Section III.6 example 1 SECTION III.6 EXAMPLE 1
For each of the constant-shear-flow web sections shown, determine:
• (a) the magnitude of the resultant shear force,
• (b) the location of the resultant force (same as shear center), with respect to point 'O' shown with each section.
• (c) the direction of the resultant force.
Let q=100 lb/in in each section.

EQUATIONS USED

SOLUTION

At a glance all of the three sections look very different, but once we do the analysis, we will be able to see the similarities.

Maybe this is a good place to make an educated guess. With respect to magnitude, location, and direction of the resultant force how similar are these sections?

Section 1:

The straight distance between the stringers at the ends of the web is

The magnitude of the force is therefore given as

The direction of the force is parallel to 'h'. The location of the force from point 'O' is

The area A is shown by the hashed region.

Notice that the line of action of the resultant force passes through the shear center of the section.

You may ask the question, how would I know by using the scalar equation for 'e' which side of line 'h' the force should be shown?

The answer is simple. Remember that the resultant force should produce the same moment about point 'O' as the components. Looking at the shear flow direction shown for section 1, the shear flow produces a counter clockwise moment about point 'O'. Therefore, the resultant shear force must do the same. The arrow on the resultant force vector is consistent with the direction of the shear flow along the web, or the shear force in each segment of the web (i.e., the vertical parts and the horizontal part).

Section 2:

The magnitude of the force is given as

The direction of the force is parallel to 'h'. The location of the force measured from point 'O' is

The area 'A' is shown by the hashed region.

Notice that the same reasoning as before can be used to show the correct direction and location of the resultant shear force.

Section 3:

The magnitude of the force is given as

The direction of the resultant shear force is parallel to 'h'. The location of the resultant shear force requires the use of equation for 'e'.

Note that for the previous two sections when we drew straight lines from each stringer to reference point 'O', the resulting region produced a single enclosed area. With the shape we have here, when we connect the ends of the web to point 'O' by two straight lines, the result is two instead of one area. Hmmm! Interesting! What should we do now?

Well we don't give up this easily. We just have to interpret equation we use for 'e' a bit more carefully. In situations such as this, when more than one cell is created, the direction of moment of shear flow in the two areas may be opposite. As is the case here. So we look at the expanded form of equation for 'e'. We calculate 'e' associated with each cell from the equation e=2A/h, then we add them together by paying attention to the direction of moment the shear flow in each cell produces about point 'O'.

Here, the bigger cell produces a counter clockwise moment, whereas the smaller one produces a clockwise moment. The location of the resulant force is, hence, determined as

Now, when we compare the three sections, we realize in all three, the magnitude of the resultant shear force was exactly the same. The difference in these sections appeared only in magnitude of 'e'. Also note that in all sections, the shear flow was directed in northwest direction, and as can be seen so was the direction of the resultant shear force. Finally, the resultant force in each section passes through the shear center of the section.

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