A horizontal beam (I = 8.616 × 107mm4) carries a uniformly distribute...
Area of each rod
= A1 =
= 490.87 mm
2Area of the middle rod
= A2 =
= 706.86 mm
2Let, P1 = Tension in each end rod
P2 = Tension in the middle rod
f1 = Stress in the rods
f2 = Stress in the middle rod
δ1 = Extension of each end rod
δ2 = Extension of the middle rod
Deflection of the midpoint of the beam
δ2 − δ1 =
EI(δ2 − δ1) =
Deflection of beams
8.616 × 107 × 1800(f2 − f1)
=
2205.696(f2 − f1) = 250 − 5654.88f2
2205.696f2 − 2205.696f1 = 250 − 5654.88f2
7860.576f2 − 2205.696f1 = 250 ⋯ ①
2f1A1 + f2A2 = 50
2f1(490.87) + f2 (706.86) = 50
981.74f, + 706.86f, = 50 ⋯ ②
From equation ① and ②,
7860.576f2 − 2205.696f1 = 5(981.74f,1 + 706.86f2)
7860.576f2 − 2205.696f1 = 4908.70f1 + 3534.3f2
4326.76f2 = 7114.396f1
∴ f2 = 1.644f1
Substituting in equation ②
981.74f1 + 706.86 × 1.644f1 = 50
2143.8178f1 = 50
f1 = 0.02332 kN/mm2 = 23.32 N/mm2
f2 = 1.644 × 23.32 = 38.33 N/mm2
Deflection of the midpoint of the beam
δ
2 − δ
1 =
=
=0.135 mm
Question_Type: 5