1001Ferramentas
🧱 Calculators

Column Compression Stress

Computes compression stress on a column from axial load and section area.

Compression stress in a short column

When a short column carries an axial load, the normal compression stress comes out to σ = P/A. Here P is the axial load and A the cross-sectional area. You check it against σ_max ≤ σ_allowable, and because the column is short there's no buckling reduction to worry about. Take P = 300 kN and A = 400 cm². That gives σ = 300·10³ / (400·10⁻⁴) = 7.5 MPa. With concrete f_ck = 25 MPa, the admissible stress lands around 18 MPa (roughly 0.85·f_ck/γ_c, taking γ_c = 1.4).

A column counts as short when its slenderness λ = K·L/r stays below the limit set by NBR 6118 §15 (for unbraced concrete columns that's usually λ < 35). Once you go past that, second-order effects and buckling need their own separate check. Keep the units lined up: P in N, A in m², σ in Pa.

Applications

Use it to dimension short reinforced-concrete columns under NBR 6118, to size pedestals and short steel posts, and to verify precast slab supports and bearing walls. It also works for a quick axial-load check before you bring in moment and slenderness effects.

FAQ

When is a column considered short? When λ = K·L/r stays below the code limit. NBR 6118 §15.8 puts that around 35 for typical concrete columns. Go past it and you have to add the second-order moment.

Does this include the rebar contribution? No. The full capacity of an RC column is N_d = 0.85·f_cd·A_c + f_yd·A_s. What this tool reports is the gross-concrete stress, meant as a quick check.

Why is the admissible stress so much lower than f_ck? NBR 6118 applies γ_c = 1.4 and then a 0.85 factor for sustained loading (the Rüsch effect). So σ_allow works out to roughly 0.85·f_ck/1.4 ≈ 0.6·f_ck, and that's before any safety margin on the load side.

Related Tools

📊

Pile Structural Stress

Calculate the structural compression stress in a pile shaft, σ = Q ÷ (π·D²/4), from the applied load Q (kN) and the pile diameter D (m); the result is in MPa. Besides the SOIL having capacity to support the pile (geotechnical capacity), the pile itself, as a STRUCTURAL element (concrete, steel or timber), must resist the load without failing or deforming excessively — this is the pile's STRUCTURAL check. The shaft compression stress is simply the load over the cross-sectional area. It must be below the pile material's allowable stress: codes limit cast-in-place pile concrete working stress to conservative values (typically 5-8 MPa, less than the concrete strength, due to subsurface execution uncertainties — blind concreting, possible defects, eccentricities). This check often GOVERNS the minimum pile diameter (the pile may have ample geotechnical capacity, but structural stress limits the load). Pile design is always the SMALLER of geotechnical (soil) and structural (material) capacity — both must be checked. Enter the applied load and the pile diameter.

📰

Column Width CPL

Computes ideal column width in characters per line CPL for comfortable reading.

✂️

Shear Stress Calculator

Compute shear stress τ = F/A. Output in Pa, kPa and MPa.

⬆️

Cable Tension in Accelerated Lift

Calculate the dynamic tension in a cable while lifting a load with acceleration, T = W·(1 + a/g), from the load weight W (N), the vertical lift acceleration a (m/s²) and gravity g. When a load is lifted with ACCELERATION (at lift start, when accelerating the rise), the cable must provide not only the force to support the weight (W) but ALSO the force to accelerate the mass upward — by Newton's second law, the total tension is the weight times the factor (1 + a/g). This means the DYNAMIC tension is GREATER than the static weight: an acceleration of g/2 (5 m/s²) raises the tension by 50%! That is why ABRUPT lifts (fast start, or worse, lifting an already-moving load or stopping abruptly) generate dangerous dynamic OVERLOADS in the cable, which can break it even with the static load within capacity. The effect is worse in abrupt STOPS and in loads 'snatching off the ground' (cable slack suddenly removed, generating an impact). So experienced operators lift SMOOTHLY (low acceleration), and the cable safety factors (5 or more) exist precisely to cover these inevitable dynamic overloads. This calculation quantifies the tension increase due to acceleration, essential in the safety analysis of dynamic lifts. Enter the load weight and the lift acceleration.

📐

Bolt Tensile Stress Area (Metric)

Calculate the tensile stress area of a metric-thread bolt, A_t = (π/4)·(d − 0.9382·p)², from the nominal diameter d (mm) and the thread pitch p (mm). The tensile stress area is the EFFECTIVE cross-section resisting tension in a threaded bolt — and it is NOT the nominal-diameter area (the smooth cylinder) nor the root-diameter area (the thread bottom). Because of the helical thread geometry, tensile rupture occurs at an intermediate section, and tests showed it corresponds to an effective diameter equal to the average of the pitch and root diameters, leading to the formula with the 0.9382·p term (a geometric constant of the ISO metric thread, 60° triangular profile). The tensile area is the fundamental parameter for all bolt strength calculations: preload, tensile stress, proof load and ultimate strength are all found by multiplying A_t by the corresponding material stress. Using the wrong area (the larger nominal-diameter one) would overestimate strength and lead to undersized joints. Bolt tables list A_t for each diameter-pitch combination; this formula computes it for any metric thread. Enter the nominal diameter and the thread pitch.

📐

Sling Leg Tension

Calculate the tension in each leg of a multi-leg inclined sling, T = W ÷ (n·cos α), from the load weight W (N), the number of legs n and the angle of each leg from vertical α (degrees). When a load is lifted by a multi-leg sling (ropes or chains from the hook spreading to the attachment points on the load), the tension in each leg is NOT simply the weight divided by the number of legs — because the legs are INCLINED. The more OPEN the angle (more horizontal legs), the HIGHER the tension in each leg, possibly MULTIPLYING the load several times! This happens because, with inclined legs, part of each leg's force is 'spent' on the horizontal component (which cancels between opposite legs, compressing the load), and only the vertical component supports the weight — so the total tension must be higher for the vertical components to sum to the weight. This is one of the most dangerous and common rigging errors: using slings with very open angles overloads the legs, possibly breaking them even with a load 'apparently' within capacity. So codes LIMIT the leg angle (typically 60° max from vertical, ideally less) and sling WLL tables give the REDUCED capacity per angle. Enter the load weight, the number of legs and the angle.

The results provided by this tool are for general informational and educational purposes only and do not constitute professional, financial, medical, legal, tax or accounting advice. Always confirm important decisions with a qualified professional and official sources.