A steel plate, connnected to a fixed channel using three identical bolts $A, B$ and $C$ , carries a load of $6 k N$ as shown in the figure. Considering the effect of direct load and moment,the magnitude of resultant shear force (in kN) on bolt $C$ is

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Solution

The correct option is B $17$
$P_{P} = \frac{P}{n} = \frac{P}{3} = 2 k N$
$r_{A} = r_{C} = 50; (P_{S})_{A} = (P_{S})_{c}$
$r_{B} = 0 \Rightarrow (P_{S})_{B} = 0$
$\frac{(P_{S})_{C}}{r_{C}} [r_{C}^{2} + r_{B}^{2} + r_{A}^{2}] = P \times e$

$(P_{S})_{C} \times 2 \times r_{C} = P \times e$
$(P_{S})_{C} = \frac{6 \times 250}{2 \times 50} = 15 k N$
$θ_{A} = 180^{\circ}; θ_{C} = 0^{\circ}$
Resultant shear force $(R_{C}) = P_{p} + (P_{s})_{C} [∵ θ_{C} = 0^{\circ}]$
$R_{c} = 2 + 15 = 17 k N$
$R_{A} = (P_{s}) A - P_{p} = 13 k N$
$R_{B} = P_{p} = 2 k N$

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A steel plate, connnected to a fixed channel using three identical bolts A,B and C, carries a load of 6 kNas shown in the figure. Considering the effect of direct load and moment,the magnitude of resultant shear force (in kN) on bolt C is

The correct option is B 17PP=Pn=P3=2 kN rA=rC=50;(PS)A=(PS)c rB=0⇒(PS)B=0 (PS)CrC[r2C+r2B+r2A]=P×e (PS)C×2×rC=P×e (PS)C=6×2502×50=15kN θA=180∘;θC=0∘ Resultant shear force (RC)=Pp+(Ps)C [∵θC=0∘] Rc=2+15=17kN RA=(Ps)A−Pp=13 kN RB=Pp=2 kN

A steel plate, connnected to a fixed channel using three identical bolts $A, B$ and $C$ , carries a load of $6 k N$ as shown in the figure. Considering the effect of direct load and moment,the magnitude of resultant shear force (in kN) on bolt $C$ is