Engineering Mechanics

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(^238) „„„„„ A Textbook of Engineering Mechanics



  1. Complete the parallelogram OLMN, with OM as diagonal.

  2. Measure ON and OL. The length OL gives the magnitude and direction of the reaction RA.
    The length of OL gives the magnitude of the reaction RB.

  3. By measurement , we find that
    RA = 3.0 kN; RB = 1.2 kN and θ = 41° Ans.
    Example 12.16. Fig. 12.32 represents a north-light roof truss with wind loads acting on it.


Fig. 12.32.
Determine graphically, or otherwise, the reaction at P and Q.
Solution. Given: Span = 6.92 m
Let RP = Reaction at P, and
RQ = Reaction at Q.
Since the truss is freely supported on rollers at P, therefore the reaction at this end will be
vertical (because of horizontal support). Moreover, it is hinged at Q, therefore the reaction at this end
will be the resultant of horizontal and vertical forces, and inclined with the vertical.
The example may be solved either analytically or graphically. But we shall solve it analytically
only.
Taking moments about Q and equating the same.
RP × 6.92 = (2 × 3) + (1 × 6) = 12


12
1.73 kN
P 6.92
R == Ans.

We know that total wind load
= 1 + 2 + 1 = 4 kN
∴ Horizontal component of total wind load
= 4 cos 60° = 4 × 0.5 = 2 kN
and vertical component of total wind load
= 4 sin 60° = 4 × 0.866 = 3.46 kN
∴ Balance vertical reaction at Q
VQ = 3.46 – 1.73 = 1.73 kN
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