Study Companion: MM06 — Allowable Stress and Design for Axial Load and Direct Shear

A tension rod will break at a stress of 600\,\mathrm{MPa}, and it will carry 40\,\mathrm{kN} in service. What diameter should it have? If we size it to break at exactly 40\,\mathrm{kN} it will fail the first time anything goes slightly wrong: the load is a little more than planned, the steel a little weaker than the sample, a notch a little sharper. Every real design leaves a margin, and the whole subject of this lesson is how that margin is stated, how big it should be, and how it turns the stresses of Lessons MM01–MM05 into the sizes of rods, pins and plates.

Sources: Gere and Goodno, Mechanics of Materials, 9th edition (SI), sections 1.9–1.10 (the syllabus for this lesson); Beer, Johnston, DeWolf and Mazurek, 8th edition, section 1.5; Hearn, Mechanics of Materials 1, sections 1.16–1.17; Muvdi and Elhouar, section 1.4.5. Hearn's two sections and Muvdi's single page are definitions only, with no worked example; the load factor (Hearn 1.17) and load-and-resistance-factor design (Beer 1.5D) appear in one book each. The strength and stress values used in the examples are the references' data, in SI. The average stresses of shear and bearing (Lesson MM05) and of tension (Lesson MM02) are the stresses being limited.

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 rod must carry 40\,\mathrm{kN}. Its steel has an ultimate strength of 600\,\mathrm{MPa} and the design calls for a factor of safety of 3.3. What diameter? The plausible answer is "the diameter that makes \sigma = 600\,\mathrm{MPa}", 9.2\,\mathrm{mm}, a rod that would break at exactly the service load. The right answer is 17\,\mathrm{mm} (stock size); see Nugget 04.

Nugget 01: A Design Needs a Margin: the Factor of Safety

Claim. The factor of safety n is how many times stronger the structure is than it needs to be: actual strength over required strength. It must exceed 1, and the margin of safety is the part above 1.

The picture. A bar showing the service load and the failure load on the same scale.

Strength may mean a load capacity or a stress, and failure may mean fracture or any loss of function, such as a deflection too large to use, which can happen well below collapse. The ratio is the same idea in each case.

Check with numbers. A structure that fails at 240\,\mathrm{kN} and is meant to carry 150\,\mathrm{kN} has n = 240/150 = 1.6 and a margin of 0.6 (60 per cent). Gere's example: a strength 1.75 times the required one has n = 1.75 and a margin of 0.75. A margin of zero or less means the structure is presumed to fail. Beer measures the strength of a material in a test: a bolt in double shear that fails at a load P_U has an ultimate shear stress \tau_U = P_U/(2A); a 12\,\mathrm{mm} bolt failing at 90\,\mathrm{kN} in double shear gives \tau_U = 90\,000/(2\times113.1) = 397.9\,\mathrm{MPa}.

The idea. Gere (Eq. 1-24), Beer (Eqs. 1.25, 1.26), Hearn (Eq. 1.9) and Muvdi (Eq. 1.24) all define it as a ratio, in slightly different forms. Beer's load form is \text{F.S.} = P_{\text{ultimate}}/P_{\text{allowable}} and his stress form is \text{F.S.} = \sigma_{\text{ultimate}}/\sigma_{\text{allowable}}. The two agree only when stress is proportional to load; near the ultimate load the relation is not linear, so the stress form is a convention, but it is the widely used one ("allowable-stress method"). Hearn defines the stress ratio with the maximum (failure) stress, and notes that because plastic deformation is seldom acceptable the yield or proof stress is often used instead; if a quoted factor does not say which, assume the ultimate version. This book says which each time. The "margin of safety" is Gere's term (also in a Beer footnote); Hearn and Muvdi do not use it.

The equation (definition):

n = \frac{\text{actual strength}}{\text{required strength}},\qquad \text{margin of safety} = n - 1.

Worked number. n = 3.0 and margin 2.0 for the failure and service loads in the picture; n = 1.6, margin 0.6 for the 240/150 structure.

