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WSM stresses under 50 kNm moment

Short answer

With neutral axis 165.72 mm and lever arm 494.76 mm, the extreme concrete stress is 4.07 N/mm2 and tension steel stress is 125.66 N/mm2, both below the permissible limits, so the section is safe.

A rectangular beam of size 300 mm x 550 mm effective depth is reinforced with 4-16 mm diameter tension bars. Determine the stress induced in the top compression fibre of the concrete and tension steel when it is subjected to a moment of 50 kN-m. Take M20 concrete and Fe 415 grade steel with the permissible stress in concrete and steel not exceeding 5 MPa and 140 MPa respectively.

Given: b=300b=300 mm, d=550d=550 mm, 4-16 mm dia, Ast=4×π4(16)2=804.2A_{st}=4\times\frac{\pi}{4}(16)^2=804.2 mm2^2, M=50M=50 kN·m. Permissible: σcbc=5\sigma_{cbc}=5, σst=140\sigma_{st}=140 N/mm2^2. For M20, modular ratio m=13.33m=13.33 (IS 456, Annex B-1.3).

Step 1: Neutral axis (elastic cracked section)

bx22=mAst(d−x) ⇒ 300x22=13.33×804.2(550−x)\frac{b x^2}{2}=m A_{st}(d-x)\ \Rightarrow\ \frac{300x^2}{2}=13.33\times 804.2(550-x) x=165.72 mmx=165.72\ \text{mm}

Step 2: Lever arm

z=d−x3=550−165.723=494.76 mmz=d-\frac{x}{3}=550-\frac{165.72}{3}=494.76\ \text{mm}

Step 3: Stresses

σc=2Mb x z=2×50×106300×165.72×494.76=4.07 N/mm2σs=MAst z=50×106804.2×494.76=125.66 N/mm2\begin{aligned} \sigma_{c}&=\frac{2M}{b\,x\,z}=\frac{2\times 50\times10^6}{300\times 165.72\times 494.76}=4.07\ \text{N/mm}^2\\ \sigma_{s}&=\frac{M}{A_{st}\,z}=\frac{50\times10^6}{804.2\times 494.76}=125.66\ \text{N/mm}^2 \end{aligned}

Both are less than the permissible values (5 and 140 N/mm2^2), so the section is safe.

Answer: stress in extreme compression concrete =4.07=4.07 N/mm2^2; stress in tension steel =125.66=125.66 N/mm2^2.

✓ Answer checked by Claude Opus 5.5 (high effort)

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