Study Companion: MM05 — Shear Stress and Strain

A 12\,\mathrm{mm} steel pin holds two plates together and the plates pull on it with 30\,\mathrm{kN}. Nothing is being stretched; the plates try to slide past each other and the pin is in the way. Cut the pin between the plates and the two halves would slip. How hard is the pin's material working against that slipping, and what can go wrong besides the pin cutting in two? This lesson is about stresses that act along a surface instead of through it: shear stress, the shear strain it causes and the several checks every bolted or pinned connection needs.

Sources: Gere and Goodno, Mechanics of Materials, 9th edition (SI), section 1.8 (the syllabus for this lesson); Beer, Johnston, DeWolf and Mazurek, 8th edition, section 1.2 (the shear, bearing and pin parts; the complementary-shear derivation it uses is Beer's 1.4) and section 2.7 (shearing strain); Hearn, Mechanics of Materials 1, sections 1.12–1.15; Venkataraman, Essential Mechanics, section 3.4; Muvdi and Elhouar, section 1.3.3. Hearn's four sections run to about a page and a half and do not treat bearing or complementary shear; Muvdi has only the single- and double-shear equations. The relation G = E/[2(1+\nu)] is stated here and derived in Lesson MM16. All working is in SI; the US examples of Beer and Gere are converted.

How to use this companion

Each nugget is one idea. It opens with a claim, then a picture shows why it holds, a number checks it, and an equation states it, labelled with the kind of statement it is. A worked example with units, a real use, the most common wrong answer and a check question follow. A bridge says why the next idea is needed, and the Narration spine summarises the video beat. The synthesis, computed views, traps and self-check close the lesson.

The puzzle

A 12\,\mathrm{mm} pin carries 30\,\mathrm{kN} from one plate to another through its single shear plane. How large is the stress that tries to cut it? The plausible answer is "the same as in a 12\,\mathrm{mm} rod pulled by 30\,\mathrm{kN}", 265\,\mathrm{MPa}; the number is right but it is a different kind of stress, with a different failure and a different allowable value. See Nugget 01 and Nugget 02.

Nugget 01: Shear Stress Acts Along the Surface Instead of Through It

Claim. When two equal and opposite forces act across a body so that one part tries to slide past the other, the internal force on the cut lies in the plane of the cut. Its intensity, per area, is the shear stress \tau. The average is \tau_{\text{aver}} = V/A.

The picture. Two plates joined by a pin, loaded to slide against each other.

Pull a normal stress through a section and the force is perpendicular to it. Here the force is parallel to it. The same word, force per area, but the material resists in another way, and when it fails it fails by sliding along the plane, not by pulling apart.

Check with numbers. The puzzle's pin: d = 12\,\mathrm{mm}, A = \pi(6\,\mathrm{mm})^2 = 113.1\,\mathrm{mm^2}, V = P = 30\,000\,\mathrm N, one plane. \tau = 30\,000/113.1 = 265.3\,\mathrm{MPa}. Cutting a 12\,\mathrm{mm} rod in tension by the same load gives the same arithmetic but the allowable value is different: shear strengths are about half the tensile ones (Nugget 06), so this pin is far more stressed in relation to its capacity.

The idea. Gere (1.8) and Beer (1.2B) describe the same picture: the transverse forces P cut the member between them and the section carries a shear V = P. The V/A value is an average. The true stress is not uniform: it is zero at a free surface and larger toward the middle, with peaks around holes and edges. Beer points ahead to the real distribution in the beams chapter (Lesson MM31); Venkataraman and Muvdi idealise it as uniform. Shear arises in bolts, pins, rivets, keys, welds and glued joints, in a blade cutting a carrot (Venkataraman) and in the cut of a bicycle brake block (Hearn): the smaller the area under the load, the larger the stress for the same force.

The equation (definition, average shear stress; valid for the idealised uniform distribution):

Worked number. \tau_{\text{aver}} = 265.3\,\mathrm{MPa} for the puzzle's pin. A sharp knife on a carrot concentrates a few newtons on a fraction of a square millimetre, which is why it cuts.

Real example (one you can see). Cut paper with scissors: the two blades pass each other with the sheet between them, a pure shear on the plane between the blades. Dull scissors have a larger area and need a larger force for the same stress.