Real example (one you can see). A lift cable's rated load is a fraction of its breaking load: the plate in the cabin states 1000\,\mathrm{kg} although the cables would carry several times that before breaking.

Plausible wrong answer. "A factor of safety of 3 means the structure is three times as safe." It means the capacity is three times the required load; how safe that is depends on how well the loads and strengths are known (Nugget 05).

Check. A pin fails in a test at 50\,\mathrm{kN} and is to carry 20\,\mathrm{kN}. Find n and the margin. (2.5 and 1.5.)

Narration spine. A load scale; service load and failure load are marked; the gap is shaded and the ratio written.

Bridge. The factor applies to a load; engineers work in stresses, so we need the stress that corresponds.

Nugget 02: Allowable Stress: the Strength Divided by the Factor

Claim. The allowable stress (working or design stress) is a material strength divided by the factor of safety: the highest stress we permit at any point in service. The strength used is the yield strength for ductile materials and the ultimate strength for brittle ones.

The picture. A stress-strain curve with the allowable stress below the yield point.

Why yield, for ductile metals? Yielding begins when the yield stress is reached at any point, and beyond it the member does not return to its shape. Keeping the stress below it everywhere keeps the member elastic, so the allowable stress must not be exceeded anywhere. Why ultimate, for brittle materials? Concrete, some plastics, wood and high-strength steels have no clear yield point; their strength is their ultimate stress, and brittle failures give no warning, so the factors applied to it are larger.

Check with numbers. Mild steel with \sigma_Y = 250\,\mathrm{MPa} and n_1 = 1.67 (the factor Gere quotes for yielding in tension in building design): \sigma_{\text{allow}} = 250/1.67 = 149.7\,\mathrm{MPa} \approx 150\,\mathrm{MPa}. The same 1.67 on yield corresponds to about 2.8 on the ultimate strength for a steel whose ultimate is 1.68 times its yield, since 1.67\times1.68 = 2.81. For shear, the yield strength in shear is about 0.5–0.6 of the tensile yield, so with \tau_Y = 0.6\sigma_Y = 150\,\mathrm{MPa}: \tau_{\text{allow}} = 150/1.67 = 89.8\,\mathrm{MPa}. A brittle example: cast iron with a compressive ultimate of 800\,\mathrm{MPa} and a factor of 4 has \sigma_{\text{allow}} = 200\,\mathrm{MPa}.

The idea. Gere's Eqs. 1-26 to 1-28 and Beer's 1.26 give the same rule: on yield, \sigma_{\text{allow}} = \sigma_Y/n_1 and \tau_{\text{allow}} = \tau_Y/n_2; on the ultimate, \sigma_{\text{allow}} = \sigma_U/n_3 and \tau_{\text{allow}} = \tau_U/n_4. Muvdi adds that the words "allowable", "working" and "service" attached to a stress, strain or deformation always mean a permissible value with the safety factor already built in, and that such values are codified.

The equation (design rule):

\sigma_{\text{allow}} = \frac{\sigma_Y}{n_1}\ \text{or}\ \frac{\sigma_U}{n_3},\qquad \tau_{\text{allow}} = \frac{\tau_Y}{n_2}\ \text{or}\ \frac{\tau_U}{n_4}.

Worked number. \sigma_{\text{allow}} = 149.7\,\mathrm{MPa} (steel on yield); 200\,\mathrm{MPa} (cast iron in compression on ultimate, n = 4). For Beer's rod: \sigma_U = 600\,\mathrm{MPa}, n = 3.3 gives 600/3.3 = 181.8\,\mathrm{MPa}.

Real example (one you can see). A road sign says "max 7.5 t" on a bridge: the posted value is the allowable load; the bridge would carry more before it was damaged.

Plausible wrong answer. "Use the ultimate strength with the yield factor." The factors differ because the strengths do: 1.67 goes with yield, about 2.8 with ultimate; mixing them is unsafe in one direction and wasteful in the other.

