Active forum topics
- Hydraulics: Rotating Vessel
- Inverse Trigo
- General Solution of $y' = x \, \ln x$
- engineering economics: construct the cash flow diagram
- Eliminate the Arbitrary Constants
- Law of cosines
- Maxima and minima (trapezoidal gutter)
- Special products and factoring
- Integration of 4x^2/csc^3x√sinxcosx dx
- application of minima and maxima
New forum topics
- Hydraulics: Rotating Vessel
- Inverse Trigo
- General Solution of $y' = x \, \ln x$
- engineering economics: construct the cash flow diagram
- Integration of 4x^2/csc^3x√sinxcosx dx
- Maxima and minima (trapezoidal gutter)
- Special products and factoring
- Newton's Law of Cooling
- Law of cosines
- Can you help me po to solve this?
Recent comments
- Determine the least depth…1 week 2 days ago
- Solve mo ang h manually…2 weeks 6 days ago
- Paano kinuha yung height na…2 weeks 6 days ago
- It's the unit conversion…1 month ago
- Refer to the figure below…3 weeks 5 days ago
- Yes.4 months 2 weeks ago
- Sir what if we want to find…4 months 2 weeks ago
- Hello po! Question lang po…5 months 1 week ago
- 400000=120[14π(D2−10000)]
(…6 months 1 week ago - Use integration by parts for…7 months 1 week ago
Given:
Given:
Mu = 1100 kN-m
f'c = 27.6 MPa
fy = 276 MPa
Step 1 - Calculate β1 and ρmax
β1 = 0.85
ρmax = β1 * 0.85 * f'c * 3/ fy * 7
ρmax = 0.85*0.85*276*3/ 276 * 7
ρmax = 0.030964
Step 2 - Calculate Rmax
Rmax = ρmax * fy *(1-ρmax* fy/1.7*f'c)
Rmax = 0.0300964* 276*(1-(0.00309*276/(1.7*27.6))
Rmax = 6.9895 MPa
Step 3 - Calculate Mmax and compare with Mu
Mmax = Rmax * φ * b * d^2
Mmax = 6.9895 * 0.90 * 350 * 725^2
Mmax = 1,157.265 kN-m
Mu = 1100 kN-m < Mmax = 1,157.626 kN-m
Hence, The beam can be designed as Singly-Reinforced Beam (SRB)