Mechanics of Materials II
<Home Assignment #1>
2019121114
신정빈
(Deadline: September 26, 2023)
Steeldensixy
Problem 1.
⇒
γ
= 7nuo / m
3
A cantilever beam AB with a circular cross section and length 𝐿 = 750 mm supports
a load 𝑃 = 800 N acting at the free end (see figure). The beam is made of steel with
an allowable bending stress of 120 MPa.
!
a. Determine the required diameter 𝑑min (figure part a) of the beam.
b. Repeat part (a) if the beam is hollow with wall thickness 𝑡 = 𝑑/8 (figure part
b); compare the cross-sectional areas of the two designs.
di
:
d 2A d
4
=
-
차
sd
=
A
ad d- ( 1
4-
=
5
@a)
intesity /oad :
+eel
M
하중
d3 γ
-
qw
8w 일
:
(
qwL
내
까.
다
터가
Figure 1
Z
)
2
양
…
미
"
8 pv +
)
nad
r
=
6
-
단
냐퍼
-디
+
사
4 d -( π
4
=
64
-
*
nta ?=
n
에
∴
- 0 . d 다g D
5
81 O
d
=
37 55 세래
.
y
-
64
n
2
)
I
8800 5 o + 7 m 10
x
rA
=
Qa Lv
86에1~
=
qw
-
=
(b )
-
r
d
195 π d 4
-
16384
~ 판
리에
5
=더
또
pV
oallow
=
42 39 m 래
.
다
"
+ r
+
800750+ ( nn10-
6
.
madn
64-
=
120
-
- nd따
64
.
n 5이
)=
π
4
adz
4
=
- PL -
=
=
4
=
4
'
yad
6
a
Problem 2.
A cantilever beam AB is loaded by a couple 𝑀0 at its free end (see figure). The length
of the beam is ㅇ
𝐿 = 2 m, and the longitudinal normal strain at the top surface is ㅇ
𝜀=
ㅇ. The distance from the top surface of the beam to the neutral surface is ㅇ
0.001
𝑐=
85
mm.
C
a. Calculate the radius of curvature 𝜌, the curvature 𝜅 at the end of the beam.
b. If allowable strain 𝜀𝑎 = 0.0008, what is the maximum acceptable depth (height)
0
of the beam? [Assume that the curvature is unchanged from part (a)].
Figure 2
L 2m
:
Emax
C
=
=
O ool
.
o 5 mm
ac)
85
=
-
85000 mm
=
0 001
=
85m
.
K 85
=
0011
C
n 0
=
bl )
Emax
=
0
0008
.
85
=⇒ 0 0008
=
-
10m
.
2
SMθ=
S
0
=
=>
sin
'
2
-
=
0
PX ( tcos θ )
=
θ
=
o
.
201 rad
@
13 cn
). %
c0
X ( 1
2
=
-
s
0
2
.1
0
.
Problem 3.
A circular pole is subject to linearly varying distributed force with maximum intensity
𝑞0 . Calculate the diameter 𝑑0 of the pole if the maximum allowable shear stress for
the pole is 75 MPa.
ㅇ
Figure 3
C
imak
4 Umux
3A
6 vnax
sko
max
=
=
1
'
=
4
A
=
)
-
d
=
1buka×
.
3π c
5
Z .2 =100w
:
do
=
16 X 100 × 103
=
3대
75
nsnobmn
@
4
Problem 4.
Determine the ratios of the weights of four beams that have the same length, are made
of the same material, are subjected to the same maximum bending moment, and have
the same maximum bending stress if their cross sections are (1) a rectangle with height
equal to twice the width, (2) a square, (3) a circle, and (4) a pipe with outer diameter d
and wall thickness 𝑡 = 𝑑/8 (see figure).
