Question 1.11: In Figure 1.31(a), the rod of uniform cross-section is rigid...
In Figure 1.31(a), the rod of uniform cross-section is rigidly fixed at both ends and it is subjected to loads P_1 \text { and } P_2 \text { as shown. If } P_1=30 kN , P_2=60 kN \text { and } E=200 GPa , find out the reactions at each end and change in length of each portion. Take diameter of the rod as d = 25 mm.
The problem is statically indeterminate as the number of unknown quantities are two, and only the equation of equilibrium is along the horizontal direction. Let us consider that reaction at the left end (end A) be P and it is compressive as shown in Figure 1.31(b). For segment AC, load is compressive; for segment CD, load is compressive and for segment DB, load is tensile. If after solving we get ‘P’ to be negative, then we will definitely be able to understand the exact nature of loading on segments. It may change (say after calculation, ‘P’ comes out to be negative; obviously segment AC will be in tension).
Now for segment AC as per free-body diagram, contraction \Delta L_1 is given by
\Delta L_1=\frac{-P \times 275}{A E}
negative sign is because of contraction. Similarly for segment CD, contraction is given by
\Delta L_2=\frac{-(30+P) \times 150}{A E}
and for segment DB, elongation is given by
\Delta L_3=\frac{(30-P) \times 375}{A E}
As the supports are rigid, the total change in length is zero (compatibility equation). Thus,
\Delta L_1+\Delta L_2+\Delta L_3=0
or \frac{-P \times 275}{A E}+\frac{-(30+P) \times 150}{A E}+\frac{(30-P) \times 375}{A E}=0
or -275 P-(30+P) × 150+(30-P) × 375=0
This gives P = 8.4375 kN . So, on segment AC, load is compressive as we assumed.
For segment CD, load is compressive and its magnitude is = (30 + 8.4375) kN = 38.4375 kN.
For segment DB, load is (30 – 8.4375) kN or 21.5625 kN and it is tensile in nature.
Therefore, support reaction at A is 8.4375 kN (compressive) and support reaction at B is 21.5625 kN (tensile).
Now, for segment AC,
\text { Contraction }=\frac{8.4375 \times 275}{(\pi / 4) \times 25^2 \times 200 \times 10^3}=0.0236 mm
For segment CD,
\text { Contraction }=\frac{38.4375 \times 150}{(\pi / 4) \times 25^2 \times 200 \times 10^3}=0.0587 mm
For segment DB,
\text { Elongation }=\frac{21.5625 \times 375}{(\pi / 4) \times 25^2 \times 200 \times 10^3}=0.0823 mm
Therefore, changes in length are 0.0236 mm (contraction), 0.0587 mm (contraction) and 0.0823 mm (elongation), respectively.