Check. A steel with \sigma_Y = 340\,\mathrm{MPa} and n_1 = 1.67: allowable stress? (203.6\,\mathrm{MPa}.)

Narration spine. The curve, the yield line, a lower dashed line for the allowable stress, and the ultimate line above both; the arithmetic is shown as a division.

Bridge. From an allowable stress to an allowable load takes the area, and there is more than one stress to check.

Nugget 03: Allowable Load: Check Every Way the Connection Can Fail

Claim. Each mode of failure has its own allowable stress and its own area; the allowable load for each is the product; the smallest of them is the allowable load of the connection.

The picture. A hanger loaded through a bolt, with the failure modes listed.

Allowable loads follow from the allowable stresses of Nugget 02 times the right area. In tension the area is the net area (the gross area minus any hole); in compression, when a bolt fills the hole, the gross area may be used; in direct shear it is the pin cross-section in single shear and twice that in double; in bearing it is the projected area t\,d. The relations assume a uniform stress over each area and a compression member that does not buckle (Chapter 11 of Gere).

Check with numbers. Gere's Example 1-12. Hanger 38\,\mathrm{mm}\times13\,\mathrm{mm}, widened to 75\,\mathrm{mm} at bolt 3 (25\,\mathrm{mm} diameter, double shear); bolts 1 and 2 are 12\,\mathrm{mm} with 28\,\mathrm{mm} washers on 9.5\,\mathrm{mm} clip angles. Allowables (MPa): hanger 110 (and 75 at the hole, lower because of the stress concentration there), bearing on bolt 3 180, shear in bolt 3 45, bolts 1 and 2 in tension 160, bearing under washers 65, shear in the angle 35.

Mode	Formula	Allowable load
(a) hanger tension	110\times(38\times13)	54.3\,\mathrm{kN}
(b) net section at bolt 3	75\times(75-25)\times13	48.8\,\mathrm{kN}
(c) bearing on bolt 3	180\times25\times13	58.5\,\mathrm{kN}
(d) shear of bolt 3 (double)	2\times45\times\frac{\pi}{4}25^2	44.2\,\mathrm{kN}
(e) tension of bolts 1, 2	160\times2\times\frac{\pi}{4}12^2	36.2\,\mathrm{kN}
(f) bearing under washers	65\times2\times\frac{\pi}{4}(28^2-12^2)	65.3\,\mathrm{kN}
(g) shear through angle	35\times2\times\pi\times28\times9.5	58.5\,\mathrm{kN}

The smallest is (e): P_{\text{allow}} = 36.2\,\mathrm{kN}. The load that the connection may carry is the load that its weakest mode may carry.

The idea. The seven checks come from three of Lesson MM05's stresses plus tension. They are averages; a refined analysis would add the weight of the hanger (Lesson MM02) and the local effects near holes (Lesson MM12). Gere frames the example in four steps (conceptualise, categorise, analyse, finalise), as every example in this book does.

The equation (design rule):

P_{\text{allow}} = \min\left\{\sigma_{\text{allow}}A_{\text{net}},\ \tau_{\text{allow}}A_{\text{shear}},\ \sigma_{b,\text{allow}}A_b\right\}.

Worked number. P_{\text{allow}} = 36.2\,\mathrm{kN}, controlled by the tension of the two small bolts. Using a larger bolt for 1 and 2 would let mode (d), 44.2\,\mathrm{kN}, take over.

Real example (one you can see). A chain is as strong as its weakest link, and a connection is as strong as its weakest check. Picture a seat belt anchorage: the webbing, the stitching, the bolt, the sheet metal around the bolt must each hold; the smallest one sets the limit.

Plausible wrong answer. "The weakest part governs." The weakest check governs; two checks may apply to the same part (bolt 3 in shear and bearing).

Check. Which mode would govern if the washers were reduced to 20\,\mathrm{mm} diameter? (Mode (f): 65\times2\times\frac{\pi}{4}(20^2-12^2) = 26.1\,\mathrm{kN}, which is below 36.2; the washers now govern.)

