The largest permissible vertical shearing force that can be applied to the reinforced beam without exceeding the allowable shearing stress in the bolts is approximately 5.157 kN.
To determine the largest permissible vertical shearing force that the rolled-steel beam with the attached plates can withstand, we need to consider the shearing capacity of the bolts and ensure it does not exceed the allowable shearing stress.
Let's calculate it step by step:
1. Calculate the Total Shearing Area of the Bolts (Atotal):
- The area of a single bolt can be calculated using the formula for the area of a circle: A = π * (radius)^2.
- Given that the diameter of the bolts is 18 mm, the radius (r) is 9 mm (0.009 m).
- The area of one bolt is A = π * (0.009 m)^2 ≈ 0.0002545 square meters.
- There are 2 bolts per spacing interval (125 mm), so the total shearing area of bolts per spacing interval is 2 * 0.0002545 = 0.000509 square meters.
2. Calculate the Number of Bolts in One Spacing Interval (N):
- To determine the number of bolts in one spacing interval, we divide the width of the plates (175 mm) by the spacing interval (125 mm).
- N = 175 mm / 125 mm = 1.4 bolts. Since we can't have a fraction of a bolt, we'll use 1 bolt in our calculations.
3. Calculate the Total Shearing Area of Bolts in One Spacing Interval (Ainterval):
- Ainterval = N * Atotal = 1 * 0.000509 square meters ≈ 0.000509 square meters.
4. Calculate the Shearing Force That Can Be Withstood by One Spacing Interval (Finterval):
- The shearing force (F) that can be resisted by the bolts in one spacing interval is given by the formula: F = τ * Ainterval, where τ is the allowable shearing stress in the bolts.
- τ = 105 MPa = 105 N/mm² = 105 N/(0.000001 m²) = 105,000,000 N/m².
- Finterval = 105,000,000 N/m² * 0.000509 m² ≈ 53,745 N (Newtons).
5. Calculate the Total Shearing Force (Ftotal):
- Since there are multiple spacing intervals along the length of the plates, we need to calculate the total shearing force by multiplying Finterval by the number of intervals.
- Ftotal = Finterval * (Total Length of Plates / Spacing Interval)
- The total length of the plates is 12 mm (0.012 m).
- Ftotal = 53,745 N * (0.012 m / 0.125 m) ≈ 5,157 N (Newtons).
6. Convert Newtons to kiloNewtons (kN):
- 1 kN = 1000 N, so Ftotal = 5,157 N / 1000 ≈ 5.157 kN.