Two prismatic bars are rigidly fastened
together and support a vertical load of 40 kN.
- The upper
bar is of steel, having: - Unit weight
γs=760 kN/m3 - Length L=10 m
- Cross-sectional
area As=80.7 cm2 - The lower
bar is of brass, having: - Unit weight
γb=800 kN/m3 - Length L=6 m
- Cross-sectional
area Ab=64.5 cm2 - Modulus of
elasticity: - For steel,
Es=207 GPa - For brass,
Eb=90 GPa
Determine:
(a) Extension of free end due to
self-weight only
(b) Maximum stress in each material
Solution:
(a) Extension of Free End Due to Self-Weight
The extension Δ
will be due to the following:
- Axial
deflection of BC (brass) due to its own weight - Axial
deflection of AB (steel) due to its own weight - Axial
deflection of AB (steel) due to the weight of BC
(i) Axial deflection of BC due to self-weight
Δ1=γbLb22Eb
Δ1=800×103×622×0.9×1011=0.16 mm
(ii) Axial deflection of AB due to self-weight
Δ2=γsLs22Es
Δ2=760×103×1022×2.07×1011=0.1836 mm
(iii) Axial deflection of AB due to weight of BC
Weight of BC:
WeightBC=γb×Ab×L=800×6.45×10–3×6=30.96 kN
Axial deflection:
Δ3=PLAsEs=30.96×103×10×1038070×2.07×105=0.185 mm
Total Extension at Free End:
Δ=Δ1+Δ2+Δ3
Δ=0.16+0.1836+0.185=0.5286 mm
(b) Maximum Stress in Each Material
(i) Maximum Stress in Brass
Occurs at junction (B), and is due to:
- External
load 40 kN - Self-weight
of brass bar 30.96 kN
P=40+30.96=70.96 kN
σbmax=PAb=70.96×1036450=11.00 N/mm2
(ii) Maximum Stress in Steel
Occurs at the fixed support (A), and is
due to:
- External
load = 40 kN - Self-weight
of steel = γs×10×As
=760×10×8.07×10–3=61.33 kN
- Weight of BC
transferred = 30.96 kN
So, total force at support:
P=40+30.96+61.33=132.29 kN
σsmax=PAs=132.29×1038070=16.39 N/mm2
Final Answers:
- Extension of
free end: Δ=0.5286 mm - Maximum
stress in brass: 11.00 N/mm2 - Maximum
stress in steel: 16.39 N/mm2