Narration spine. A load path is traced through the connection; seven labelled checks pop up with their numbers; the shortest bar of a bar chart is highlighted.

Bridge. Analysis asks "how much can it carry"; design asks "what size must it be".

Nugget 04: Design Runs the Check Backwards: Required Area and Stock Sizes

Claim. Design inverts analysis: from the load and the allowable stress, find the area, and from the area, the dimension. The required area is A_{\text{req}} = \text{load}/\text{allowable stress}, then rounded up to a stock size. Order matters: a pin is sized before the plate it bears on.

The picture. A bracket with a control rod and a pin, with the loads that fix the sizes.

The sequence follows the load path. First the free body gives the forces (equilibrium of the bracket gives the rod force and the pin reaction). Then each member: rod from tension, pin from shear, bracket from bearing. Rounding up to stock sizes is part of the design, not an afterthought.

Check with numbers. Beer's Sample Problem 1.3, with the rod strength \sigma_U = 600\,\mathrm{MPa}, the pin strength \tau_U = 350\,\mathrm{MPa}, allowable bearing 300\,\mathrm{MPa} and n = 3.3. The loads are 50\,\mathrm{kN} at 0.3\,\mathrm m and 15\,\mathrm{kN} at 0.6\,\mathrm m from the pin line; the rod pulls at 0.6\,\mathrm m. Moments about C: P(0.6) - 50(0.3) - 15(0.6) = 0, so P = 40\,\mathrm{kN}; the pin reaction has components C_x = 40\,\mathrm{kN} and C_y = 65\,\mathrm{kN}, resultant \sqrt{40^2 + 65^2} = 76.3\,\mathrm{kN}.

Rod: \sigma_{\text{all}} = 600/3.3 = 181.8\,\mathrm{MPa}, A_{\text{req}} = 40\,000/181.8 = 220.0\,\mathrm{mm^2}, d = \sqrt{4\times220.0/\pi} = 16.74\,\mathrm{mm}; stock size 17\,\mathrm{mm}. (The puzzle: 9.2\,\mathrm{mm} would give \sigma = 600\,\mathrm{MPa}, the break stress; 17\,\mathrm{mm} gives 176\,\mathrm{MPa}.)
Pin C in double shear: \tau_{\text{all}} = 350/3.3 = 106.1\,\mathrm{MPa}; each plane carries 76.3/2 = 38.15\,\mathrm{kN}; A_{\text{req}} = 38\,150/106.06 = 360\,\mathrm{mm^2}; d = 21.4\,\mathrm{mm}; use 22\,\mathrm{mm}.
Bearing on the supports: each of the two supports takes 76.3/2 over 22\,\mathrm{mm}\times t: A_{\text{req}} = 38\,150/300 = 127.2\,\mathrm{mm^2}, so t = 127.2/22 = 5.78\,\mathrm{mm}; use 6\,\mathrm{mm}.

The pin diameter was needed to get the thickness, which is why the pin comes first.

A second example in the cable-and-pin style of Gere's Example 1-13. A cable of allowable normal stress 125\,\mathrm{MPa} carries T = 9.527\,\mathrm{kN} (from the statics of a canopy, which is Gere's problem and not repeated here): A = 9527/125 = 76.2\,\mathrm{mm^2}, d = 9.85\,\mathrm{mm}. Pins in double shear with \tau_{\text{allow}} = 80\,\mathrm{MPa} and forces 9.527, 13.96 and 18.53\,\mathrm{kN} need A = F/(2\tau_{\text{allow}}) = 59.54, 87.25 and 115.8\,\mathrm{mm^2}, so d = 8.71, 10.54 and 12.14\,\mathrm{mm}.

The idea. Gere (1.10) states the contrast: in analysis the structure is known and the response is found; in design the response is prescribed and the structure is found. Design is usually longer and often involves repeated analysis. Strength (Eq. 1-33, A_{\text{req}} = \text{load}/\sigma_{\text{allow}}) is only one of the criteria; others are stiffness (limits on stretching, bending and twisting, next parts), stability (buckling of slender compression members, Part 10) and optimisation (least weight or cost). Loads are active forces from outside, known in advance; reactions are passive forces at supports, found in the analysis.

