Skip to content
CivilSolve

Statics & Mechanics of Materials

Axially Loaded Stepped Bar

Find internal force, normal stress, and elongation in a stepped or composite bar fixed at one end, with axial loads and optional temperature change.

Direct solving is free and needs no account. Have a word problem instead? Submit it as text.

Inputs

Load a sample problem:
segmentsBar segments in order starting AT the fixed support: segment 1 touches the wall, the last segment ends at the free end. Nodes are numbered 0 at the support, 1 at the end of segment 1, and so on to the free end.
segments 1
node loadsConcentrated axial loads applied at nodes (may be empty for a thermal-only problem)
thermalOptional uniform temperature change. The bar is fixed at ONE end only, so it expands freely: ΔT adds elongation but produces no stress.
alphaCoefficient of thermal expansion of the bar material

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

SI stepped composite bar: steel 500 mm² then aluminum 250 mm², two loads

Given

segments
  1. 1.length 0.5 marea 500 mm^2modulus 200 GPa
  2. 2.length 0.75 marea 250 mm^2modulus 70 GPa
node loads
  1. 1.node 1force 30 kNdirection toward-support
  2. 2.node 2force 50 kNdirection away-from-support

Assumptions

  • Linear elastic behavior, uniaxial stress, and loads applied along the bar axis; self-weight neglected. Tension is positive; a positive end deflection means the free end moves away from the support.

Solution steps

  1. Internal force in each segment (method of sections)

    Cut inside a segment and sum the applied axial loads between the cut and the free end; loads pulling away from the support put the segment in tension (positive).

    N(segment) = Σ F applied beyond the cut, tension positive

    segment 1: N = 20 kN; segment 2: N = 50 kN

  2. Normal stress in each segment

    Axial stress is the internal force divided by the segment's cross-sectional area.

    σ = N/A

    segment 1: σ = 40 MPa; segment 2: σ = 200 MPa

    = 200 MPa

  3. End deflection by superposition of segment deformations

    Each segment stretches by NL/AE; the free-end deflection is the sum over the segments (plus the thermal term when present).

    δ = Σ(N·L/(A·E))

    δ = 0.1 mm + 2.143 mm = 2.243 mm

    = 2.243 mm

Results

Internal axial force in segment 1 (tension +)

20kN

Normal stress in segment 1 (tension +)

40MPa

Internal axial force in segment 2 (tension +)

50kN

Normal stress in segment 2 (tension +)

200MPa

Deflection of the free end (positive = away from the support)

2.243mm

Where this answer was checked
source
δ = Σ NL/AE for a stepped bar (NCEES FE Reference Handbook) — two segments, loads at both nodes, internal forces by sections
verified by
hand-recomputed
derivation
Loads: 30 kN toward the support at node 1, 50 kN away at node 2 (free end). Segment 2 (beyond node 1): N2 = +50 kN; segment 1: N1 = 50 − 30 = 20 kN. σ1 = 20e3/500e-6 = 40 MPa; σ2 = 50e3/250e-6 = 200 MPa. δ = 20e3·0.5/(500e-6·200e9) + 50e3·0.75/(250e-6·70e9) = 1.0e-4 + 2.142857e-3 = 2.242857e-3 m = 2.2429 mm.

Example 2

US single steel rod: 2 in², 6 ft, 20 kip pull

Given

segments
  1. 1.length 6 ftarea 2 in^2modulus 29000 ksi
node loads
  1. 1.node 1force 20 kipdirection away-from-support

Assumptions

  • Linear elastic behavior, uniaxial stress, and loads applied along the bar axis; self-weight neglected. Tension is positive; a positive end deflection means the free end moves away from the support.

Solution steps

  1. Internal force in each segment (method of sections)

    Cut inside a segment and sum the applied axial loads between the cut and the free end; loads pulling away from the support put the segment in tension (positive).

    N(segment) = Σ F applied beyond the cut, tension positive

    segment 1: N = 20 kip

  2. Normal stress in each segment

    Axial stress is the internal force divided by the segment's cross-sectional area.

    σ = N/A

    segment 1: σ = 10 ksi

    = 10 ksi

  3. End deflection by superposition of segment deformations

    Each segment stretches by NL/AE; the free-end deflection is the sum over the segments (plus the thermal term when present).

    δ = Σ(N·L/(A·E))

    δ = 0.02483 in = 0.02483 in

    = 0.02483 in

Results

Internal axial force in segment 1 (tension +)

20kip

Normal stress in segment 1 (tension +)

10ksi

Deflection of the free end (positive = away from the support)

0.02483in

Where this answer was checked
source
δ = PL/AE (NCEES FE Reference Handbook) — single segment in tension
verified by
hand-recomputed
derivation
N = 20 kip; σ = 20/2 = 10 ksi; δ = 20·72/(2·29000) = 1440/58000 = 0.024828 in.

Example 3

SI rod with load and heating: 20 kN + 40°C rise

Given

segments
  1. 1.length 2 marea 400 mm^2modulus 200 GPa
node loads
  1. 1.node 1force 20 kNdirection away-from-support
thermal
delta t 40 degCalpha value 0.000012per degC

Assumptions

  • Linear elastic behavior, uniaxial stress, and loads applied along the bar axis; self-weight neglected. Tension is positive; a positive end deflection means the free end moves away from the support.

Solution steps

  1. Internal force in each segment (method of sections)

    Cut inside a segment and sum the applied axial loads between the cut and the free end; loads pulling away from the support put the segment in tension (positive).

    N(segment) = Σ F applied beyond the cut, tension positive

    segment 1: N = 20 kN

  2. Normal stress in each segment

    Axial stress is the internal force divided by the segment's cross-sectional area.

    σ = N/A

    segment 1: σ = 50 MPa

    = 50 MPa

  3. Free thermal expansion

    The bar is restrained at only one end, so it expands freely: the temperature change adds elongation over the full length but induces no stress.

    δT = α · ΔT · L

    δT = 1.20e-5 /K × 40 K × 2 m = 0.96 mm

    = 0.96 mm

  4. End deflection by superposition of segment deformations

    Each segment stretches by NL/AE; the free-end deflection is the sum over the segments (plus the thermal term when present).

    δ = Σ(N·L/(A·E)) + α·ΔT·L

    δ = 0.5 mm + 0.96 mm = 1.46 mm

    = 1.46 mm

Results

Internal axial force in segment 1 (tension +)

20kN

Normal stress in segment 1 (tension +)

50MPa

Deflection of the free end (positive = away from the support)

1.46mm

Elongation from the temperature change alone

0.96mm

Where this answer was checked
source
δ = NL/AE + αΔT·L, free thermal expansion (NCEES FE Reference Handbook) — mechanical + thermal elongation, one-end-fixed bar
verified by
hand-recomputed
derivation
L=2 m, A=400 mm², E=200 GPa, α=12e-6/°C, ΔT=+40°C, P=20 kN away. N=20 kN, σ=20e3/400e-6=50 MPa. δ_mech=20e3·2/(400e-6·200e9)=5.0e-4 m; δ_T=12e-6·40·2=9.6e-4 m; δ=1.46e-3 m=1.46 mm.