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Flexural rigidity; T-beam bending stresses (60 kN loads)

Short answer

Flexural rigidity EI resists bending. M_max=60 kN·m. Section centroid 13.21 cm from bottom, I=5900.3 cm⁴. Maximum tensile stress (bottom) 134.4 MPa, compressive stress (top) 69.0 MPa.

Explain flexural rigidity and its importance. A beam is loaded as shown in figure. Determine maximum tensile and compressive bending stress developed in the beam. [Figure: simply supported beam A-B-C-D with a hinge at A and a roller at D; two 60 kN downward loads at B and C; A to B 1 m, B to C 1.5 m, C to D 1 m. Cross-section: inverted T-shape with a top flange 7.5 cm + 7.5 cm either side of the web (web width 5 cm), web height 15 cm and flange thickness 5 cm as marked.]

Flexural rigidity

Flexural rigidity is the product EIEI of Young's modulus and the moment of inertia of the section about the neutral axis. It measures the resistance of a beam to bending: from MEI=1R=d2ydx2\dfrac{M}{EI} = \dfrac{1}{R} = \dfrac{d^2y}{dx^2}, the curvature and deflection of a beam are inversely proportional to EIEI.

Importance: a larger EIEI gives smaller deflection and curvature for the same loading. It is used to check deflection limits, to compare sections (deep sections have large II), and it appears in the deflection and column formulas (δ∝WL3EI\delta \propto \dfrac{WL^3}{EI}, Pcr=π2EILe2P_{cr} = \dfrac{\pi^2EI}{L_e^2}).

Bending stresses in the beam

Reading of the figure: simply supported beam with span AD=1+1.5+1=3.5AD = 1 + 1.5 + 1 = 3.5 m and two 60 kN loads at B (1 m from A) and C (1 m from D). The T-section has a flange 7.5+5+7.5=207.5 + 5 + 7.5 = 20 cm wide and 5 cm thick, and a web 5 cm wide and 15 cm deep (total depth 20 cm). The flange is taken at the top (if the flange is at the bottom, the tensile and compressive values below are interchanged).

Reactions and maximum moment: by symmetry RA=RD=60R_A = R_D = 60 kN. The maximum moment is constant between B and C:

Mmax=60×1=60 kN⋅m=60×106 N⋅mmM_{max} = 60\times1 = 60\ \text{kN·m} = 60\times10^6\ \text{N·mm}

Centroid (from the bottom of the web):

yˉ=(20×5)(17.5)+(5×15)(7.5)100+75=13.214 cm\bar y = \frac{(20\times5)(17.5) + (5\times15)(7.5)}{100 + 75} = 13.214\ \text{cm}

Moment of inertia about the neutral axis:

I=[20(5)312+100(17.5−13.214)2]+[5(15)312+75(13.214−7.5)2]=5900.30 cm4I = \left[\frac{20(5)^3}{12} + 100(17.5-13.214)^2\right] + \left[\frac{5(15)^3}{12} + 75(13.214-7.5)^2\right] = 5900.30\ \text{cm}^4

Stresses (σ=My/I\sigma = My/I, M=60×105M = 60\times10^5 N·cm):

σtension (bottom, y=13.214 cm)=60×105×13.2145900.30=13437.6 N/cm2=134.38 N/mm2\sigma_{tension}\ (\text{bottom},\ y = 13.214\ \text{cm}) = \frac{60\times10^5 \times 13.214}{5900.30} = 13437.6\ \text{N/cm}^2 = 134.38\ \text{N/mm}^2 σcompression (top, y=6.786 cm)=60×105×6.7865900.30=6900.4 N/cm2=69.00 N/mm2\sigma_{compression}\ (\text{top},\ y = 6.786\ \text{cm}) = \frac{60\times10^5 \times 6.786}{5900.30} = 6900.4\ \text{N/cm}^2 = 69.00\ \text{N/mm}^2

Answer: maximum tensile stress =134.38= 134.38 N/mm² (bottom of the web); maximum compressive stress =69.00= 69.00 N/mm² (top of the flange).

✓ Answer checked by Claude Opus 5.5 (high effort)

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