Plausible wrong answer. "The pin is in tension because the plates pull on it." The plates pull on the plates; the pin is pushed sideways at two places, and its cut sees a transverse force.

Check. What is the average shear stress in a 10\,\mathrm{mm} rivet carrying 15\,\mathrm{kN} in single shear? (15\,000/78.54 = 191\,\mathrm{MPa}.)

Narration spine. Two plates are pulled apart; a plane slices the pin between them; arrows appear lying in the cut, equal to P; the area is shaded and the ratio written.

Bridge. The count of cut planes changes the answer.

Nugget 02: A Bolt in Single or Double Shear: Count the Planes

Claim. The shear force on a plane is the load divided by the number of planes that share it. A pin joining a bar to the two prongs of a fork is cut on two planes at once, so each carries half.

The picture. A single-shear lap joint beside a double-shear clevis joint.

Check with numbers. The same 12\,\mathrm{mm} pin, 30\,\mathrm{kN}. Single shear: V = P, so \tau = 265.3\,\mathrm{MPa}. Double shear: V = P/2 = 15\,\mathrm{kN} per plane over the same 113.1\,\mathrm{mm^2}, so \tau = 15\,000/113.1 = 132.6\,\mathrm{MPa}. The second layout halves the stress by doubling the planes. Hearn's Example 1.3 is a double-shear coupling: a 50\,\mathrm{kN} pull on a 15\,\mathrm{mm} bolt gives \tau = 50\,000/(2\times176.7) = 141.5\,\mathrm{MPa}. The same layout with 60\,\mathrm{kN}, a 16\,\mathrm{mm} bolt and 50\times8\,\mathrm{mm} plates: bolt \tau = 60\,000/(2\times201.06) = 149.2\,\mathrm{MPa}, centre-plate tension 60\,000/400 = 150\,\mathrm{MPa}, each fork 30\,000/400 = 75\,\mathrm{MPa}.

The idea. Gere (Fig. 1-49) and Beer (1.2B, Figs. 1.16–1.19) both draw the free body of the bolt: each plane of the bolt carries the shear, and the bolt halves in double shear where the force splits equally. The split is equal only if the arrangement is symmetrical; a multi-plane pin whose planes are loaded unequally needs a free body for each plane. Beer's pin B in the boom frame (Lesson MM01) has 25\,\mathrm{kN} on one plane and 15\,\mathrm{kN} on the other.

The equation (definition with a geometric relationship: the number of planes):

Worked number. Beer's boom frame: pin C (25\,\mathrm{mm}, 50\,\mathrm{kN}) in single shear: A = 490.9\,\mathrm{mm^2}, \tau = 101.9\,\mathrm{MPa} (the book prints 102.0). Pin A in double shear with 40\,\mathrm{kN}: 20\,000/490.9 = 40.7\,\mathrm{MPa}. Pin B: the greater plane force is 25\,\mathrm{kN}: 25\,000/490.9 = 50.9\,\mathrm{MPa}.

Real example (one you can see). A bicycle chain pin sits between inner and outer plates, shearing on two planes; a door hinge pin is in single or double shear depending on the hinge.

Plausible wrong answer. "Double shear doubles the stress" or "P/2 over the whole area". The load splits over two planes, each with the full cross-section A: V = P/2, A unchanged.

Check. A 20\,\mathrm{mm} pin in double shear carries 60\,\mathrm{kN}. Stress? (30\,000/314.16 = 95.5\,\mathrm{MPa}.)

Narration spine. Two layouts; a dashed plane through the bolt in the first; two planes in the second with half-sized arrows; the stress is written under each.

Bridge. The pin also presses on the plates around it.

Nugget 03: Bearing Stress: Where the Bolt Pushes on the Plate

Claim. Besides cutting the bolt, the load presses it against the walls of the hole. That contact stress is the bearing stress \sigma_b, computed on the projected area, a rectangle of plate thickness times bolt diameter.

The picture. A bolt pressed against the wall of a hole.

The real pressure on the half-cylinder varies from a maximum at the middle of the contact to zero at its edges, and it is complicated to compute. The nominal value, force over the projected area, is what is conventionally compared with an allowable bearing stress (Gere: customary, and uniform is assumed). Any friction from tightening the bolt is neglected: that is conservative, as friction would carry some load.

