3]
Given: E =2200 t/cm2
=2200 x 1000 = 2.2 x 106
kg/cm2
Required:
∆ = ?
Solution
P= σ A ,
E
l×
=∆
σ
, w (h+∆) = 0.5 (σ A) (σl/E)
])(211[
WL
AE
h
A
W
++=σ
2
22
cm01.2
4
6.1
4
d
A =
×π
=
×π
=
2
6
/65.6646])
30060
102.201.2
(100211[
01.2
60
cmkg=
×
××
××++=σ
cm906.0
102.2
30065.6646
6
=
×
×
=∆∴
******
245.
4]
Given:
A2 = 2A1, σy1 = σy2
Required:
w1 / w2 = ?
Solution
EAl
U
18
2
σ
=
Since σy1 = σy2, E1 = E2, L1 = L2
2
1
2
1
2
1
22
2
11
1
==⇒=∴
A
A
U
U
LA
U
LA
U
)(22
1
)(
)(
2
2
2
1
22
11
2
1
∆+
∆+
=⇒=
∆+
∆+
=∴
h
h
w
w
hw
hw
U
U
∆1 and ∆2 are very small compared to h. Thus ∆1 and ∆2
can be neglected.
2
1
w
w
2
1
≅∴
**********
246.
5]
Given:
σy = 3400kg/cm2
, E = 660x1000 kg/cm2
= 6.6x105
kg/cm2
Required:
Y =? P/w =? Y/∆ =?
Solution
cm
Ey
l
431.1
3106.612
1003400
12 5
22
=
×××
×
==∆
σ
I
ym×
=σQ ,
4
pl
m =
I
ylp
×
××
=∴
4
σ
yl
I
p
×
××
=∴
σ4
kgp 4896
3100
12
6
34004
4
=
×
××
=∴
اﻟﺪاﺧﻠﻴﺔ اﻟﻄﺎﻗﺔ=اﻟﺨﺎرﺟﻴﺔ اﻟﻄﺎﻗﺔ
)(
2
1
∆+=∆ Ywp
cmYY 877.95)431.1(36431.14896
2
1
=⇒+×=××
136
36
4896
==
w
p
93.66
431.1
877.95
==
∆
Y
*******
247.
6]
Given:
σy = 3600kg/cm2
, E = 2100x1000 kg/cm2
= 2.1x106
kg/cm2
Required:
w =? ∆max =?
Solution
L
ba
EI3
p
,
L
Pab
M
22
×=∆=
aby
LI
P
LI
Paby
L
Pab
M,
I
yM σ
=∴=σ⇒=
×
=σ
4
4
33.341
12
8
cmI ==
kg3840P
4240120
33.3413603600
P =⇒
××
××
=∴
cm11.4
2401203840bap
22
6
22
=
×
×=×=∆
kg0.146w)114.450(w114.43840
1
≅⇒+=××
********
)h(wP
1
∆+=∆
ب-اﻟﻜﺴﺮ ﺷﻜﻞ ﺕﻮﺽﻴﺢﻣﻊ اﻟﻤﺘﻮﻗﻊ اﻟﺤﻤﻞ:
1- For 15 cm cube in compression
fc (28 days) = 1.5 x fc (7 days)
fc ( 7 days) = 54000/15x15 = 240 kg/cm2
fc (28 days) = 1.5 x 240 = 360 kg/cm2
P = 360 x15 x 15 = 81 ton
اﻟﻤﻜﻌﺐ ﻻﻥﻬﻴﺎر اﻟﻤﺘﻮﻗﻊ اﻟﺤﻤﻞ=٨١ﻃﻦ
2- For cylinder 15 x30 in compression
fc (28 days) = 0.8 x 360 = 288 kg/cm2
P = 288 x π x (7.5)2
= 50.9 ton
اﻟﻤﻜﻌﺐ ﻻﻥﻬﻴﺎر اﻟﻤﺘﻮﻗﻊ اﻟﺤﻤﻞ=٥٠٫٩ﻃﻦ
3- For 10 cm cube in compression
fc (cube 10) = 360/0.97 = 371.1 kg/cm2
P = 371.1 x 10 x 10 =37.11 ton
اﻟﻤﻜﻌﺐ ﻻﻥﻬﻴﺎر اﻟﻤﺘﻮﻗﻊ اﻟﺤﻤﻞ=٣٧٫١١ﻃﻦ
4- For cylinder 15 x 30 in indirect tensile strength
ft = 0.1 x 360 = 36 kg /cm2
ton45.25P
P2
36
P2
ft =⇒
×
===
اﻟﻤﻜﻌﺐ ﻻﻥﻬﻴﺎر اﻟﻤﺘﻮﻗﻊ اﻟﺤﻤﻞ=٢٥٫٤٥ﻃﻦ
308.
5- For beam10 x 10 x 70 cm in bending
fflex = 0.14 x 360 = 50.4 kg/cm2
3
4
flex cm33.833
12
10
I,P15
4
pl
M,
I
yM
f ====
×
=
ton560.0P
33.833
5P15
4.50fflex =⇒
×
==
ااﻟﻤﻜﻌﺐ ﻻﻥﻬﻴﺎر اﻟﻤﺘﻮﻗﻊ ﻟﺤﻤﻞ=٥٦٠آﺞ
ب-ﺏﻌﺪ اﻟﻤﺮوﻥﺔ ﻟﻤﻌﺎﻳﺮ اﻟﻤﺘﻮﻗﻌﺔ اﻟﻘﻴﻤﺔ٢٨ﻳﻮم:
2
cm/t6.26536014000 === cuc f14000E
**********
309.
اﻷﻗﺼﻰ اﻟﺤﻤﻞ
اﻟﺸﺪ ﻣﻘﺎوﻣﺔاﻟﻤﺒﺎﺷﺮ=ـــــــــــــــــــــــــآﺞ/ﺳﻢ٢
اﻟﻤﻘﻄﻊ ﻣﺴﺎﺡﺔ
٢×اﻷﻗﺼﻰ اﻟﺤﻤﻞ
اﻟﺸﺪ ﻣﻘﺎوﻣﺔاﻟﻤﺒﺎﺷ ﻏﻴﺮﺮ)اﻟﺒﺮازﻳﻠﻲ= (ــــــــــــــــــــــــــــــــآﺞ/ﺳﻢ٢
ط×اﻟﻄﻮل×اﻟﻘﻄﺮ
P
P
وﺳﺎدة
وﺳﺎدة
ﺽﻐﻂ إﺟﻬﺎدات
ﺽﻐﻂ إﺟﻬﺎدات
ﺷﺪ إﺟﻬﺎدات
2P
πDL
=
D
12
D
12
+
-
-
5 D
6
D
P
P L