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Question: A Steel rod of cross-sectional area 1m² is acted upon by forces as shown in the figure. Determine th...

A Steel rod of cross-sectional area 1m² is acted upon by forces as shown in the figure. Determine the total elongation of the bar. Take Y = 2.0 × 10¹¹N/m².

A

1.3 µm

B

1.7 µm

C

0.9 µm

D

1.2 µm

Answer

1.3 µm

Explanation

Solution

  1. Identify segments and forces: The rod is divided into three segments: AB (length 1.5m), BC (length 1m), and CD (length 2m). Forces are applied at the ends and interfaces.

    • Force at the left end of AB: FA=60 kNF_A = 60 \, kN (left)
    • Force at B: FB=10 kNF_B = 10 \, kN (left)
    • Force at C: FC=20 kNF_C = 20 \, kN (right)
    • Force at the right end of CD: FD=50 kNF_D = 50 \, kN (right)

  2. Calculate internal forces in each segment: We use the method of sections. For equilibrium, the sum of external forces is 60 kN+10 kN60 \, kN + 10 \, kN (left) =70 kN= 70 \, kN (left) and 20 kN+50 kN20 \, kN + 50 \, kN (right) =70 kN= 70 \, kN (right). The system is in equilibrium.

    • Segment AB: Consider a cut in AB. The force to the left of the cut is 60 kN60 \, kN (left). Thus, the internal tensile force in AB is FAB=60 kN=60×103 NF_{AB} = 60 \, kN = 60 \times 10^3 \, N.
    • Segment BC: Consider a cut in BC. The forces to the left of the cut are 60 kN60 \, kN (left) and 10 kN10 \, kN (left). The net force to the left is 60+10=70 kN60 + 10 = 70 \, kN. Thus, the internal tensile force in BC is FBC=70 kN=70×103 NF_{BC} = 70 \, kN = 70 \times 10^3 \, N.
    • Segment CD: Consider a cut in CD. The forces to the left of the cut are 60 kN60 \, kN (left), 10 kN10 \, kN (left), and 20 kN20 \, kN (right). The net force to the left is 60+10−20=50 kN60 + 10 - 20 = 50 \, kN. Thus, the internal tensile force in CD is FCD=50 kN=50×103 NF_{CD} = 50 \, kN = 50 \times 10^3 \, N.

  3. Calculate elongation in each segment: The elongation of a segment is given by ΔL=FLAY\Delta L = \frac{FL}{AY}.

    • Given: A=1 m2A = 1 \, m^2, Y=2.0×1011 N/m2Y = 2.0 \times 10^{11} \, N/m^2.
    • ΔLAB=FABLABAY=(60×103 N)×(1.5 m)(1 m2)×(2.0×1011 N/m2)=90×1032×1011 m=45×10−8 m=0.45×10−6 m=0.45 μm\Delta L_{AB} = \frac{F_{AB} L_{AB}}{A Y} = \frac{(60 \times 10^3 \, N) \times (1.5 \, m)}{(1 \, m^2) \times (2.0 \times 10^{11} \, N/m^2)} = \frac{90 \times 10^3}{2 \times 10^{11}} \, m = 45 \times 10^{-8} \, m = 0.45 \times 10^{-6} \, m = 0.45 \, \mu m.
    • ΔLBC=FBCLBCAY=(70×103 N)×(1 m)(1 m2)×(2.0×1011 N/m2)=70×1032×1011 m=35×10−8 m=0.35×10−6 m=0.35 μm\Delta L_{BC} = \frac{F_{BC} L_{BC}}{A Y} = \frac{(70 \times 10^3 \, N) \times (1 \, m)}{(1 \, m^2) \times (2.0 \times 10^{11} \, N/m^2)} = \frac{70 \times 10^3}{2 \times 10^{11}} \, m = 35 \times 10^{-8} \, m = 0.35 \times 10^{-6} \, m = 0.35 \, \mu m.
    • ΔLCD=FCDLCDAY=(50×103 N)×(2 m)(1 m2)×(2.0×1011 N/m2)=100×1032×1011 m=50×10−8 m=0.50×10−6 m=0.50 μm\Delta L_{CD} = \frac{F_{CD} L_{CD}}{A Y} = \frac{(50 \times 10^3 \, N) \times (2 \, m)}{(1 \, m^2) \times (2.0 \times 10^{11} \, N/m^2)} = \frac{100 \times 10^3}{2 \times 10^{11}} \, m = 50 \times 10^{-8} \, m = 0.50 \times 10^{-6} \, m = 0.50 \, \mu m.

  4. Calculate total elongation: The total elongation is the sum of elongations of all segments.

    • Total Elongation = ΔLAB+ΔLBC+ΔLCD=0.45 μm+0.35 μm+0.50 μm=1.30 μm\Delta L_{AB} + \Delta L_{BC} + \Delta L_{CD} = 0.45 \, \mu m + 0.35 \, \mu m + 0.50 \, \mu m = 1.30 \, \mu m.