The equation (design rule, valid where the stress is uniform over the area):

A_{\text{req}} = \frac{\text{load}}{\text{allowable stress}},\qquad d = \sqrt{\frac{4A_{\text{req}}}{\pi}}.

Worked number. Rod 16.74\to17\,\mathrm{mm}, pin 21.4\to22\,\mathrm{mm}, support thickness 5.78\to6\,\mathrm{mm}.

Real example (one you can see). Look at the pin of a tow hitch: its diameter is a stock size (e.g. 19 or 25\,\mathrm{mm}), larger than a strict calculation for the tow load would need.

Plausible wrong answer. "Round to the nearest millimetre." Round up to the next standard size; rounding down reduces the factor of safety below the design value.

Check. A cable carries 9.5\,\mathrm{kN} at \sigma_{\text{allow}} = 125\,\mathrm{MPa}. Minimum diameter? (A = 76.0\,\mathrm{mm^2}, d = 9.84\,\mathrm{mm}.)

Narration spine. Free body, then three size calculations in the order rod, pin, bearing; each result snaps to a stock size with an animation of rounding up.

Bridge. All of this took n as given. Where does the number come from?

Nugget 05: Choosing the Factor: What the Number Pays For

Claim. The factor of safety pays for what is not known: the loads, the material, the construction, the analysis. A larger factor buys protection and costs weight and money; codes fix many values, but they come from the same considerations.

The picture. A balance between the cost of a failure and the cost of material.

Too low a factor is likely to fail. Too high wastes material and may produce a structure too heavy to use (an aircraft). The values are set by groups of experienced engineers on a probabilistic basis and written into codes, some enacted into law.

Check with numbers. Typical values met in the references: 1.67 for yield in tension in building design (Gere), about 2.8 on the ultimate for the same steel, 2.5 for low-consequence static loads up to 10 for shock loading or high risk to life (Hearn), 3.0 in Beer's chapter problems, 3.3 in his Sample Problem 1.3 and 4.0 for cast iron in compression on ultimate (Gere). Take a rod of 40\,\mathrm{kN} with \sigma_U = 600\,\mathrm{MPa} (the sweep in the MATLAB view): the diameter grows from 11.3\,\mathrm{mm} at n = 1.5 to 20.6\,\mathrm{mm} at n = 5, and the mass per metre rises from 0.78 to 2.62\,\mathrm{kg}: every extra unit of n costs material.

The idea. Beer lists seven things the factor must cover:

variation in material properties and dimensions, and residual stresses from manufacture;
the number of load cycles (fatigue);
the type of loading and the chance of change in use (larger for dynamic, cyclic or impulsive loads);
the type of failure (brittle collapse and buckling give no warning, yielding does: use more where there is none);
uncertainty in the analysis;
deterioration from poor maintenance or corrosion;
the importance of the member to the whole (secondary bracing may use less).

Gere adds the probability of accidental overload, the accuracy of construction and the consequences of failure. Hearn emphasises that no realistic basis exists for one value, and that it is chosen from the social, human-safety and economic consequences. Muvdi stresses that the actual lifetime factors are impossible to foresee. Codes: AISC for steel buildings, ACI for concrete, AF&PA for timber, AASHTO for highway bridges.

The equation (judgement, no formula): n is chosen from the list; it is not calculated from it.

Worked number. n from 1.5 to 5 for the same rod: d = 11.28 to 20.60\,\mathrm{mm} (MATLAB view).

Real example (one you can see). A footbridge over a motorway carries a lower probability of a full crowd than a stadium stand; a stand over a crowd has the larger factor.

Plausible wrong answer. "A bigger factor is always safer." A heavier structure may carry its own extra weight badly, and a design that is hard to build may be built worse.

Check. Which would take the larger factor: a lifting hook, or a shelf bracket for books? (The hook: dynamic load, serious consequence, fatigue.)

