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Mock Q2: full solution and invented marking guide

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MOCK Q2(a) · double-cover splice, plan; all geometry repeated in the adjacent question. — Invented mock-question diagram: double-cover splice in plan. Four bolts per half, with one cover plate on each main-plate face. The dashed line is the main-plate butt joint; each nearest bolt centre is $40 mm$ from the joint. Longitudinal pitch $60 mm$ . The cover plate projects $40 mm$ beyond the outermost bolt centre. Transverse dimensions $30$ , $80$ , $30 mm$ . Main-plate thickness $12 mm$ ; cover-plate thickness $8 mm$ ; width $140 mm$ . All diagram dimensions are in $mm$ . Not to scale; dimension labels govern.Open full-size image.

(a) Follow one splice half and every requested failure mode

Simple explanation: Count where the bolt can be sheared

The plate interfaces are the places trying to cut across the bolt.

Count actual loaded shear planes, not merely visible plates.
Choose shank or threaded area for the plane concerned.
Compare group resistance with the force that group transfers.

Remember: Bolts on opposite sides of a splice do not all act in parallel.

Related concept and full method

Simple explanation: Why holes reduce tension resistance

The pull must squeeze through the steel left beside the holes.

Choose a possible fracture line across the member.
Subtract the holes crossed by that line using hole diameter.
Apply the course effective-area rule and its gross-area limit.

Remember: Do not subtract every hole anywhere in the connection.

Related concept and full method

Each transfer half carries the full $300 kN$ . Four bolts share it: $75 kN$ per bolt; each cover/plane takes $37.5 kN$ . Eight bolts exist overall, but halves transfer load in series; do not divide $300$ by $8$ .

M20 lookup

A_{t} = 245 {mm}^{2}

p_{s} = 375

p_{bb} = 1000 \frac{N}{{mm}^{2}}

. Shear resistance per plane:

\frac{245 \times 375}{1000} = 91.875 kN

Two shear planes per bolt give

183.75

; four bolts per half give

735 kN > 300

. Bolt bearing at the main plate:

\frac{20 \times 12 \times 1000}{1000} = 240 kN

The two cover-plate contacts of each bolt:

\frac{2 \times 20 \times 8 \times 1000}{1000} = 320 kN

Governing bearing resistance for four bolts:

4 \times \min (240, 320) = 960 kN > 300

Connected-part bearing uses the least of all applicable terms. Main and cover end distances are $40$ ; clear hole ligament $60 - 22 = 38 mm$ . Main web/plate load per bolt $75 kN$ ; cover load per bolt $37.5 kN$ .

Main-plate diameter term:

\frac{20 \times 12 \times 550}{1000} = 132 kN

Main-plate end-distance term:

\frac{0.5 \times 40 \times 12 \times 550}{1000} = 132 kN

Main-plate net-clearance term and cap:

\frac{\min (1.5 \times 38 \times 12 \times 510, 2 \times 20 \times 12 \times 800)}{1000}

= \min (348.84, 384) = 348.84 kN

Main plate, per bolt:

\min (132, 132, 348.84) = 132 > 75

Half-joint resistance

528 kN

. For each

8 mm

cover plate, reduce each term by

\frac{8}{12}

88

88

\min (232.56, 256) = 232.56

, giving the governing per-bolt value

88 > 37.5 kN

. Bearing resistance of one half of each cover plate:

4 \times 88 = 352 kN > 150

A critical transverse tension section crosses TWO holes, one in each bolt line. Longitudinally separated holes are not all deducted at the same cut.

Main plate:

A_{g} = 140 \times 12 = 1680 {mm}^{2}

A_{n} = (140 - 2 \times 22) \times 12 = 1152 {mm}^{2}

A_{e} = \min (1.1 \times 1152, 1680) = 1267.2 {mm}^{2}

12 mm

S355 uses

p_{y} = 355

;

P_{t} = \frac{1267.2 \times 355}{1000} = 449.856 kN > 300

Each cover plate:

A_{g} = 140 \times 8 = 1120 {mm}^{2}

A_{n} = 96 \times 8 = 768 {mm}^{2}

A_{e} = \min (844.8, 1120) = 844.8 {mm}^{2}

P_{t, each} = \frac{844.8 \times 355}{1000} = 299.904 kN > 150

M20 minimum edge distance for rolled edges

26 mm

; end distance

40

and side edge distance

30

both pass.
Longitudinal pitch

60 \geq 2.5 \times 20 = 50

; transverse spacing

80 \geq 3 \times 20 = 60

.
Governing maximum spacing

\min (12 \times 8, 150) = 96 mm

;

60

／

80

both pass.
Maximum cover-plate edge distance

11 \times 8 \times \sqrt{\frac{275}{355}} = 77.452848 mm

;

30

passes. The thicker main plate permits a larger maximum edge distance.
Width check:

30 + 80 + 30 = 140 mm

All requested checks pass. Main net tension $449.856 kN$ is the lowest calculated full-half resistance. This scoped result does not claim an unrequested complete block-shear design.

