Shear in beams

The previous lesson derived bending stress under the fiction of pure bending, a constant moment with no shear. Almost no real beam is in that state: the moment varies along the span, and the third lesson showed that its rate of change is exactly the shear force. Working out what that shear does inside the section explains why beams have webs and why a stack of planks is not a beam.

Why there has to be horizontal shear

Stack three loose planks and load them across a span. They bend together, and their ends slide relative to one another, leaving three visible steps. Glue the same three planks into one block and the ends stay flush, so the glue must be carrying the force that the sliding was previously relieving. The glued beam is also far stronger, because its depth is three times greater and $I$ goes as depth cubed.

The same argument works on a single beam. Take a slice of beam of length $d x$ and consider only the material above some horizontal level. On the left face the bending stresses come from a moment $M$ ; on the right face they come from $M + d M$ . The stresses on the right are larger, so the horizontal forces on the two faces do not balance, and the imbalance has to be carried by a shear force on the horizontal plane at the bottom of the piece being considered. There is no other surface available.

So bending that varies along the span implies horizontal shear inside the section, always. It is not an extra load; it is a consequence of the bending already there.

Complementary shear

Shear stresses cannot exist alone on one plane. Take a small square element with a shear stress $τ$ on its top and bottom faces. Those two form a couple, and nothing balances it unless equal shear stresses act on the vertical faces as well, forming an opposing couple. Taking moments about the centre of a square element of side $a$ and thickness $t$ shows the two magnitudes are equal.

τ_{horizontal} = τ_{vertical}

This is the complementary shear result, and it is why the shear force $V$ that the diagrams give, which is plainly a vertical action, can be used to find a horizontal stress. It also explains why a free surface always has zero shear stress parallel to it, since there is no material outside to supply the complement, and that boundary condition is what pins the distribution at the top and bottom of a beam.

The shear stress formula

Return to the slice of length $d x$ and take everything above a level $y_{1}$ . The bending stress at a height $y$ is $M y / I$ , so the total horizontal force on the left face of the piece is $\int (M y / I) d A$ over the part of the section above $y_{1}$ , and on the right face the same with $M + d M$ . The difference is

\frac{d M}{I} \int y d A = \frac{d M}{I} Q

where $Q = \int y d A$ is the first moment of area of the part of the section above the level considered, taken about the neutral axis. It is easiest computed as area times the distance from the neutral axis to that area's own centroid.

That difference is resisted by a shear stress $τ$ acting on the horizontal cut, whose area is $t d x$ with $t$ the width of the section at that level. Balancing, $τ t d x = Q d M / I$ , and using $d M / d x = V$ ,

τ = \frac{V Q}{I t}

Read the formula through $Q$ . At the very top of the section there is no area above the cut, so $Q = 0$ and the shear stress is zero, as the free surface demands. Moving down, $Q$ grows, and it is largest at the neutral axis, where the whole of one half of the section lies above the cut. So shear stress is largest at the neutral axis and zero at the extreme fibres, which is precisely the opposite of the bending stress distribution. A beam's most highly bent material carries no shear, and its most highly sheared material carries no bending stress.

The rectangular section

For a rectangle of width $b$ and depth $d$ , the area above a level $y$ is $b (d / 2 - y)$ and its centroid sits at $(d / 2 + y) / 2$ , so

Q = \frac{b}{2} (\frac{d^{2}}{4} - y^{2})

Substituting with $I = b d^{3} / 12$ and $t = b$ gives a parabola in $y$ , zero at both faces and maximum at $y = 0$ :

τ_{\max} = \frac{3 V}{2 A}

Fifty per cent more than the average $V / A$ , and worth remembering because the average is what a first guess uses.

Example. A timber beam $100$ mm wide and $250$ mm deep carries a shear force of $30$ kN. Find the maximum shear stress.

The area is $25 000$ mm², so the average is $30 000 / 25 000 = 1.2$ MPa and the maximum is $1.5 \times 1.2 = 1.8$ MPa at the neutral axis. That sounds trivially small next to the bending stresses of the previous lesson, and it is not, because timber's shear strength parallel to the grain is only a few megapascals: an order of magnitude below its bending strength, since the failure is a split running between the fibres rather than across them.

Now you. A timber beam $75$ mm wide and $200$ mm deep carries $20$ kN of shear. Find the maximum shear stress.

Answer

$A = 15 000$ mm², so the average is $1.33$ MPa and the maximum is $1.5 \times 1.333 = 2.0$ MPa.

Timber beams are therefore checked for shear as a matter of routine, and they are notched at their ends only with great care, because a notch on the tension face puts a re-entrant corner exactly where the shear is trying to split the beam.

The I-beam, and where the shear actually goes

For a section whose width changes, the $t$ in the denominator produces a discontinuity, and in an I-beam it is dramatic.

