The beam AB is attached to the wall in the xz plane by a fixed support at A. A f
ID: 1870662 • Letter: T
Question
The beam AB is attached to the wall in the xz plane by a fixed support at A. A force of F=(-133i^+94.0j^+392k^) N is applied to the end of the beam at B. The weight of the beam can be modeled with a uniform distributed load of intensity w = 80.0 N/m acting in the negative z direction along its entire length. Find the support reactions at A.
Values for dimensions on the figure are given in the following table. Note the figure may not be to scale.
A = ( i^ + j^+ k^) N
MA = ( i^ + j^+ k^) N-m
Explanation / Answer
applied force F=-133i+94j+392k N
weight of beam = 80*5.9=472 N = -472k N
Total load acting on beam = -133i+94j+392k -472k=-133i+94j-80k N
Reaction force at A will be equal and opposite to the total force = 133i-94j+80k N
vector AB=5.9j
moment about A due to F=5.9j x (-133i+94j+392k)=2312.8i+784.7k Nm
moment about A due to weight of beam = (5.9/2)jx(-472k)=-1392.4i Nm
Total moment acting about A=2312.8i+784.7k -1392.4i = 920.4i+784.7k Nm
Reaction moment at A=-920.4i-784.7k Nm
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