A simple truss supports a force F at joint A, as shown in the figure below. The force has a magnitude F and is inclined by an angle 0 = 60° from the horizontal. The lengths for all horizontal members is s = 8 m and for all vertical members is h = 3 m. The truss is held in equilibrium by a pin at point E and a roller at point F. Using the method of joints, determine the largest applied force Fmaz so that the force in each truss member does not exceed +1600 kN. Note: Express tension forces as positive (+) and compression forces as negative (-). F Fmaz S number (rtol-0.01, atol-1e-05) S kN S E

Structural Analysis
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ISBN:9781337630931
Author:KASSIMALI, Aslam.
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Chapter2: Loads On Structures
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A simple truss supports a force F at joint A, as shown in the figure below. The force has a magnitude F and is
inclined by an angle 0 = 60° from the horizontal. The lengths for all horizontal members is s = 8 m and for all
vertical members is h = 3 m. The truss is held in equilibrium by a pin at point E and a roller at point F. Using
the method of joints, determine the largest applied force Fmaz so that the force in each truss member
does not exceed +1600 kN.
Note: Express tension forces as positive (+) and compression forces as negative (-).
F
0
Fmaz=
+
A
B
D
NA
F
S
number (rtol-0.01, atol-1e-05)
S
kN
S
E
h
Transcribed Image Text:A simple truss supports a force F at joint A, as shown in the figure below. The force has a magnitude F and is inclined by an angle 0 = 60° from the horizontal. The lengths for all horizontal members is s = 8 m and for all vertical members is h = 3 m. The truss is held in equilibrium by a pin at point E and a roller at point F. Using the method of joints, determine the largest applied force Fmaz so that the force in each truss member does not exceed +1600 kN. Note: Express tension forces as positive (+) and compression forces as negative (-). F 0 Fmaz= + A B D NA F S number (rtol-0.01, atol-1e-05) S kN S E h
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