Example. Take the I-section of the previous lesson: flanges $200$ mm by $15$ mm, web $8$ mm thick, overall depth $300$ mm, with $I = 1.351 \times 10^{8}$ mm⁴. It carries a shear force of $250$ kN. Find the shear stress at the neutral axis, and just above and just below the flange-to-web junction.

At the neutral axis, $Q$ is the flange plus the upper half of the web: $200 \times 15 \times 142.5 + 8 \times 135 \times 67.5 = 427 500 + 72 900 = 500 400$ mm³. With $t = 8$ mm,

τ = \frac{250 000 \times 500 400}{1.351 \times 10^{8} \times 8} = 115.8 MPa

At the junction, $Q$ is the flange alone, $427 500$ mm³. Taken in the web, where $t = 8$ mm, $τ = 98.9$ MPa. Taken in the flange one millimetre higher, where $t = 200$ mm, the same $Q$ gives $τ = 4.0$ MPa. The stress drops by a factor of twenty-five across a boundary where nothing physical changes except the width available to carry it.

Now you. The same section carries $150$ kN of shear. Find the stress at the neutral axis, and compare it with the crude estimate $V$ divided by the web area.

Answer

$τ = 150 000 \times 500 400 / (1.351 \times 10^{8} \times 8) = 69.5$ MPa. The crude estimate is $150 000 / (300 \times 8) = 62.5$ MPa, which is $10$ per cent low, and that is the usual size of the error.

Two conclusions follow, and both are standard practice. First, the web carries essentially all the shear in an I-beam, the flanges almost none, so the rough check $τ \approx V / (d t_{w})$ using the full depth times the web thickness is close enough for design and errs on the unsafe side by about ten per cent, which codes correct for. Second, the flanges carry the bending and the web carries the shear, so the two checks are almost independent and a section can be chosen for moment and then verified for shear.

The web has its own failure modes that the stress formula does not see. A slender web buckles diagonally under shear long before it yields, which is why plate girders carry vertical stiffeners. And a web can crush locally where a concentrated load or a support reaction is delivered into it, which is called web bearing or crippling and is checked separately at every point load.

Shear flow, and the spacing of connectors

For a built-up member the useful quantity is not the stress but the force per unit length along the joint, the shear flow

q = \frac{V Q}{I}

in newtons per millimetre, where $Q$ now refers to the area of the piece being attached. Multiply by the spacing of the fasteners and you get the force each one has to carry, which is how nail, screw, bolt and weld spacing is decided in every built-up beam.

Example. A timber T-beam is made by nailing a flange $200$ mm wide and $50$ mm thick on top of a web $50$ mm wide and $200$ mm deep, giving an overall depth of $250$ mm. The shear force is $4$ kN, and the nails have a capacity of $1200$ N each, used in pairs across the joint. What spacing is needed?

Both pieces have area $10 000$ mm², with centroids $225$ mm and $100$ mm above the bottom, so the neutral axis is at $162.5$ mm. The parallel axis theorem gives $I = 1.135 \times 10^{8}$ mm⁴. For the flange, $Q = 10 000 \times (225 - 162.5) = 625 000$ mm³. So

q = \frac{4000 \times 625 000}{1.135 \times 10^{8}} = 22.0 {N mm}^{- 1}

A pair of nails carries $2400$ N, so the spacing is $2400 / 22.0 = 109$ mm. Round down to $100$ mm and specify that.

Now you. The same beam carries $6$ kN of shear. What spacing is now needed?

Answer

$q = 6000 \times 625 000 / 1.135 \times 10^{8} = 33.0$ N mm⁻¹, so the spacing is $2400 / 33.0 = 73$ mm. In practice, nails would be closer near the supports where $V$ is largest and further apart near midspan, since $q$ follows the shear diagram.

When shear governs, and what the formula misses

Compare the two demands on a simply supported beam of span $L$ under a uniform load. The moment is $w L^{2} / 8$ and the shear is $w L / 2$ , so the ratio of bending demand to shear demand grows with the span. Long beams are governed by bending, short deep ones by shear, and the crossover for a rectangular timber section is somewhere around a span of ten to fifteen times the depth. A lintel over a doorway, a bracket, and a beam with a heavy load close to a support are all cases where shear decides, and where a designer who checked only bending would be wrong.

The formula has real limits. It assumes the shear stress is uniform across the width $t$ at any level, which is a good approximation for a narrow web and a poor one for a wide flange, where the true distribution varies across the width and the flange also carries shear flowing horizontally rather than vertically. That horizontal flange shear is what makes a channel section twist unless it is loaded through a particular point, the shear centre, offset outside the web.

It also assumes elastic behaviour and slender geometry, so it does not describe a deep beam, and it says nothing about the buckling failures that usually get a thin web first.

With bending stress and shear stress both in hand, a beam can be checked for strength. It can still be entirely unusable, because a floor that bounces or a lintel that sags visibly has failed in the eyes of everyone who uses it while remaining comfortably below every stress limit. Deflection is next.