Check with numbers. A 20\,\mathrm{mm} bolt pressing a 10\,\mathrm{mm} plate with 40\,\mathrm{kN}: A_b = td = 10\times20 = 200\,\mathrm{mm^2} and \sigma_b = 40\,000/200 = 200\,\mathrm{MPa}. In a double-shear joint, which thickness and which force? Each outer plate takes P/2 over its own thickness; the central bar takes P over its own (Gere Fig. 1-49). With a washer or nut, the bearing area is the ring \frac{\pi}{4}(d_w^2 - d_b^2).

The idea. Gere (Eq. 1-15) and Beer (Eq. 1.11) agree. Bearing is a compressive contact stress, a different failure mode (crushing or ovalising the hole) from shear (cutting the bolt) and from tension in the plate (Nugget 07).

The equation (definition, nominal bearing stress):

Worked number. \sigma_b = 200\,\mathrm{MPa} for the 20\,\mathrm{mm} bolt in a 10\,\mathrm{mm} plate. Beer's boom pin A: boom t = 30\,\mathrm{mm}, 40\,\mathrm{kN} on a 25\,\mathrm{mm} pin: 40\,000/(30\times25) = 53.3\,\mathrm{MPa}; on the double bracket (t = 2\times25 = 50\,\mathrm{mm}): 40\,000/(50\times25) = 32.0\,\mathrm{MPa}.

Real example (one you can see). An old bolted wooden joint with an oversized, ovalled hole: the bolt has crushed the wood grain where it bore, and the bolt itself is still straight.

Plausible wrong answer. "Use the full surface of the half-cylinder." The convention uses the projected area td, so allowable values are calibrated for it.

Check. The plate in the example is doubled to 20\,\mathrm{mm}. Bearing stress? (100\,\mathrm{MPa}.)

Narration spine. The bolt presses; pressure arrows fan out around the half-cylinder, then a dashed rectangle replaces them: t\cdot d.

Bridge. If the load is big enough, the bolt does not shear in two planes but a plug of the plate shears out.

Nugget 04: Punching a Hole Shears a Cylinder

Claim. A punch pushing through a plate shears a cylindrical surface whose area is the circumference of the hole times the plate thickness. The shear stress is the load over that area.

The picture. A punch cutting a disc from a plate.

Check with numbers. Gere's Example 1-9: a 20\,\mathrm{mm} punch through an 8.0\,\mathrm{mm} plate with 110\,\mathrm{kN}. The sheared area is A_s = \pi d t = \pi(20)(8) = 502.7\,\mathrm{mm^2}, so \tau = 110\,000/502.7 = 218.8\,\mathrm{MPa}. The punch itself carries 110\,000/314.16 = 350\,\mathrm{MPa} in compression. The plate shear is the smaller stress, but a material's shear strength is only half its compressive or tensile strength, so the plate fails first. The averages ignore impact (an actual punch strikes) and stress concentrations.

The idea. The same shear area governs a bolt with a washer pulled through a bracket (Gere Fig. 1-60e): the cylinder under the washer is \pi d_w t.

The equation (geometric relationship plus the definition of \tau):

Worked number. 502.7\,\mathrm{mm^2}, 218.8\,\mathrm{MPa}, punch 350\,\mathrm{MPa}. Required force to punch a 16\,\mathrm{mm} hole in a 6\,\mathrm{mm} plate with shear strength 300\,\mathrm{MPa}: P = \pi(16)(6)(300) = 90.5\,\mathrm{kN}.

Real example (one you can see). A paper punch cuts a clean disc with the circumference as its edge; a thicker stack needs a stronger hand because t grew.

Plausible wrong answer. "The shear area is the hole's circle, \pi d^2/4." The disc moves relative to the plate across its edge; the area is the edge, \pi d t.

Check. Doubling the plate thickness changes the punching force by what factor? (Doubles.)

Narration spine. The punch descends and a disc is pushed out; the cylindrical edge is unrolled into a rectangle \pi d \times t.

Bridge. Shear stresses always appear in pairs. The next idea explains why.

Nugget 05: Shear Stresses Come in Perpendicular Pairs

Claim. A shear stress on one face of an element requires equal shear stresses on the other three faces: two on the opposite faces with opposite directions, two on the perpendicular faces such that both point toward, or both away from, their common edge. This is complementary shear.

The picture. A small block with shear on its vertical faces and then on all four.

