Step 1: Formula for limiting moment of resistance.
The limiting moment of resistance for a singly reinforced beam is given by:
\[
M_{lim} = 0.36 \, f_{ck} \, b \, x_{u,max} \left(d - 0.42x_{u,max}\right)
\]
where:
- \( f_{ck} = 20 \, \text{N/mm}^2 \) (M-20 concrete),
- \( b = 200 \, \text{mm} \),
- \( d = 300 \, \text{mm} \),
- \( x_{u,max} = 0.48d = 0.48 \times 300 = 144 \, \text{mm}. \)
Step 2: Substitution.
\[
M_{lim} = 0.36 \times 20 \times 200 \times 144 \times (300 - 0.42 \times 144)
\]
First compute:
\[
300 - 0.42 \times 144 = 300 - 60.48 = 239.52 \, \text{mm}.
\]
Now:
\[
M_{lim} = 0.36 \times 20 \times 200 \times 144 \times 239.52
\]
\[
M_{lim} = 498,99000 \, \text{Nmm} \approx 5.0 \times 10^7 \, \text{Nmm}
\]
Convert to kNm:
\[
M_{lim} = \frac{5.0 \times 10^7}{10^6} = 50 \, \text{kNm}.
\]
Step 3: Conclusion.
The limiting moment of resistance is approximately \(\, 50 \, \text{kNm}.\)
A weight of $500\,$N is held on a smooth plane inclined at $30^\circ$ to the horizontal by a force $P$ acting at $30^\circ$ to the inclined plane as shown. Then the value of force $P$ is:
A steel wire of $20$ mm diameter is bent into a circular shape of $10$ m radius. If modulus of elasticity of wire is $2\times10^{5}\ \text{N/mm}^2$, then the maximum bending stress induced in wire is: