Draw the envelope of the bending moment and shear force diagrams from the different load cases showing the corresponding extreme values.

Assignment Brief

CE5013 REINFORCED CONCRETE RETAKE CW– Aug 2020

ID: The ‘n’ values below are based on your KU ID, i.e. K n₁ n₂ n₃ n₄ n₅ n₆ n₇

A simply supported reinforced concrete beam with a cantilever of rectangular cross section 250 mm width X 525 mm height and concrete grade C35/40 is an interior support of a one way slab with uniformly distributed characteristic loads from

• Tiles and bedding (0.5+0.1*n₃) kN/m²
• 50 mm layer of screed with unit weight of 22 kN/m³
• 180 mm deep slab with RC unit weight of 25 kN/m³
• Suspended ceiling of 0.15 kN/m²
• Services of 0.30 kN/m²

and a variable load q_k = (2+0.1*n₇) kN/m².The distance c/c of the beams is (4+0.1*n₆) m. Design the section as a rectangular beam to satisfy the ULS and SLS requirements of EuroCode 2. 

Assessment criteria: 

Sample Answer

Design of Shear Reinforcement

1. Assumed Beam Data

• Beam width (b): 250 mm

• Beam height (h): 525 mm

• Effective depth (d): 475 mm (assuming 50 mm cover + 20 mm bar diameter)

• Concrete grade: C35/40

• Characteristic compressive strength (f_ck): 35 MPa

• Partial safety factor for concrete (γ_c): 1.5

• Steel grade: B500 (f_yk = 500 MPa)

• Variable design shear force (V_Ed): 110 kN (assumed max)

• Span: 6.0 m

2. Design Shear Resistance without Shear Reinforcement (V_Rd,c)

Eurocode 2 Clause 6.2.2:

VRd,c=[CRd,c⋅k⋅(100⋅ρ⋅fck)1/3]⋅b⋅dV_{Rd,c} = [C_{Rd,c} cdot k cdot (100 cdot ho cdot f_{ck})^{1/3}] cdot b cdot dVRd,c​=[CRd,c​⋅k⋅(100⋅ρ⋅fck​)1/3]⋅b⋅d

Where:

• C_Rd,c = 0.18 / γ_c = 0.18 / 1.5 = 0.12

• k = 1 + √(200/d), with d in mm, max 2.0

  k=1+200/475=1+0.648=1.648k = 1 + sqrt{200/475} = 1 + 0.648 = 1.648k=1+200/475​=1+0.648=1.648

• ρ = A_s / (b·d), assume A_s = 1256 mm² (4T20)

  $$\rho =1256/(250\cdot 475)=0.0106$$

Now,

VRd,c​=[CRd,c​⋅k⋅(100⋅ρ⋅fck​)1/3]⋅b⋅d

Where:

• CRd,c=0.18γc=0.181.5=0.12C_{Rd,c} = frac{0.18}{gamma_c} = frac{0.18}{1.5} = 0.12CRd,c​=γc​0.18​=1.50.18​=0.12

• k=1+200d=1+200475=1+0.648=1.648k = 1 + sqrt{frac{200}{d}} = 1 + sqrt{frac{200}{475}} = 1 + 0.648 = 1.648k=1+d200​​=1+475200​​=1+0.648=1.648

• ρ=Asb⋅d=1256250⋅475=0.0106 ho = frac{A_s}{b cdot d} = frac{1256}{250 cdot 475} = 0.0106ρ=b⋅dAs​​=250⋅4751256​=0.0106

• fck=35 MPaf_{ck} = 35 , ext{MPa}fck​=35MPa

• $b=250\text{mm}$

• $d=475\text{mm}$

Now compute step by step:

100⋅ρ⋅fck=100⋅0.0106⋅35=37.1100 cdot ho cdot f_{ck} = 100 cdot 0.0106 cdot 35 = 37.1100⋅ρ⋅fck​=100⋅0.0106⋅35=37.1 (100⋅ρ⋅fck)1/3=37.13=3.33(100 cdot ho cdot f_{ck})^{1/3} = sqrt[3]{37.1} = 3.33(100⋅ρ⋅fck​)1/3=337.1​=3.33 VRd,c=(0.12⋅1.648⋅3.33)⋅250⋅475V_{Rd,c} = (0.12 cdot 1.648 cdot 3.33) cdot 250 cdot 475VRd,c​=(0.12⋅1.648⋅3.33)⋅250⋅475 $$=0.658\cdot 250\cdot 475=78.4\text{kN}$$

Final Answer:

VRd,c=78.4 kNoxed{V_{Rd,c} = 78.4 , ext{kN}}VRd,c​=78.4kN​

Since V_Ed = 110 kN > V_Rd,c = 78.4 kN, shear reinforcement is required.

3. Design Shear Reinforcement (V_Rd,s)

According to Eurocode 2 Clause 6.2.3:

VRd,s=Asw⋅fywd⋅zsV_{Rd,s} = frac{A_{sw} cdot f_{ywd} cdot z}{s}VRd,s​=sAsw​⋅fywd​⋅z​

Assume:

• z = 0.9·d = 0.9·475 = 427.5 mm

• f_ywd = 500 / 1.15 = 435 MPa

• Use 2-legged T8 stirrups → A_sw = 2 × 50.3 = 100.6 mm²

Rearranged to solve for spacing s:

s=Asw​⋅fywd​⋅zVEd​​

Substitute the values:

s=100.6×435×427.5110×103s = frac{100.6 imes 435 imes 427.5}{110 imes 10^3}s=110×103100.6×435×427.5​ s=18,683,092.5110,000=169.85 mms = frac{18,683,092.5}{110,000} = 169.85 , ext{mm}s=110,00018,683,092.5​=169.85mm

Let’s adopt s = 150 mm c/c to meet requirements.

Continued...