"
bb
□
"
"
디
3
ㆍ
6
"
d
하편
'
Figure 4
G,
/umak 가 동일 ⇒ Sectin modulus ( s ) 가 전부 동일하다
Si
S=
sr
=
ss
=
b
bkb ) :
S
=
4
위 '
'
52 ⇒ b ( 23 1
S=
-
8
=
j
A=
a
(6,
a=
'
3
=
.
dsk
3
A=
t
Ia
30195252
Se
권t=
=
tI
:
AlAr
As : H 4
=
2
.
620n 4 : 3 3019
=⇒ : W : W : Wz : W 4
2
:
1
:
3
-
( =)
.
'
69054 ss
띤
←
3
zcIu
=
3
i
( -)
4
1 845252
.
3
j
3s
π
πd
=
리,
69054 : 1 8452
.
=
d
8 64
=
AE
=
-
4
d
:
=⇒
☆
π3%
따
대 4%
.0
1
d-
…
(
A kh =
b
.
1256 : 14082 : 09041
.
2
∴
Problem 5.
The simply supported wood beam with overhang is subjected to uniformly distributed
load q. The beam has a rectangular cross section with width 𝑏 = 200 mm and height
ℎ = 250 mm. Determine the maximum permissible value q if the allowable bending
stress is 𝜎a = 11MPa, and the allowable shear stress is 𝜏a = 1.2MPa.
Nmax
=
sheaf force
maximum
k By
'
fbt
s
=
3 Vacx
Ymanz
-
2A
↑
( 4 q 124
/
8q
=
3
By
⇒
BY
=
-
원
Figure 5
9 54
52
.
원
0 88889
=
.
( tmg
(0 .21
:
=
쯤
0
µ
6unx
5
-
-
-
tbt
0
548
Maax Ay 54
=
=
49
:
=
S
388
:
AyfBy - 4
Ay
By
3
=
..
.
max= 480( 0. 88889 ㆍ )1m
'
5
q =25 . 7838K 101
m
⇒
A
oo5 m
=
VA
=
54q
3
VB hegutive
=
-
,
VB PositTve
-
=
fy-3 E
-
=
558
55 8 +By
=
N
1 시00
9
VC AyfBy
=
vmax
incx
=
58
-
3(
,
3 Vma*
:
48
ㅇ
=
∴
-
-
:
2
zA
⇒ q
t
551
-
%
- ( .0 위
-
=
0
24 Ku1m
1
= . 2 X 06
1
Problem 6.
A cantilever beam of length 𝐿 = 2 m supports a load 𝑃 = 8.0 kN (see figure). The
beam is made of wood with cross-sectional dimensions 120 mm × 200 mm.
OO
Calculate the shear stresses due to the load P at points located 25 mm, 50 mm, 75 mm,
and 100 mm from the top surface of the beam. From these results, plot a graph showing
the distribution of shear stresses from top to bottom of the beam.
,
평
ㅡ
스
Figure 6
V
p
=
bh '
I
=
5
200
=120 .
음
810 x gmm
=
h:
다 2
I l
)
4 y
Ys
95
46000 2002
[
,
-
Y5
4
=
o
4
2 . 8 시onm
8000
=
)
m
-
200 -50
(4
)
2 8 Xlonany
-
=지 q 4Pa
=
39s4pa
.
M5=
4694 p와
distance
,
100mm
싶
.
=
Ymak
⇒
500kpa
=
MPa
500
'
bh
2
=
9
일에
조
y
31 ) 즈
.
]
Problem 7.
A bridge girder AB on a simple span of length 𝐿 = 14 m supports a distributed load of
maximum intensity q at mid-span and minimum intensity q/2 at supports A and B that
includes the weight of the girder (see figure). The girder is constructed of three plates
welded to form the cross section shown.
Determine the maximum permissible load q based upon (a) an allowable bending stress
𝜎a = 110 MPa, and (b) an allowable shear stress 𝜏a = 50 MPa.
