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Design considerations for prismatic beams

Short answer

Design finds required section modulus Z = M_max / σ_all, then checks shear, principal stress, deflection and stability. Ductile materials use symmetric sections and yield stress; brittle need unsymmetric sections with larger tensile flange. Long beams are governed by bending; short beams by shear. Procedure: compute M_max, Z_req, select section, verify shear and deflection.

Explain the basic considerations in the design of prismatic beams. Distinguish between the design of beams of ductile and brittle materials, and between short and long beams. Outline the design procedure.

A prismatic beam has a constant cross-section along its length. Design means choosing a section that keeps the maximum stresses within allowable limits and, where needed, the deflection within the permitted value.

Basic considerations

  1. Bending governs most designs. The required section modulus is
Zreq=MmaxσallZ_{req} = \frac{M_{max}}{\sigma_{all}}
  1. Shear must be checked, especially for short, heavily loaded beams, and for thin webs: τmax≤τall\tau_{max} \le \tau_{all}.
  2. Principal stresses may be critical at points where both σ\sigma and τ\tau are large, such as the web-flange junction of an I-beam.
  3. Deflection limits (for example span/325 to span/250 for many structural members) and stability against lateral buckling may govern.
  4. Economy: choose the section with the least area satisfying Z≥ZreqZ \ge Z_{req}, with depth >> width so that the material is placed away from the neutral axis.

Ductile versus brittle materials

AspectDuctile (mild steel, aluminium)Brittle (cast iron, concrete)
Strength basisYield stress σy\sigma_yUltimate stress σu\sigma_u
Tension and compressionEqualCompression much larger than tension
SectionSymmetrical about NA (I-section)Unsymmetrical, bigger flange on the tension side (T or inverted T)
Failure criterionMax shear stress or distortion energyMax principal stress

Short versus long beams

  • Long beams (L/dL/d greater than about 10 to 15): bending governs; the shear stress is small.
  • Short beams (L/dL/d small): the shear force is large relative to the moment; the shear stress, bearing at supports and web crippling may govern.

Design steps

  1. Find reactions, draw the shear force and bending moment diagrams, get MmaxM_{max} and VmaxV_{max}.
  2. Compute Zreq=Mmax/σallZ_{req} = M_{max}/\sigma_{all}.
  3. Select a section (standard section from tables or fix the proportions h=2bh = 2b for timber).
  4. Check shear stress at the NA, principal stresses at critical points, and deflection.
  5. Revise the section if any check fails.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Strength of Materials chapter 9 questions →