Take a block a \times b \times c (width a, height b, depth c) with shear stress \tau_1 on its two vertical faces. The forces are \tau_1 bc each, acting a distance a apart: a couple of moment \tau_1 abc. Nothing on the vertical faces can cancel it, since their forces are all vertical. The couple must be balanced by the horizontal faces: forces \tau_2 ac a distance b apart, a couple \tau_2 abc in the opposite sense. Setting the two moments equal gives \tau_1 = \tau_2. This is the step Beer takes with moments about the z axis (\tau_{xy} = \tau_{yx}) and that Venkataraman builds up one face at a time.

Check with numbers. With \tau_1 = 40\,\mathrm{MPa} on the vertical faces of a block 30\,\mathrm{mm}\times20\,\mathrm{mm}\times10\,\mathrm{mm}, the couple is 40\times(20\times10)=8000\,\mathrm N per face over a distance of 30\,\mathrm{mm}: 240\,\mathrm{N\,m}. The horizontal pair of faces (30\times10) needs \tau_2(300)(20)=240\,000\,\mathrm{N\,mm}, so \tau_2 = 40\,\mathrm{MPa} ✓.

The idea. Gere (Eq. 1-17) states two facts: (1) shear stresses on opposite faces are equal and opposite; (2) on adjacent faces they are equal and either both toward or both away from the shared edge. A state with shear stresses only is called pure shear. The result also holds when normal stresses are present, because the normal stresses on opposite faces are equal and opposite and have no moment.

Signs. Gere calls a face positive if its outward normal points along a positive axis; a shear stress is positive on a positive face when it points along a positive axis, or on a negative face when it points along a negative axis. Venkataraman, Muvdi and Beer agree on the subscript meaning: the second subscript names the direction of the stress, the first the face's normal. In pure shear all four stresses on the element are positive together.

The equation (physical law: moment equilibrium of the element):

Worked number. The torque-free block above: \tau_2 = \tau_1 = 40\,\mathrm{MPa}, regardless of the block's dimensions.

Real example (one you can see). Slide a deck of cards so the top moves sideways: the deck leans, and it will stay leaning only if something holds the bottom — the pair of horizontal and vertical shears. Twist a shaft: the same pairs exist on its surface (Part 3).

Plausible wrong answer. "A shear on the vertical faces is balanced by the shear on the opposite vertical face." They form a couple that is not balanced by themselves; the other faces must help.

Check. If the shear stress on the horizontal faces of an element is 25\,\mathrm{MPa}, what is it on the vertical faces? (25\,\mathrm{MPa}, directed so the four arrows meet or leave at corners.)

Narration spine. A block under vertical shears begins to rotate; arrows appear on the top and bottom to stop it; the four arrows cycle round the corners.

Bridge. Shear also deforms: an element changes shape.

Nugget 06: Shear Strain and the Shear Modulus

Claim. Shear deforms a right angle; the change in the angle is the shear strain \gamma. Within the elastic range the shear stress is proportional to it, \tau = G\gamma, with G the shear modulus (modulus of rigidity).

The picture. An element before and after shear.

The side lengths do not change: a pure shear element changes shape, not size. The angle is measured in radians, so \gamma is dimensionless. Gere and Beer sign it positive when the angle between two positive faces decreases. Beer is explicit that \gamma is the change of angle between the faces and not the rotation of one face: a rigid rotation of the element produces none.

The shear stress-strain curve. Direct-shear or torsion tests give a \tau–\gamma curve shaped like the tension curve of Lesson MM03, with a straight start of slope G. Shear yield and ultimate values are about half the tensile values (Gere: the shear yield of structural steel is 0.5–0.6 of its tensile yield).

Check with numbers. Venkataraman's Example 3.7: a brick-like block 200\,\mathrm{mm}\times60\,\mathrm{mm} in plan and 50\,\mathrm{mm} high has its top pushed with 120\,\mathrm{kN} and moves 2\,\mathrm{mm} relative to the fixed bottom. \gamma = \delta/h = 2/50 = 0.04\,\mathrm{rad}. \tau = 120\,000/(200\times60) = 10\,\mathrm{MPa}. G = \tau/\gamma = 10/0.04 = 250\,\mathrm{MPa}.

