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In Fig 1.1a we have a sample problem to illustrate the general approach. A beam of length L is anchored solidly in a wall with a force P acting on its free end. What is holding the beam together are the molecular forces acting between adjacent atoms/molecules/cristalls. Lets say we want to determine these forces in a cross-section a distance "a" away from the free end.
Equilibrium analysis is our big friend here. In our mind we sever the right end of the beam at the cross-section in mind and consider the forces/moments which would be necessary to act on the cut in order to hold the severed part in static equilibrium. These forces/moments would have to be exerted by the atoms to the left of our cut onto the atoms to the right of our cut while the beam is still one unit !!!!!
In Fig. 1.1b you see the forces/moments necessary to achieve equilibrium.
How do we come up with them ?
First of all, it is customary to split the forces acting on the cut surface
into a normal force, N, and parallel force V. Demanding zero for the sum
of the forces in horizontal direction would actually allow you to determine
the value of N for given P. The internal force V, which we often address
as the shear force, is determined in the same way from the sum of the forces
invertical direction. If you enter both N and V into your Free-Body-Diagram
one or the other might come out zero of course.
So, why did I put a moment M into the free-body diagram as well ? and
how do atoms exert a moment onto each other ?
First : the moment had to be inserted, because the sum of the moments with
respect to any point of rotation is certainly not zero if only the
forces P, N, and V are present. Right ?
Second : It is correct that a single atom has a hard time to exert a moment
onto another atom. What we are talking about here though is the action of
all atoms to
the left onto all atoms to the right of the cut. If in our example the atoms
near the top of the beam would be pulling to the left and those near the
bottom would be pushing to the right we would have a moment acting on the cut.
Could we have considered the equilibrium of the part of the beam to the right of the cut ? Sure we could, but in order to get a complete Free-Body-Diagram we would have to determine the action of the wall onto the beam first. Try it out and you will see that the results are identical for either approach.
Of course, the determination of the internal force system consisting in general of a normal force, a shear force, and a moment is only the first but absolutely necessary step to judge whether a proposed beam with given loads holds up. So keep on reading .......
Klick here to do the test.
Prob. 7.9a :
Addition/subtraction of vectors
Prob. 7.9b : Multiplication by a scalar
Prob. 7.9c : Magnitude of Unit vector
Prob. 7.9d : Value of dot product
Prob. 7.9e :
Dot product -- Angle between vectors
Prob. 7.9f : Evaluate cross product
Prob. 7.9g : Proof on cross product
Prob. 7.9h :
Cross product -- Angle between vectors