6
m
했고
I
:
5
=
NFB
-
12
c)
-
사FA
'
bh
:
.,
,
x
드미스
Y
( 차5설 + ( 2 24
FAL+ ( 2 -t )( *L ) +( 조
r
*
와
⇒
FA
qla )
FA
,
,
83
원
=
=
-
q
tx
-
29 . - qla )
x
2
V
( 5 조0
=
FA
-
8x
-
x
- Zgcx A
)
-
2
x
*
ql atx
-
=
-
V
⇒ Fy
=
0
=
Figure 7 이때
N
Mmaa
8 L)
U=
1)
]
FAX - rx
2
tx
qf 6
-
=
(b )
Y
=
bIt
2
=
×
'
.
)
-
4
5
- 3×
8
)
E
( bh -bhifth )
.
8
.
=
,
와
∴
83 bhz
-
=
64
az -
q
q
3 14000( 450 (8642
-
4S⑤ 1800
~
16
+ 16 18002 )
=
24어 11N1mm
,
f
l' lli)
.
4
4
3
18643
)
-
al
485
(450
( -을 )1800
450
.-
12
을 18003
=
3 .423 x 07 mms
5ql
=
12
19 X 1010mm 4
S =⇒
'
=
486a 10W
s
X 103
48 bal10 ws48110 3423
=
:
50. 319 X 1010.
.
Hol0nn
319
.
=
3
=
50 3 19 X 1010 16
bhirfthi
3
i
4
6
-
ㅁ
.
ql =
차이므로
8
-
I3
(0w
=
3
=
'
umaol
A d
-
I
450
=
- 2 q(x x 5 x
x
ql
=
I IT -
x ( 3tx) +M
xA (zq x) (x + 2 위 ( A
oF
-
Umak
o
=
5 22
Sx / 400w
2
C
=
184 4101mm
.
)
Problem 8.
A simply supported wood beam AB with a span length 𝐿 = 4 m carries a uniform load
of intensity 𝑞 = 5.8 𝑘N/m (see figure).
a. Calculate the maximum bending stress 𝜎max due to the load q if the beam has
a rectangular cross section with width 𝑏 = 140 mm and height ℎ = 240 mm.
b. Repeat part (a) but use the trapezoidal distributed load shown in the figure (b).
(a)
-
FAL
/+ g /
lz
=
.
FA
=
)
L
크
FA - A
V
0
0
=
FAx + ( q - x )xXM
-
V
=
-
FaX
n:
q( 2 x 7
umux
-
qxcL x)
=
=
⇒ N
0
-
-
5
.
8 X 40002
=
스
8
sumax
Gmax
s
=
=
11600000N
=
bmax
=
=
8 G 3 | MPa
.
L . 소+
← ( 202
=
i
+
.
48
힘에
tir vr
FA
V
-
=
X
Zxqcx)
FA
zr* - ixqx)
8 -
-
V
=
0
' 원
5
:
V = al
:
-
tix
x
-
=
+
-
M
=
FAX
=
청
5
-
r
Z
FAx
-
x x + 2 a (x)
0
x
.
5x+U
=
o
02 :A talx
)A
~
Z
-
-
=
52)
*
=
N
22 nL 5
al
=
m 에
Figure 8
0
FAL + 2
=
.
mmg
1344000
C
∴
FA
장
.
-
-
8ac
)
s
f
( b)
2 MB
2
( 1600000
.
-
=
=
140 24
bhz
S=
-
2
q z ( 24
=
o
=
Mman
=⇒
o
ax
2
-
IMFB
IKFA
- x)
q(
=
N
-
x?
=
x
o
.
-
5x
-
4
ql pt
-
3
x
+
)
=
0
서
s295 L
2 52n Lm
.
S0
…
Mhax
q (
-
5
0
쁨
2
=
0 09403%(
=
.
6maxz
Mmal
5
=
n 딸)
- ( . 52
- ( 0.52 n
+ 0
팀
89259840
만
@
25984 = .4 a3Mp
-6 n
.
1344.00
6