A bearing pad (Gere's Example 1-10): an elastomer pad, a = 75\,\mathrm{mm} by b = 60\,\mathrm{mm}, thickness h = 20\,\mathrm{mm}, G_e = 1.25\,\mathrm{MPa}, sheared by V = 0.8\,\mathrm{kN}. \tau = 800/(75\times60) = 0.178\,\mathrm{MPa}, \gamma = 0.178/1.25 = 0.1422\,\mathrm{rad}, and the top moves \delta = h\tan\gamma = 2.86\,\mathrm{mm} (small-angle: h\gamma = 2.84\,\mathrm{mm}, agreeing within 0.7\,\%). The model assumes \tau uniform through the thickness, which is only reasonable when a \gg h, since the stress is zero at the free edges.

The idea. For steel G \approx 75–80\,\mathrm{GPa}, for aluminium about 26–28\,\mathrm{GPa} (Gere's Appendix I gives 75 and 28; Venkataraman copies 79.3 and 25.5 from a different source; use a stated value). The three elastic constants E, \nu and G of an isotropic material are not independent: G = E/[2(1+\nu)]. With 0 < \nu < \frac{1}{2} this puts G between E/3 and E/2. Gere states the relation here; Beer derives it in his 2.8 and so does this book, in Lesson MM16. Beer's general Hooke's law (Eq. 2.29) adds three shear relations \gamma_{xy} = \tau_{xy}/G and so on to the three normal ones, and the shear stresses do not change the normal strains.

The equation (geometric relationship for \gamma, material model for Hooke's law in shear):

Worked number. G = 250\,\mathrm{MPa} (Venkataraman's block); pad deflection 2.86\,\mathrm{mm}. For steel (E = 200\,\mathrm{GPa}, \nu = 0.30): G = 200/2.6 = 76.9\,\mathrm{GPa}; aluminium (70\,\mathrm{GPa}, 0.33): 26.3\,\mathrm{GPa}. Beer's rubber block, G = 620.5\,\mathrm{MPa}, top plate moved 1.016\,\mathrm{mm} over 50.8\,\mathrm{mm}: \gamma = 0.020, \tau = 12.4\,\mathrm{MPa}.

Real example (one you can see). A rubber engine mount between two metal plates deflects sideways under a load by exactly this shear: a soft low-G pad, a large \delta. An elastomeric bridge bearing lets a girder slide a few millimetres with temperature.

Plausible wrong answer. "Shear strain is the length that the top moved." It is the angle \delta/h, not \delta.

Check. A block 40\,\mathrm{mm} high has its top displaced 0.5\,\mathrm{mm} with G = 80\,\mathrm{GPa}. What shear stress? (\gamma = 0.0125, \tau = 1000\,\mathrm{MPa}, a stress well above the elastic limit of any metal: the displacement in the question is already enormous for steel.)

Narration spine. A square leans into a parallelogram; \gamma is marked at one corner; a \tau–\gamma line is drawn with slope G; three constants E, \nu, G join by a formula.

Bridge. Real connections combine shear, bearing and tension.

Nugget 07: A Connection Has Several Failure Modes: Check Each

Claim. A bolted or pinned connection can fail in the bolt by shear, in the plate by bearing, in the plate by tension at the net section, by shear-out of a plug. Design means computing each stress and comparing it with its own allowable value; the smallest margin governs.

The picture. The five places a bolt in a bracket can fail.

Check with numbers. Take Gere's Example 1-11, the bracket of a basketball rim, reduced to one bolt with tension T = 6.708\,\mathrm{kN} and shear V = 437.5\,\mathrm N. Bolt d_b = 10\,\mathrm{mm}, washer d_w = 16\,\mathrm{mm}, bracket thickness t = 5\,\mathrm{mm}. Areas: bolt A_b = 78.54\,\mathrm{mm^2}, washer ring A_w = \frac{\pi}{4}(16^2 - 10^2) = 122.52\,\mathrm{mm^2}, bearing on the hole td = 50\,\mathrm{mm^2}, bracket shear \pi d_w t = 251.33\,\mathrm{mm^2}. Then:

(Gere's T = 6.708\,\mathrm{kN} comes from a moment of 1.006\,\mathrm{kN\,m} divided by the bolt pair's 0.15\,\mathrm m spacing, V = 1.75/4 = 437.5\,\mathrm N from the load shared by four bolts. In his second load case the bolt shear and bearing triple while the others change by about a quarter.) Each is compared with its own allowable: the five numbers are different kinds of stress and cannot be added.

