Class notes BS-Civil Engineering

FEBRUARY 10 , 2025




CHAPTER 1




ULATE THE AXIAL STRESS IN THE CABLE .




D




3 ft



20
80'
4 . 5 .
C
&


A



T
189 C

>NTPHBY5x5= 243x5=
RCX .
4 83




600 it RCY


2 Mc = 0 5 =
i
5 2857 1516
T5 2 60005 0
, 89 10
T

+
.

.
-
=



= 0 .
502


3 psi
T =
2857 15 16.




8 =
14551 .




PROBLEM 6


DETERMINE THE LARGEST WEIGHT I THAT CAN BE SUPPORTED BY TWO WIRES AS SHOWN IN THE FIGURE .
THE STRESS IN EITHER "*RE IS NOT


TO EXCEED 35 Ksi.
B C
GIVEN : E35ks :

AAB =
0 40.
in?

Arc = 0 50. in ?




300 A 500




W


@ JOINT A

ALLOWABLE STRESS O 35ksi
IF =

ABTLAC
,




ABsin30 + ACsin50 -
I = 0 Eg . 1 Gab = JAC = PA
EFx 0
=
=



W =in2
-



35ks1 35ksi
in2
-

PABCOS 30 +
PACCOS50 =
0
PAB =
14 kips Pac = 17 5 .
kips
PAC =
1 3473
. PAB -
Eq. 2




SATISFYING THE CONDITION SUBSTITUTE PAB AND PAC IN EQUATION I


WHEN PAB =
14 kips WHEN PAC =
17 5 kips
.
#sim #Asin50 =
I = O


from Ea. 2 ; PAC =
1 3473
. PAB from Ea. 2 ; PAC =
1 3473 . PAB 13sin30 + 17 5 sin 50
.
-
W = 0



Pac =
18 86.
kips PAB =
13 kips W =
19 91
.
kips

* *
DID NOT SATISFY THE ALLOWABLE STRESS USE PAB =
13 KipS & PAC =
17 5.
kips




ALTERNATIVE SOLUTION




DOINT A



#B #sim3 #Asin50 =
I =
0 . AB =
0 653 W
.




= 0 #c =
0 879 W
.


N
N/
-
ABCOs30 + ACCOs50 =
0 . 2




& WEIGHT AB & WEIGHT AC


GAB =
PPAB Jac =
W =
19 91 .
kips
<
SAFE WEIGHT




5 #

, FEBRUARY 12 , 2025




EARING STRESS IN THE CONTACT PRESSURE BETWEEN TWO SEPARATE BODIES .




IN SYMBOLi On =
Ph

OBBEARSTRESSNP
WHERE OR si




Ab =
PROJECTED AREA mm2 or in
?




CONSIDER THE PLATES FASTENED TOGETHER BY A RIVET ;



P = V =
Pp =
=
D




in


t


-
P = V =
Pb



5 =
P WHERE AD = DE NO . OF RIVETS




NOTE FOR DIFFERENT PLATE THICKNESS
>
USE THE THINNER PLATE




PROBLEM I


THE LAP JOINT SHOWN IN THE FIGURE IS FASTENED BY FOUR . in
3 4 DIAMETER RIVETS . CALCULATE THE MAXIMUM SAFE LOAD P THAT CAN


BE APPLIED TO SATISFY THE FOLLOWING REQUIREMENTS : 0 20ksi ,
I I 14 ksi ,
AND Op =
18 ksi . ASSUME THE APPLED LOAD IS UNIFORMLY


DISTRIBUTED AMONG THE FOUR RIVETS .




PC t = 1 .
5"

> P




P

·
& AXIAL STRESS & SHEARING STRESS & BEARING STRESS


P
5 =

A
i
= Op =
20ksi =
-Pi . 5
14k si =

+3 18 ksi =
354
P = 90 kips p = 24 74.
kips p = 81 kips




USE P = 24 74
.
kips

, IT IS TO CARRY A GAS AT A PRESSURE OF 1400 psi .
THE DIAMETER


OF THE VESSEL IS 2ft AND THE STRESS IS LIMITED TO 12ksi .




* ks i psi




TANGENTIAL STRESS ↓ONGITUDINAL STRESS



ot =
P % =
P
12 " 12
psi2x 1400psi2x
1400
1200 psi = 1200 psi =




t =
1 4 in
. t = 0 7 in
.




PROBLEM 2


THEGTHISIkiP CACULATE THE MIN DAMETER OF THE




HEREOFTONEREAS
FOR




GIVEN :
Y = 33 kip f


-

= 16 kip f




N


GIRTH JOINT




TANGENTIAL STRESS




↑
runnur e
PN 33 kip ft 2P PDL
=
=




2p F F PDL
I PD
= =
; =

~

T
33 kip f+ x
1000lb xiT
=
150lb in
↓
D = 36 67 in
.




ONGHUD L
L #
P =
=




F
16 kipf


=
pπz
+

D =
=




16


35 56 in
14
kiP/f+

.
x 1000lbIf=150lbn
Se




PROBLEM 3


THE TANK SHOWN IS FABRICATED FROM I8"STEEL PLATE . CALCULATE THE MAXIMUM HOOP AND LONGITUDINAL STRESS CAUSED BY AN


INTERNAL PRESSURE OF 125 psi .




- V




↑
~+
V

2


& SECTION 1 - T

F
[Fy = 0 2P = F




·
2P F 0 p
Ez
-
= =




2P = F
=
525011b in
2
F =
PAT ; A
= 3 .
5'xL

↑ = 26251 16 i
= 125 MPa 3 5 ++
.

xin x +
FROM Of =




·
F =
5250116 in

Ap =
of =

2625bin s
Ot = 21000 psi




& SECTION 2-2

* IT 0 752 TWO SEMI-CIRCLE


752 x 12 in
>
.




F =
PAFi AF = Ift 1 5 ft
.
+ 0 .




i =
PD ; zp =
F
- S
AF = 57 21 in.
257 . 20575041




a PPp
1 3
FROML
.
=
F 125 MPa AF
-
N =




i = p = F
# 7150 72 16 7950 718801
5 x"X
=
. .


Ap =
2 + 21 + + 1 .

&

21
[Fx =
0
Ap = 13 07 in
.
2 13 . 06858347


P F 7950 721b

4150Tb
= =
.




51 =




01 =
547 17 .
psi 547 1686214
.




PERINTERFORNONG