--- title: "Design of 250x300 beam on masonry walls" subject: "Design of RCC Structure" chapter: 4 key: "semester_7/design_of_rcc_structure:design-of-250x300-beam-on-masonry" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/design-of-250x300-beam-on-masonry site: BCT Engineering --- # Design of 250x300 beam on masonry walls Design of RCC Structure · Chapter 4: Design of beams: Behavior in Flexure - Asked in: 2081 Bhadra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/design-of-250x300-beam-on-masonry - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-4 ## Question A beam of size 250 mm x 300 mm is simply supported on masonry walls of thickness 300 mm and 4 m apart to support a live load of 8 kN/m and dead load of 10 kN/m including its self-weight. Design the beam in flexure. Use M20 grade concrete and Fe 415 steel. ## Short answer Mu=61.25 kNm. Section doubly reinforced: required Ast=806 mm² (2-20+1-16 mm) and Asc=200 mm² (2-12 mm). ## Answer **Given:** $b=250$ mm, $D=300$ mm, clear span 4 m between walls 300 mm thick, DL = 10 kN/m (including self weight), LL = 8 kN/m, M20, Fe415. Assume an effective cover of 40 mm at top and bottom, so $d=300-40=260$ mm and $d'=40$ mm. **Step 1: Effective span** (IS 456 cl. 22.2): the lesser of the centre-to-centre distance of supports ($4+0.3=4.3$ m) and clear span $+\,d$ ($4+0.26=4.26$ m). Take $L_{eff}=4.26$ m. **Step 2: Factored load and moment** $$ w_u=1.5(10+8)=27.0\ \text{kN/m},\qquad M_u=\frac{w_uL^2}{8}=\frac{27.0\times4.26^2}{8}=61.25\ \text{kN·m} $$ **Step 3: Limiting moment of the section** $$ M_{u,lim}=0.138f_{ck}bd^2=0.138\times20\times250\times260^2=46.63\ \text{kN·m} $$ Since $M_u>M_{u,lim}$ and the depth is fixed, the beam must be **doubly reinforced**. **Doubly reinforced design** $$ M_{u2}=M_u-M_{u,lim}=61.25-46.63=14.62\ \text{kN·m} $$ Limiting depth $x_{u,max}=0.48d=124.8$ mm. Strain in compression steel (similar triangles): $$ \varepsilon_{sc}=0.0035\left(1-\frac{d'}{x_{u,max}}\right)=0.0035\left(1-\frac{40}{124.8}\right)=0.00238 $$ From the design stress-strain curve of Fe415 (IS 456 Fig. 23): $f_{sc}=341.6$ N/mm$^2$. Stress in the concrete displaced by the steel: $f_{cc}=0.446f_{ck}=8.92$ N/mm$^2$. $$ A_{sc}=\frac{M_{u2}}{(f_{sc}-f_{cc})(d-d')}=\frac{14.62\times10^6}{(341.6-8.92)(260-40)}=200\ \text{mm}^2 $$ $$ A_{st1}=\frac{0.36f_{ck}\,b\,x_{u,max}}{0.87f_y}=622\ \text{mm}^2,\qquad A_{st2}=\frac{A_{sc}(f_{sc}-f_{cc})}{0.87f_y}=184\ \text{mm}^2 $$ $$ A_{st}=A_{st1}+A_{st2}=806\ \text{mm}^2 $$ **Provide:** tension steel 2-20 mm + 1-16 mm dia ($829$ mm$^2$) and compression steel 2-12 mm dia ($226$ mm$^2$). (Much more tension steel than required, such as 5-16 mm, should not be provided, because it would push $x_u$ beyond $x_{u,max}$ and make the section over-reinforced.) **Check:** with the bars provided, $x_u=\dfrac{0.87f_yA_{st}-A_{sc}(f_{sc}-f_{cc})}{0.36f_{ck}b}=\dfrac{299.3-75.2}{1.8}\approx124.5$ mm $\le x_{u,max}=124.8$ mm, OK. $p_t=100\times829/(250\times260)=1.28\%$ $<4\%$, OK. Two-legged 8 mm stirrups at a suitable spacing are provided for shear and to hold the compression bars (spacing $\le 16\phi$ of the compression bar). Shear design to be done separately. **Answer:** $M_u=61.25$ kN·m; $A_{st}=806$ mm$^2$ (2-20 + 1-16 mm) and $A_{sc}=200$ mm$^2$ (2-12 mm). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the effective span of a beam on masonry walls determined? A. Least of c/c of supports and clear span + d B. Always clear span C. Greatest of the two D. Clear span + wall thickness Answer: A. Least of 4.3 m and 4.26 m gives 4.26 m. (Key point: Effective span) 2. Why is the beam designed as doubly reinforced? A. Mu = 61.25 kN·m exceeds Mu,lim = 46.63 kN·m with fixed size B. Mu is below Mu,lim C. Because span exceeds 5 m D. Because steel is Fe415 Answer: A. Since the section cannot be enlarged, compression steel is added. (Key point: Doubly reinforced reason) 3. How is the strain in compression steel computed? A. 0.0035(1 - d'/xu,max) = 0.00238 B. 0.0035(d'/xu,max) C. 0.002 D. 0.87 fy/Es Answer: A. Similar triangles with xu,max = 0.48d = 124.8 mm and d' = 40 mm. (Key point: Compression steel strain) 4. How is total tension steel found? A. Ast1 (for Mu,lim) + Ast2 (balancing compression steel force) B. Ast1 only C. Asc x 2 D. Mu/(0.87 fy d) Answer: A. Ast = 622 + 184 = 806 mm^2. 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