Narration spine. The balance diagram; a list of seven considerations slides in; the chosen number settles at the crossing.

Bridge. Two refinements of the factor appear in the references: one for failure by buckling, one that separates the uncertainty of the loads from that of the structure.

Nugget 06: Load Factor, and Load and Resistance Factor Design

Claim. The factor of safety can be stated on the load (the load factor: failure load over working load) when no stress limit describes the failure, and the uncertainty in the loads can be separated from that in the structure's strength (load and resistance factor design, LRFD).

The picture. The two ways of putting the margin in.

Check with numbers. Load factor (Hearn, Eq. 1.10): a strut collapses under 90\,\mathrm{kN} and the working load is 30\,\mathrm{kN}: load factor = 90/30 = 3.0. It is the same ratio as Beer's load-form factor of safety; Hearn keeps the name "factor of safety" for the stress ratio and uses "load factor" for the load ratio, which is the right measure for a strut, where buckling cannot be described by a yield stress, and for plastic limit design.

LRFD (Beer, Eq. 1.27): a design is acceptable if

\gamma_D P_D + \gamma_L P_L \le \phi\,P_U.

The dead load P_D is the structure's own weight, the live load P_L varies with time, P_U is the ultimate load; the load factors \gamma are greater than 1 (live load larger than dead) and cover uncertainty in the loads; the resistance factor \phi is less than 1 and covers uncertainty in the strength. Illustrative factors (code values differ; these are assumed here): \gamma_D = 1.2, \gamma_L = 1.6, \phi = 0.9, for P_D = 40\,\mathrm{kN} and P_L = 60\,\mathrm{kN}. Demand = 1.2\times40 + 1.6\times60 = 144\,\mathrm{kN}, so P_U \ge 144/0.9 = 160\,\mathrm{kN}. The equivalent factor of safety, ultimate load over total service load, is 160/(40 + 60) = 1.60 (Beer's problems 1.57 and 1.58 ask for exactly this comparison).

The idea. One factor n lumps all uncertainty together; LRFD assigns it separately to loads and to strength, so a design with a well-known dead load and a poorly known live load can be treated fairly. Where stress is proportional to load (linear, elastic, no buckling), the stress factor and the load factor are the same number.

The equation (design rule):

\text{load factor} = \frac{\text{load at failure}}{\text{allowable working load}},\qquad \gamma_D P_D + \gamma_L P_L \le \phi P_U.

Worked number. Load factor 3.0; LRFD requires P_U \ge 160\,\mathrm{kN}, equivalent to n = 1.60.

Real example (one you can see). A building code prints a table of load combinations like "1.2 D + 1.6 L"; each line is an instance of the LRFD inequality, with factors set by the code.

Plausible wrong answer. "LRFD is a different, less safe method." It enforces a similar margin; it distributes it by uncertainty (loads vs strength). With the assumed factors it came to 1.60 in the example, compared with 1.67 of the allowable-stress design for yielding in tension.

Check. If the live load rose to 80\,\mathrm{kN} with the same factors, what P_U? ((48 + 128)/0.9 = 195.6\,\mathrm{kN}.)

Narration spine. Two scales; on the first, one gap; on the second, the loads are pushed up and the strength pushed down by separate factors until they just meet.

Synthesis

What no single nugget shows. Every design equation in this lesson is a comparison of a stress (or load) from Lessons MM01–MM05 with a limit from Lesson MM03 reduced by a factor. What the factor means decides the design: a stress ratio on yield, a stress ratio on ultimate, a load ratio, or separate factors on loads and strength. The mechanics is the easy part; the judgement is in the number.

Central idea. A margin turns a strength into an allowable value, and a connection is as good as its weakest check.
Key relationship. n = strength / required, \sigma_{\text{allow}} = \sigma_Y/n (or \sigma_U/n), A_{\text{req}} = P/\sigma_{\text{allow}}.
Main assumption. Uniform stress over each area, no buckling, and a factor that really covers what is unknown; stresses are averages and local peaks (Lesson MM12) are not included.
What to explore next. Deformation of axially loaded members (Lesson MM07): strength is one limit, stiffness the other.

