# Complementary and Longitudinal Shear

#### Complementary Shear

In previous articles we discussed how an applied shear is resisted by an equal and opposite shear. If we take a small cut out of this beam and examine the shear stress on a small element we can see that we have a vertical shear stress acting upwards on the left hand side and a vertical shear stress acting downwards on the right hand side.

The vertical shear stresses either side of the element are in vertical equilibrium. However, when we look a moment equilibrium about the centre of the element we can that the moments are out of balance. Clearly the element isn’t rotating, so additional stresses must be acting on the element in order to bring the element into moment equilibrium. The addition stresses that are mobilised are called the complementary shear stress and they act on the top and bottom of the element to ensure moment equilibrium: Figure 2: Vertical shear stress producing out of balance moment, moment equilibrium provided by complementary horizontal shear stresses.

It can be proved that the horizontal complementary shear stress is the same magnitude as the vertical stress. Assume an element has a breadth, $$b$$, a depth, $$d$$ and a unit thickness and vertical, $$\tau_{v}$$, and complementary, $$\tau_{h}$$, shear stresses acting on it.

$$\text{The total shear force on the left and right sides is given by}$$ $$F_{v} = \tau_{v}d$$ $$\text{The total shear force on the top and bottom is given by}$$ $$F_{h} = \tau_{h}b$$ $$\text{Moment equilibrium can be calculated about any point on the element}$$ $$\text{For simplicity we will use the bottom right corner, for moment equilibrium:}$$ $$F_{h}d = F_{v}b$$ $$\tau_{h}bd = \tau_{v}d b$$ $$\tau_{h}\cancel{bd} = \tau_{v}\cancel{d b}$$ $$\text{Therefore}$$ $$\tau_{h}= \tau_{v}$$

#### Longitudinal Shear

The theory of complementary shear stress shows that at a given point within a section the vertical shear stress and longitudinal shear stress are equal. Therefore if we are able to understand the vertical shear stress distribution in a section we can also understand the longitudinal shear stress distribution.

A separate article discusses the concept of shear flow in a section and proves the following standard equation for vertical stress distribution within a section:

$$\tau_{v} = \frac{VAy}{It}$$

Where $$\tau_{v}$$ is the vertical shear stress at the point of interest in the section, $$V$$ is the applied vertical shear force, $$A$$ is the “excluded area” this is the area between the point of interest and the edge of the section, $$y$$ is the distance from the neutral axis to the centroid of the excluded area, $$I$$ is the second moment of area for the full section and $$t$$ is the thickness of the section at the point of interest. Figure 4: Diagram showing definitions of excluded area, centroid and thickness of example beams. No thickness is given for the fabricated element as typically a longitudinal force is calculated and shared between the connections.

As $$\tau_{h} = \tau_{v}$$ we can use the above equation to calculate the longitudinal stress distribution in a cross section.

$$\tau_{h} = \tau_{v} = \frac{VAy}{It}$$

Calculating the longitudinal shear force or longitudinal shear stress is important when analysing structures. Longitudinal shear checks are particularly needed when the integrity of a structure relies on a multi-part section acting as a single unit, for example:

• Glulam beams – Glue-laminated beams are timber beams that are formed of a series of separate timber layers glued together. The glue needs to be able to withstand the longitudinal shear between separate timber layers.
• Built-up or riveted beams – Built-up or riveted beams are formed of individual plates or elements connected together with bolts of rivets. The bolted or riveted connections need to be able to transfer longitudinal shear forces between the different components of a beam.
• Composite bridges – Composite bridges are formed of a steel or concrete beams with a reinforced concrete deck cast on top. The beam and deck are connected together with shear connectors formed of shear studs, channel sections or reinforcing links.

If the longitudinal shear connections (such as the adhesive for glulam, the bolts or rivets for a built-up beam or the shear connectors for a composite bridge) fail, then the beam no longer acts as a single element and instead acts as multiple weaker elements that are free to slide relative to each other. The difference in resistance between a section acting as a single composite unit versus the resistance of the individual elements is shown below.

$$\text{A glulam beam is made of 4 timber planks each 25mm deep and 100mm wide}$$ $$\text{If the planks are properly glued together they will act as one unit. }$$ $$\text{The bending properties of this single unit are found as:}$$ $$I=\frac{bd^{3}}{12} = \frac{100\times100^{3}}{12} = 8.33 \times 10^{6}$$ $$\text{If the glue fails then the planks will act as four individual elements. }$$ $$\text{The combined bending properties of this single unit are found as:}$$ $$I=4\frac{bd^{3}}{12} = 4\frac{100\times25^{3}}{12} = 0.52 \times 10^{6}$$ $$\text{The bending stiffness of the single unit is 16 times that of the individual planks acting separately.}$$ $$\text{The bending resistance will also drop significantly.}$$