P 28
FBD:
2 pts
8 ft
⋅
2.438m
=
Σ
Ma = 0
14kN 2.438
⋅
m
36kN 2.438
⋅
m
+
Cy 3
⋅
2.438
⋅
m

0
=
Cy
14kN 2.438
⋅
m
36kN 2.438
⋅
m
+
3 2.438
⋅
m
=
Cy
16.67 kN
⋅
=
3 pts
Σ
Fy = 0
Cy
14kN

A sin 135deg
(
)
⋅
+
0
=
16.67kN
14kN

A sin 135deg
(
)
⋅
+
0
=
A
2.67 kN
⋅
sin 135.0 deg
⋅
(
)

=
A
3.78

kN
⋅
=
3 pts
Σ
Fx = 0
Cx
36 kN
⋅

A cos 135deg
(
)
⋅
+
0
=
Cx
36 kN
⋅

3.78

kN
(
) cos 135deg
(
)
⋅
+
0
=
Cx
36.0 kN
⋅
3.78 kN
⋅
cos 135 deg
⋅
(
)
⋅
+
=
Cx
33.33 kN
⋅
=
3 pts
Alternate version:
FBD
2
pts
Σ
Fx = 0
Cx
36 kN
⋅

0
=
Cx
36kN
=
3 pts
Σ
Ma = 0
14kN 2.438
⋅
m
36kN 2.438
⋅
m
+
Cy 3
⋅
2.438
⋅
m

0
=
Cy
14kN 2.438
⋅
m
36kN 2.438
⋅
m
+
3 2.438
⋅
m
=
Cy
16.67 kN
⋅
=
3 pts
Σ
Fy = 0
Cy
14kN

Ay
+
0
=
16.67kN
14kN

Ay
+
0
=
Ay
14kN
16.67kN

=
Ay
2.67

kN
⋅
=
3 pts
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentP 29
Assume that
the unknowns
are in tension!
cos 22deg
(
)
0.927
=
sin 35deg
(
)
0.574
=
cos 35deg
(
)
0.819
=
sin 22deg
(
)
0.375
=
FBD:
2
pts
Σ
Fx = 0
Fab

cos 35deg
(
)
⋅
Fbc cos 22deg
(
)
⋅

0
=
Fab

0.819
⋅
Fbc 0.927
⋅

0
=
Fab
1.132

Fbc
⋅
=
1lbf
4.448N
=
Σ
Fy = 0
500lbf
2224 N
=
500

lbf
Fab sin 35deg
(
)
⋅

Fbc sin 22deg
(
)
⋅

0
=
500

lbf
1.132

Fbc
⋅
(
) 0.574
⋅

Fbc 0.375
⋅

0
=
500

lbf
0.650 Fbc
⋅
+
0.375 Fbc
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '07
 walker
 Trigraph, pts, FBC

Click to edit the document details