SET B
SET A
P)
eKN
SEN 5M
RAsin3o
KA Sin30
d
O
dr 16mm
RA COS90
RAcos9o (1ocosio)=RAsíng0(10sinio) t 5EN(sCos 1o)
RA 5./42 KN
RA 3.2I4 kN
ZFK0
ZFxo
RAsin 30 x
3.214 sin30 x
EFy =o
RAcos 30 fEy
Bx=2.5|KN
EFy0
KACOs3o + By
5tN
BkN
5.142 cos30 t By BKN
By 3.547 kN
3.21cos9o +By = SKN
By= 2.217 KN
1-60
+B
+ 2.2
2.s+3.543
Ro 4.381 KN
Rg =2.738 KN
T
=
sin 3o Bx
KA
5.142 Sin30 = Bx
1607 kN
Rg /
Regd: Te= {
Me0
9
RAcos 3o (1oxosto) =KAsín3o(10sínno)+ 5kN(scis19)
Sx
5
20 mM
Ke d: Tp?
M
Bx
Sm
2.799(10) N
I (e6) mm
T4.35g MP
4.381(oN
double
shear
sheor stress at
B
2
T
J0.89S Ma
sher stess at
B
SETA
SET
6m
6m
om
Given
6m
OBC50MPa
4m
Agc
1o0 mm
4m
Keg'd: max W =
PBo
W
Bar FBD:
Ay
Solve for Pec
FBe Opc Agc
5,(10mm)
Fsc 5kN
From cylínder FED:
OBc
60 Mra
ABc 120 mm
Kead: max W ?
W
CyiínderFBD:
N
Na
Ax
Given
Na
N
Ax
8
TAy
Solve for Psc
PBc OBCAsc
60 Mm3 Hao mm)
Foc
7.2kN
From cylínder FED0:
Fy=6
N
WN W
Frm bar FBD:
N,(4)
Psc (8)
W()5kN(8)
W GkNF max W
,(TO)=W
N, W
Fom bor FB0
N. (4) = Fsc (8)
W()= 7.akN(8)
W= 8.c4 kNmox W
N