Animation labCount the bolt shear planes5 concepts · 5 source expressions

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

Load must cross an interface between the connected plates.
A lap joint gives one shear plane through a bolt.
A symmetric double-cover joint may provide two shear planes. Count load-transfer interfaces, not just visible plates.
Use the source’s area and shear strength. Then check bearing, plate resistance and detailing separately.

MOCK Q2(b) · closed rectangular weld; all geometry repeated in the adjacent question. — Invented mock-question diagram: closed rectangular weld. $B = 200 mm$ is width; $H = 300 mm$ is height; dashed lines intersect at the centroid. The downward design force $P = 200 kN$ acts at eccentricity $e = 250 mm$ from the centroid. Green lines are the continuous effective weld centrelines. Plate thickness $12$ and $18 mm$ . Not to scale; stated dimensions govern.Open full-size image.

(b) Direct shear plus torsion at the reinforcing corner

Simple explanation: Only the weld lines resist as weld

Imagine a wire rectangle: its empty middle is not more wire.

Count only the weld runs actually shown.
Use their total length and line second moments.
Combine direct and torsional forces at the critical location.

Remember: Line second moments have units ${mm}^{3}$ ; plate-area moments use ${mm}^{4}$ .

Related concept and full method

Simple explanation: Add arrows before taking the magnitude

Walking east and walking north do not point in the same direction.

Choose signed horizontal and vertical directions.
Add contributions along each direction separately.
For perpendicular components, use the right-triangle resultant.

Remember: Check the corner where direct and torsional components reinforce each other.

Related concept and full method

Use the weld-length centroid at the rectangle centre. Corner $x = \pm 100$ , $y = \pm 150 mm$ . Direct downward shear adds to torsional vertical shear on one side; that side governs. Weld-line inertias have units of ${mm}^{3}$ .

L = 2 (200 + 300) = 1000 mm

q_{s} = \frac{P}{L} = \frac{200}{1000} = 0.2 \frac{kN}{mm}

M = P e = 200 \times 250 = 50000 kN \cdot mm

I_{x} = \frac{300^{3}}{6} + \frac{200 \times 300^{2}}{2} = 1.35 \times 10^{7} {mm}^{3}

I_{y} = \frac{200^{3}}{6} + \frac{300 \times 200^{2}}{2} = 7333333.333333 {mm}^{3}

J = I_{x} + I_{y} = 20833333.333333 {mm}^{3}

q_{v} = 0.2 + \frac{50000 \times 100}{J} = 0.440000000 \frac{kN}{mm}

| q_{h} | = \frac{50000 \times 150}{J} = 0.360000000 \frac{kN}{mm}

q_{\max} = \sqrt{q_{v}^{2} + q_{h}^{2}} = 0.568506816 \frac{kN}{mm}

q_{cap} = \frac{0.7 s \times 250}{1000} = 0.175 s

Required weld leg:

s \geq \frac{0.568506816}{0.175} = 3.248610 mm

The strength requirement alone would permit a small leg, but the $18 mm$ thicker plate gives minimum $6 mm$ . At the $12 mm$ thinner edge maximum= $12 - 2 = 10 mm$ . Choose $6 mm$ : qcap 1.05 kN/mm, which exceeds the calculated demand. This closed effective-line mock does not use two-end deductions on each rectangle side; the geometry was expressly defined as continuous.

Selected

6 mm

weld-leg utilisation

\frac{0.568506816}{1.05} = 0.541435 < 1

; size

6

lie on opposite sides of the

6

10 mm

is within the range.

Animation labA weld group is a set of lines2 concepts · 5 source expressions

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

The centre is empty: resistance comes from the weld lines, not a solid plate filling the group.
Use line lengths for centroid weighting. Line second moments have units $mm^{3}$ ; add the parallel-axis terms.
Direct force per length is $\frac{P}{L}$ . The eccentric moment adds tangential flow proportional to distance from the centroid.
Combine signed components at every candidate corner, then compare the maximum with throat resistance per length.

Compact script and invented marking guide

(a) Four bolts/half, double shear: $735 kN$ ; bolt bearing $960$ ; main plate bearing $528$ ; main tension $449.856$ ; each cover bearing $352$ and tension $299.904 kN$ . All exceed corresponding $300$ / $150 kN$ actions; layout passes. (b) $q_{\max} = 0.568507 \frac{kN}{mm}$ ; strength leg $\geq 3.249 mm$ ; detailing governs choose $6 mm$ .

Invented credit: (a) force sharing $2$ , bolt checks $4$ , connected-part bearing $4$ , tension $4$ , layout/conclusion $4$ . (b) centroid/eccentric moment $2$ , length/inertias $3$ , corner components/resultant $3$ , size/detail/conclusion $4$ .

Animation labFollow the calculation sequence6 concepts · 2 source expressions

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

Locate the load, supports, connection geometry and any stated assumptions.
Keep given values, table lookups and calculated values distinct; reconcile their units.
The calculation player steps through the existing expressions in their original order.
Compare demand with resistance or the relevant limit. Keep missing inputs and conditional conclusions explicit.