A bracket is attached to a vertical column by means of two identical rivets U and V separated by a distance of 2a = 100 mm, as shown in the figure. The permissible shear stress of the rivet material is 50 MPa. If a load P = 10 kN is applied at an eccentricity 𝑒 = 3√7 𝑎, the minimum cross-sectional area of each of the rivets to avoid failure is ___________ mm2 .

Correct Answer/Option: The correct option is 800.

Step-by-step Explanation:

1. Understanding the Given Data:
From the problem statement and the attached schematic diagram, we have:

• Two identical rivets U and V are aligned vertically on a column.
• The vertical distance between the two rivets is $2a=100\text{mm}$, which gives $a=50\text{mm}$.
• The centroid of the rivet group (denoted as G) lies exactly at the midpoint of the line segment UV. Therefore, the distance of each rivet from the centroid is:

  $r_U=r_V=a=50\text{mm}$

• A vertical load of $P=10\text{kN}=10,000\text{N}$ is applied downwards at an eccentricity of:

  $e=3\sqrt{7}a$

• The permissible shear stress of the rivet material is $\tau =50\text{MPa}=50{\text{N/mm}}^2$.

2. Direct (Primary) Shear Force on Rivets:
The direct vertical load $P$ is distributed equally between the two identical rivets. The primary shear force ($P_d$) on each rivet acts vertically downwards and is given by:
$P_d=\frac{P}{2}$
Substituting $P=10\text{kN}$:
$P_d=\frac{10}{2}=5\text{kN}=5,000\text{N}$

3. Secondary Shear Force due to Eccentric Moment:
The eccentric load creates a clockwise moment $M$ about the centroid $G$:
$M=P\times e$
This moment is resisted by the secondary shear forces ($F_s$) on the rivets. The secondary shear force on any rivet at distance $r$ from the centroid is:
$F_s=\frac{M\cdot r}{r_U^2+r_V^2}$
Since $r_U=r_V=a$, the secondary shear force on both rivets is equal and is given by:
$F_s=\frac{(P\cdot e)\cdot a}{a^2+a^2}=\frac{P\cdot e\cdot a}{2a^2}=\frac{P\cdot e}{2a}$
Now substitute $e=3\sqrt{7}a$ into the equation:
$F_s=\frac{P\cdot \left(3\sqrt{7}a\right)}{2a}=1.5\sqrt{7}P$

4. Determining the Direction of Secondary Shear Forces:
The eccentric load causes a clockwise moment. To resist this:

• At rivet U (located above centroid G), the secondary shear force acts horizontally to the right.
• At rivet V (located below centroid G), the secondary shear force acts horizontally to the left.

Since the primary shear force $P_d$ acts vertically downwards on both rivets, the primary and secondary shear forces are perpendicular ($\theta =90°$) at both rivet locations.

5. Calculating the Resultant Shear Force:
The resultant shear force $R$ on each rivet is:
$R=\sqrt{P_d^2+F_s^2}$
Substitute the values of $P_d$ and $F_s$ in terms of $P$:
$R=\sqrt{{\left(\frac{P}{2}\right)}^2+{\left(1.5\sqrt{7}P\right)}^2}$
$R=P\sqrt{0.25+2.25\times 7}$
$R=P\sqrt{0.25+15.75}$
$R=P\sqrt{16}=4P$
Substituting $P=10,000\text{N}$:
$R=4\times 10,000=40,000\text{N}$

6. Calculation of Minimum Cross-Sectional Area:
To avoid failure, the shear stress in each rivet must not exceed the permissible value:
$\tau =\frac{R}{A}\le {\tau }_{\text{permissible}}$
Therefore, the minimum cross-sectional area $A$ is:
$A=\frac{R}{{\tau }_{\text{permissible}}}=\frac{40,000\text{N}}{50{\text{N/mm}}^2}=800{\text{mm}}^2$