The idea. Beer's Sample Problem 1.2 shows the design sequence. A tie bar carries P = 120\,\mathrm{kN} through a pin in double shear with t = 20\,\mathrm{mm}, with allowables \tau = 100, \sigma = 175 and \sigma_b = 350\,\mathrm{MPa}. Size the pin on shear: 60\,\mathrm{kN} per plane gives A = 600\,\mathrm{mm^2}, d = \sqrt{4\times600/\pi} = 27.6\,\mathrm{mm}, round up to 28\,\mathrm{mm}. Check bearing: 120\,000/(20\times28) = 214.3\,\mathrm{MPa} < 350. OK. End width: a = 60\,000/(175\times20) = 17.14\,\mathrm{mm}, so the bar's width at the hole is b = d + 2a = 28 + 34.3 = 62.3\,\mathrm{mm}. Away from the hole the whole load goes through the bar's own section: h = 120\,000/(175\times20) = 34.3\,\mathrm{mm} deep, so use 35\,\mathrm{mm}. If the bearing check had failed, the pin diameter would be increased and the checks repeated.

The equation (design rule, each mode against its own limit):

Worked number. Pin 28\,\mathrm{mm} (shear \tau = 60\,000/615.75 = 97.4\,\mathrm{MPa}), bearing 214\,\mathrm{MPa}, bar end width a = 17.14\,\mathrm{mm}. Allowable stresses and factors of safety are Lesson MM06's subject.

Real example (one you can see). A school-desk chair's seat bolts: the bolt shank can shear, the plywood can crush where the bolt bears, the washer can pull through the plywood. The cheapest joint is the one where the three limits are close together.

Plausible wrong answer. "If the bolt passes the shear check, the joint is safe." Bearing, net-section tension and shear-out are different modes and need their own checks.

Check. The pin of the tie bar is made 24\,\mathrm{mm}. Does the shear check pass at 100\,\mathrm{MPa}? (60\,000/452.4 = 132.6\,\mathrm{MPa} > 100: it fails.)

Narration spine. A bolt in a bracket; five red arrows light up one after another with their own formula and number; the smallest margin turns the joint red.

Synthesis

What no single nugget shows. Shear is the stress state of anything that slides: pins and bolts, punched plates, glue lines, twisted shafts and beams. Normal and shear stress are the two kinds of intensity that a surface can carry; every stress state is a combination. Because shear on one face forces shear on the perpendicular face, it can never be thought of in one direction alone, and the shear strain, a change of shape, is just the elastic counterpart of the normal strain, a change of size.

• Central idea. Shear stress is force along a surface per area; it comes in perpendicular pairs and makes a right angle change.
• Key relationship. \tau = V/A (counting the shear planes), \tau_{xy} = \tau_{yx} and \tau = G\gamma.
• Main assumption. The shear is uniform over the area (really not so), friction is neglected and the material is linear below its elastic limit.
• What to explore next. How shear peaks in the middle of a beam (Lesson MM31), how twisting a shaft is shear (Lesson MM14), and the allowable stresses and safety factors that complete a connection design (Lesson MM06).

Computed views

Python (numpy). MM05_shear_joints.py computes the pin, bearing, punch, tie-bar, bracket and pad numbers above.

pin 12 mm, 30 kN, single shear: 265.3 MPa
pin 12 mm, 30 kN, double shear: 132.6 MPa
bearing 40 kN, t 10, d 20: 200 MPa
coupling 60 kN, 16 mm bolt: bolt 149.2, plate 150, fork 75 MPa
punch 20 mm in 8 mm plate, 110 kN: shear 218.8 MPa, punch 350 MPa
tie bar 120 kN, tau_allow 100 MPa: d = 27.6 mm
  bearing with d 28, t 20: 214.3 MPa
  end width a = 17.14 mm, h = 34.29 mm
bolt, tension 6708 N, shear 437.5 N: normal 85.4, shear 5.57, bearing 8.75, washer 54.7, bracket 26.7 MPa
pad: tau 0.178 MPa, gamma 0.1422
  delta 2.86 mm exact, 2.84 mm small-angle
G steel (200 GPa, 0.30): 76.9 GPa
G aluminium (70 GPa, 0.33): 26.3 GPa

MATLAB (Octave-compatible). MM05_shear_joints.m does the same joints with anonymous functions (no local functions, so the file runs unchanged in both programs).

