Step 1: Find the support reactions.
The beam is simply supported at P and S, with a 10 kN load at Q (2 m from P) and a 20 kN load at R (4 m from P). The span PS is 6 m.
Take moments about P and set the sum to zero:
\[ R_S \times 6 = 10 \times 2 + 20 \times 4 \]
\[ R_S \times 6 = 20 + 80 = 100 \]
\[ R_S = \frac{100}{6} = 16.67 \text{ kN} \]
The total downward load is \(10 + 20 = 30\) kN, so
\[ R_P = 30 - 16.67 = 13.33 \text{ kN} \]
Step 2: Work out the shear force in each segment.
Between two point loads the shear force stays constant, it only changes at a load or a reaction.
From P to Q, the only force to the left of the section is \(R_P\), so \(V_{PQ} = 13.33\) kN.
At Q the 10 kN load pulls the shear down, so from Q to R, \(V_{QR} = 13.33 - 10 = 3.33\) kN.
At R the 20 kN load pulls the shear down again, so from R to S, \(V_{RS} = 3.33 - 20 = -16.67\) kN.
Just after S, the reaction \(R_S = 16.67\) kN brings the shear back to zero, which closes the diagram correctly.
Step 3: Find where the shear force diagram crosses zero.
The shear is a constant \(+3.33\) kN right up to R and a constant \(-16.67\) kN right after R, so the sign flips exactly at R. On the shear force diagram this shows up as the vertical drop line at R crossing the zero axis, so R is the point of zero shear inside the span.
This point also separates the sagging and hogging tendency of the beam, so it marks where the bending moment is greatest.
The zero values at P and S are trivial, they are just where the diagram starts and ends at the supports, not an interior sign change.
Final Answer:
The zero shear force location is R.
\[ \boxed{R} \]
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