What is the ordinate of influence line at Q for reaction at S as shown in figure given below ? (Beam P-Q-R-S with internal hinge at Q, supports at P, R, and S. Distances: PQ=3m, QR=3m, RS=6m)
(Aromatic compound with an isobutyl side chain) $\xrightarrow[KOH, Heat]{KMnO_4}$ ?
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For statically determinate beams with hinges, ILDs are always composed of straight line segments. The hinge allows the beam to rotate freely, changing the direction of the line.
Concept:
To draw the Influence Line Diagram (ILD) for a reaction at a support, we use Müller-Breslau's Principle: remove the constraint (support) and apply a unit displacement in the direction of the reaction.
Step 1: Apply unit displacement at S.
To find ILD for $R_S$:
• Lift support S by 1 unit.
• Support R is a pin/roller, so displacement at R is 0.
• The portion RS is a straight line. Since it is 1 at S and 0 at R (distance 6m), the slope is $1/6$.
Step 2: Analyze the internal hinge at Q.
The line continues from R through the hinge Q. Since the slope is constant for the segment from S to the hinge (as there are no other supports between R and Q):
• Ordinate at R ($x=0$) is 0.
• Ordinate at Q ($x=3$ from R in the opposite direction of S) is calculated by similar triangles.
Step 3: Calculate ordinate at Q.
By similar triangles between segment RS and segment RQ:
\[ \frac{\text{Ordinate at S}}{RS} = \frac{\text{Ordinate at Q}}{RQ} \]
\[ \frac{1}{6} = \frac{y_Q}{3} \]
\[ y_Q = \frac{3}{6} = 0.5 \]
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