pin 12 mm, 30 kN, single shear: 265.3 MPa
pin 12 mm, 30 kN, double shear: 132.6 MPa
bearing 40 kN, t 10, d 20: 200 MPa
coupling 60 kN, 16 mm bolt: bolt 149.2, plate 150, fork 75 MPa
punch 20 mm in 8 mm plate, 110 kN: shear 218.8 MPa, punch 350 MPa
tie bar 120 kN, tau_allow 100 MPa: d = 27.6 mm
  bearing with d 28, t 20: 214.3 MPa
pad: tau 0.178 MPa, gamma 0.1422, delta 2.86 / 2.84 mm
G steel (200 GPa, 0.30): 76.9 GPa
G aluminium (70 GPa, 0.33): 26.3 GPa

SOLIDWORKS Simulation exercise (our own; a pinned lap joint: average shear against the local peak). The hand average is a lower bound on the real peak.

• Geometry. Two steel plates 40\,\mathrm{mm} wide and 10\,\mathrm{mm} thick, overlapping 60\,\mathrm{mm}, joined by a 12\,\mathrm{mm} pin through 12\,\mathrm{mm} holes (a sliding fit in contact, no clearance); the pin has one shear plane at the interface.
• Restraint. Fixed geometry on the far end face of one plate.
• Load. 24\,\mathrm{kN} tensile on the far end face of the other plate.
• Contact. No-penetration contact between pin and holes and between the plates; no friction (the hand method ignores it).
• Mesh. Standard solid mesh with mesh control of 1\,\mathrm{mm} on the pin and the hole edges; refine once.
• Hand numbers. Pin shear \tau_{\text{aver}} = 24\,000/113.1 = 212.2\,\mathrm{MPa}; bearing 24\,000/(10\times12) = 200\,\mathrm{MPa}; plate net-section tension 24\,000/[(40-12)\times10] = 85.7\,\mathrm{MPa}.
• What to look for. Probe the shear stress on the plane between the plates across the pin: the average of the plot over the cross-section should be about 212\,\mathrm{MPa}, and the local peak higher, often noticeably, near the pin's edges, because the load is carried unevenly by the contact. Also compare the maximum normal stress in the plate at the hole edge with 85.7\,\mathrm{MPa}: the peak is a stress concentration (Lesson MM12). Record the peak-to-average ratio.

(This exercise was written for the lesson and has not been run in SOLIDWORKS here.)

Traps

• Treating shear like tension. The force lies in the plane of the cut; allowable shear values are about half the tensile ones. (Nugget 01.)
• Forgetting to count the shear planes. V = P in single shear, P/2 in double. (Nugget 02.)
• Using the whole hole surface for bearing. The convention is the projected area td. (Nugget 03.)
• Wrong thickness in a multi-plate joint. Each plate bears with its own force over its own thickness. (Nugget 03.)
• Taking the shear area of a punch as the hole's circle. It is the cylinder \pi d t. (Nugget 04.)
• Believing the average is the peak. The real shear distribution is non-uniform. (Nugget 01.)
• Mixing up shear strain and displacement. \gamma is an angle in radians. (Nugget 06.)
• A shear stress on one face only. It always has a partner on the perpendicular faces. (Nugget 05.)
• Checking only the bolt. Bearing, net-section tension and shear-out also govern. (Nugget 07.)