Computed views

Python (numpy). MM06_allowable_stress.py computes the allowable stresses, the seven hanger checks, the rod, pin and plate sizes and the LRFD comparison.

allowable stress, yield 250 MPa, n = 1.67: 149.7 MPa
  on ultimate: 1.67 * 1.68 = 2.81
allowable stress, cast iron 800 MPa, n = 4: 200 MPa
  main hanger tension: 54.34 kN
  net section at bolt 3: 48.75 kN
  bearing on bolt 3: 58.50 kN
  shear of bolt 3: 44.18 kN
  tension of bolts 1, 2: 36.19 kN
  bearing under washers: 65.35 kN
  shear through clip angle: 58.50 kN
hanger allowable load: 36.2 kN
rod: allowable 181.8 MPa, A 220.0 mm2, d 16.74 mm
pin C: resultant 76.3 kN, tau_all 106.1 MPa, d 21.4 mm
  bearing area per support 127.2 mm2, t = 5.78 mm
cable: A 76.22 mm2, d 9.85 mm
  pin A, B: A 59.54 mm2, d 8.71 mm
  pin O: A 87.25 mm2, d 10.54 mm
  pin D: A 115.81 mm2, d 12.14 mm
LRFD required ultimate load: 160 kN
  equivalent factor of safety: 1.60
load factor: 3.0
strength 240 kN, service 150 kN: n 1.60, margin 0.60

MATLAB (Octave-compatible). MM06_allowable_stress.m sweeps the factor of safety for the 40\,\mathrm{kN} rod and repeats the hanger checks.

   n   allowable (MPa)   area (mm^2)   diameter (mm)   mass (kg/m)
 1.5         400.0          100.0          11.28         0.78
 2.0         300.0          133.3          13.03         1.05
 2.5         240.0          166.7          14.57         1.31
 3.0         200.0          200.0          15.96         1.57
 3.3         181.8          220.0          16.74         1.73
 4.0         150.0          266.7          18.43         2.09
 5.0         120.0          333.3          20.60         2.62
hanger allowable loads (kN): 54.3 48.8 58.5 44.2 36.2 65.3 58.5
controlling check 5: 36.2 kN

SOLIDWORKS Simulation: — A design rule is not a field result: allowable stress and factor of safety are choices applied to results, not something a study computes. (A study can plot a factor of safety distribution from a stress result and a chosen limit, but that is Lesson MM43's von Mises plot, and the number it divides by is still the engineer's choice.)

Traps

Sizing to the breaking stress. Size to the allowable stress, which already includes the factor. (Nugget 04.)
Mixing the yield factor with the ultimate strength (or the reverse). (Nugget 02.)
Checking only one failure mode. The smallest allowable load among tension, shear and bearing governs. (Nugget 03.)
Using the gross area in tension at a hole. Use the net area. (Nugget 03.)
Rounding a diameter down. Round up to the stock size. (Nugget 04.)
Sizing the plate before the pin. The pin diameter sets the bearing area. (Nugget 04.)
Quoting a factor without saying what it multiplies. A bare "3" may be on yield, on ultimate or on load. (Nugget 01.)
Treating a larger factor as free. It costs weight and money. (Nugget 05.)
Equating the stress factor and the load factor where stress is not proportional to load, as in buckling. (Nugget 06.)

