Two steel truss members, AC and BC, each having cross sectional area of $100 m m^{2},$ are subjected to a horizontal force F as shown in figure. All the joints are hinged.

The maximum force F in kN that can be applied at C such that the axial stress in any of the truss members DOES NOT exceed 100 MPa is

8.17

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11.15

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14.14

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22.30

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Solution

The correct option is B 11.15
Method I :
Maximum force = 0.8965 F

$∴ Max stress = \frac{0.8965 F}{100} \leq 100 M P a$

$∴ 100 \geq \frac{0.8965 \times F}{100}$

$∴ F \leq 11.154 k N$

Method II:
$A s, F_{A C} = 0.895 k N$
then

$\frac{F \times F_{A C}}{A} = 100$

$\frac{F \times 0.895}{100} = 100$

$F = 11173.18 N = 11.173 k N$

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