Self-check

1. A 12\,\mathrm{mm} pin carries 30\,\mathrm{kN} in single shear. Find the shear stress, and the stress if the joint is changed to double shear.
2. A coupling joins a 50\times8\,\mathrm{mm} centre plate to two 50\times8\,\mathrm{mm} fork plates by a 16\,\mathrm{mm} bolt. Under 60\,\mathrm{kN}, find the shear stress in the bolt and the tensile stress in the centre plate and in each fork plate (ignore the hole).
3. A 20\,\mathrm{mm} bolt presses a 10\,\mathrm{mm} plate with 40\,\mathrm{kN}. Find the bearing stress.
4. A 16\,\mathrm{mm} hole is punched in a 6\,\mathrm{mm} plate whose shear strength is 300\,\mathrm{MPa}. Find the force and the compressive stress in the punch.
5. An elastomer pad, 80\,\mathrm{mm}\times60\,\mathrm{mm}, 25\,\mathrm{mm} thick, G = 1.0\,\mathrm{MPa}, carries a shear force of 1.2\,\mathrm{kN}. Find \tau, \gamma and the deflection with and without the small-angle approximation.
6. Find G for steel (E = 200\,\mathrm{GPa}, \nu = 0.3) and aluminium (70\,\mathrm{GPa}, 0.33).
7. A block of shear stress \tau = 40\,\mathrm{MPa} on its vertical faces is 30\times20\times10\,\mathrm{mm}. Show by moments that the horizontal faces carry 40\,\mathrm{MPa}.
8. A block 150\,\mathrm{mm}\times50\,\mathrm{mm} in plan, 40\,\mathrm{mm} high, has its top plate pushed with 90\,\mathrm{kN} and moved 1.5\,\mathrm{mm}. Find \tau, \gamma and G.
9. A tie bar carries 80\,\mathrm{kN} through a double-shear pin, bar thickness 15\,\mathrm{mm}, with \tau_{\text{allow}} = 100\,\mathrm{MPa} and \sigma_{b,\text{allow}} = 350\,\mathrm{MPa}. Size the pin (to the next whole millimetre) and check bearing.
10. Why does the average shear stress underestimate the real maximum in a pinned connection? Name two causes.
11. A steel bolt has tensile yield 250\,\mathrm{MPa}. Using a shear yield of 0.6 times the tensile yield, what single-shear load can a 20\,\mathrm{mm} bolt carry before yielding in shear?

Fully worked solutions

1. A = 113.1\,\mathrm{mm^2}. Single: 30\,000/113.1 = 265.3\,\mathrm{MPa}. Double: 15\,000/113.1 = 132.6\,\mathrm{MPa}.

2. Bolt in double shear: \tau = 60\,000/(2\times201.06) = 149.2\,\mathrm{MPa}. Centre plate: 60\,000/400 = 150\,\mathrm{MPa}. Each fork: 30\,000/400 = 75\,\mathrm{MPa}.

3. \sigma_b = 40\,000/(10\times20) = 200\,\mathrm{MPa}.

4. Shear area \pi\times16\times6 = 301.6\,\mathrm{mm^2}; P = 300\times301.6 = 90.5\,\mathrm{kN}. Punch: 90\,478/201.06 = 450\,\mathrm{MPa}.

5. \tau = 1200/(80\times60) = 0.25\,\mathrm{MPa}; \gamma = 0.25/1.0 = 0.25\,\mathrm{rad}; exact \delta = 25\tan 0.25 = 6.38\,\mathrm{mm}; small-angle 25\times0.25 = 6.25\,\mathrm{mm} (the small-angle formula is 2\,\% low here, because 0.25\,\mathrm{rad} is not very small).

6. Steel: 200/(2\times1.3) = 76.9\,\mathrm{GPa}. Aluminium: 70/(2\times1.33) = 26.3\,\mathrm{GPa}.

7. Vertical-face forces \tau_1(20\times10) = 8000\,\mathrm N each, a distance 30\,\mathrm{mm} apart: couple 240\,000\,\mathrm{N\,mm}. Horizontal faces have area 30\times10 = 300\,\mathrm{mm^2} and are 20\,\mathrm{mm} apart: couple \tau_2\times300\times20. Equating: \tau_2 = 240\,000/6000 = 40\,\mathrm{MPa}.

8. \tau = 90\,000/(150\times50) = 12\,\mathrm{MPa}; \gamma = 1.5/40 = 0.0375\,\mathrm{rad}; G = 12/0.0375 = 320\,\mathrm{MPa}.

9. Each plane: 40\,\mathrm{kN}; A \ge 40\,000/100 = 400\,\mathrm{mm^2}; d \ge \sqrt{4\times400/\pi} = 22.57\,\mathrm{mm}, use 23\,\mathrm{mm}. Bearing: 80\,000/(15\times23) = 231.9\,\mathrm{MPa} < 350: passes.

10. The real distribution is zero at free surfaces and larger inside, so the peak exceeds the average; also the contact between pin and holes is uneven (the load sits at the hole edges), and stress concentrations occur at holes and corners.

11. Shear yield = 0.6\times250 = 150\,\mathrm{MPa}; A = 314.16\,\mathrm{mm^2}; P = 150\times314.16 = 47.1\,\mathrm{kN}.