Self-check

A 16\,\mathrm{mm} rod has \sigma_Y = 300\,\mathrm{MPa} and the factor on yield is 1.67. Find the allowable stress and the allowable tension.
A structure fails at 240\,\mathrm{kN} and carries 150\,\mathrm{kN} in service. Find n and the margin of safety.
A pin in double shear carries 70\,\mathrm{kN}; \tau_U = 350\,\mathrm{MPa} and n = 3.3. Find the minimum diameter (to the next whole millimetre).
A plate 80\,\mathrm{mm} wide and 10\,\mathrm{mm} thick has a 20\,\mathrm{mm} hole. With \sigma_{\text{allow}} = 140\,\mathrm{MPa}, find the allowable tension.
A 16\,\mathrm{mm} bolt in single shear (\tau_{\text{allow}} = 60\,\mathrm{MPa}) joins two 8\,\mathrm{mm} plates, each 50\,\mathrm{mm} wide with a 16\,\mathrm{mm} hole; \sigma_{b,\text{allow}} = 150\,\mathrm{MPa} and \sigma_{\text{allow}} = 120\,\mathrm{MPa} in tension. Find the allowable load and the controlling mode.
A rod carries 40\,\mathrm{kN} and \sigma_U = 600\,\mathrm{MPa}. With n = 3 find the diameter.
A cable carries 9.5\,\mathrm{kN} with \sigma_{\text{allow}} = 125\,\mathrm{MPa}. Find the minimum diameter.
A cast-iron column has a compressive ultimate of 800\,\mathrm{MPa} and n = 4. What area carries 150\,\mathrm{kN}?
Using \gamma_D = 1.2, \gamma_L = 1.6 and \phi = 0.9, find the required P_U for P_D = 40\,\mathrm{kN} and P_L = 60\,\mathrm{kN}, and the equivalent factor of safety.
Explain why the stress form and the load form of the factor of safety differ for a slender column.
Which should have a larger factor of safety, a passenger lift cable or a garden shelf bracket? Give two reasons.

Fully worked solutions

1. \sigma_{\text{allow}} = 300/1.67 = 179.6\,\mathrm{MPa}; A = \pi(8)^2 = 201.06\,\mathrm{mm^2}; P_{\text{allow}} = 179.6\times201.06 = 36.1\,\mathrm{kN}.

2. n = 240/150 = 1.60; margin = 0.60 (60 per cent).

3. \tau_{\text{allow}} = 350/3.3 = 106.1\,\mathrm{MPa}; each plane 35\,\mathrm{kN}; A = 35\,000/106.06 = 330\,\mathrm{mm^2}; d = \sqrt{4\times330/\pi} = 20.5\,\mathrm{mm}, so 21\,\mathrm{mm}.

4. Net area (80 - 20)\times10 = 600\,\mathrm{mm^2}; P = 140\times600 = 84\,\mathrm{kN}.

5. Shear: 60\times201.06 = 12.06\,\mathrm{kN}. Bearing: 150\times(8\times16) = 19.2\,\mathrm{kN}. Net tension: 120\times(50-16)\times8 = 32.6\,\mathrm{kN}. The smallest, 12.1\,\mathrm{kN}, is shear of the bolt, which governs.

6. \sigma_{\text{allow}} = 200\,\mathrm{MPa}; A = 40\,000/200 = 200\,\mathrm{mm^2}; d = \sqrt{4\times200/\pi} = 15.96\,\mathrm{mm}, so 16\,\mathrm{mm}.

7. A = 9500/125 = 76.0\,\mathrm{mm^2}; d = \sqrt{4\times76/\pi} = 9.84\,\mathrm{mm}; use 10\,\mathrm{mm}.

8. \sigma_{\text{allow}} = 800/4 = 200\,\mathrm{MPa}; A = 150\,000/200 = 750\,\mathrm{mm^2}.

9. Demand 1.2(40) + 1.6(60) = 144\,\mathrm{kN}; P_U \ge 144/0.9 = 160\,\mathrm{kN}; equivalent n = 160/100 = 1.60.

10. In a slender column the stress at failure is the buckling stress, not the material strength, and it is not proportional to the load in the usual way; dividing the material strength by n does not give the factor on the load that causes buckling. The load factor, failure load over working load, measures what actually happens (Hearn).

11. The lift cable: failure has severe consequences for life, the load is dynamic (starting and stopping), the cycles are many (fatigue) and wear and corrosion are hard to see. The shelf bracket: low consequence, static load, visible warning (it sags first).