Bsu
s==
22
242 DDA
ππ=
=
2
2
465.9
4
64
D
ss π
==
inD 614.0=
say inD8
5=
(e) Limit Design
4
2bhsM y=
For AISI C1035 steel, ksisy 55=
3
hb =
( )4
35565.21
2hh
M
==
inh 68.1=
inh
b 56.03
==
say ininh8
71875.1 == , inb
8
5=
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 27 of 64
22. A cast-iron beam, ASTM 50, as shown, is 30 in. long and supports two gradually
applied, repeated loads (in phase), one of 2000 lb. at ine 10= from the free end,
and one of 1000 lb at the free end. (a) Determine the dimensions of the cross
section if acb 3≈= . (b) The same as (a) except that the top of the tee is below.