In the three-member truss shown in the figure, AC = BC. An external force of 10 kN is applied at B, parallel to AC. The force in the member BC is
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When a load is applied parallel to a truss member, that member carries no axial force. Instead, the connected member(s) at the joint must carry the load through geometric equilibrium.
The external 10 kN force applied at point B is shown acting horizontally and is parallel to the member AC. We are required to calculate the force in the vertical member BC of the truss. Since AC = BC and AB is at a 45° angle, triangle ABC is an isosceles right triangle with AB acting as the diagonal. The applied load of 10 kN is parallel to AC, meaning the load has no component perpendicular to AC. Therefore, member AC carries no axial force due to this loading condition.
At joint B, the only members meeting are AB and BC. Since AB makes a 45° angle, equilibrium of the horizontal directions requires that AB carry the entire 10 kN load in tension. The vertical member BC must simply carry the vertical reaction of member AB, and because AB is at 45°, the horizontal component of the force in AB must equal 10 kN. For a 45° member, the force in AB produces equal horizontal and vertical components. Thus, the axial force in AB must be 10 kN.
Since AB pulls downward on joint B along its vertical component, member BC must resist the same vertical force in tension, pulling upward on joint B. Therefore, the axial force in BC equals the 10 kN applied load.
