CivilSolve

Statics & Mechanics of Materials

Normal, Shear & Bearing Stress

Chapter-one stress analysis: average normal stress σ = P/A in a member, average shear stress in bolts/pins (single or double shear, load shared over the bolt count), bearing stress on a plate σb = P/(d·t), and factors of safety against given allowable stresses. Use for: 'find the stress in the rod', 'average shear stress in the bolt', bearing checks, 'what factor of safety'. Not for: bending or combined stresses, or code-based connection design.

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Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

normal stress in a 20 mm rod under 50 kN

Given

load
50 kN
member diameter
20 mm

Assumptions

  • Average (uniform) stress distributions — the standard chapter-one idealization; stress concentrations are not included.

Solution steps

  1. Normal stress in the member

    The axial load spreads uniformly over the cross-section.

    σ = P / A

    σ = 50 kN / 314.2 mm^2 = 159.2 MPa

    = 159.2 MPa

Results

Average normal stress σ

159.2MPa

Where this answer was checked
source
σ = P/A with A = πd²/4 — circular bar normal stress
verified by
hand-recomputed
derivation
A = π·20²/4 = 314.159 mm². σ = 50 000 N / 314.159e-6 m² = 159.15 MPa.

Example 2

US single-shear bolt: 10 kip on one 0.5 in bolt

Given

load
10 kip
bolt diameter
0.5 in
shear planes
single

Assumptions

  • Average (uniform) stress distributions — the standard chapter-one idealization; stress concentrations are not included.

Solution steps

  1. Shear stress in the bolts

    Each bolt resists on one plane (single shear), and the bolts share the load equally.

    τ = P / (n · m · A_bolt), m = shear planes

    A_bolt = 0.1963 in^2; n = 1; τ = 50.93 ksi

    = 50.93 ksi

Results

Average bolt shear stress τ

50.93ksi

Where this answer was checked
source
τ = P/A_bolt, single shear — average bolt shear
verified by
hand-recomputed
derivation
A = π·0.25²... = π·0.5²/4 = 0.196350 in². τ = 10/0.19635 = 50.93 ksi.

Example 3

double shear, two 16 mm bolts, 80 kN, with bearing and FS

Given

load
80 kN
bolt diameter
16 mm
bolt count
2
shear planes
double
plate thickness
10 mm
allowable shear
150 MPa

Assumptions

  • Average (uniform) stress distributions — the standard chapter-one idealization; stress concentrations are not included.

Solution steps

  1. Shear stress in the bolts

    Each bolt resists on two planes (double shear), and the bolts share the load equally.

    τ = P / (n · m · A_bolt), m = shear planes

    A_bolt = 201.1 mm^2; n = 2; τ = 99.47 MPa

    = 99.47 MPa

  2. Bearing stress on the plate

    The bolt presses on the projected rectangle d × t of each hole.

    σb = P / (n · d · t)

    σb = 250 MPa

    = 250 MPa

  3. Factor of safety (shear)

    Capacity over demand for the bolts.

    FS = 150 MPa / 99.47 MPa = 1.508

Results

Average bolt shear stress τ

99.47MPa

Bearing stress σb

250MPa

Factor of safety (bolt shear)

1.508

Where this answer was checked
source
τ = P/(n·m·A); σb = P/(n·d·t); FS = allow/τ — full connection check
verified by
hand-recomputed
derivation
A_bolt = π·16²/4 = 201.062 mm². τ = 80e3/(2·2·201.062e-6) = 99.47 MPa. σb = 80e3/(2·0.016·0.010) = 250 MPa. FS_shear = 150/99.47 = 1.508.