Design Calculation.docx

Determination of load carrying capacity:-

Since small wheel takes up the complete load while coming in contact with edges of the stair it is much more prone to failure so load carrying capacity is found with respect to that wheel. σ y (yield strength of bolt) = 400 N/mm2 n (Factor of safety) = 5 τ y (Shear yield strength) = σ y / 2 ( From maximum shear theory ) = 400 * 0.5 = 200 N/mm2 τ ( Permissible shear stress) = τ y / n = 200/5 = 40 N/mm2 Bolt selected = M4 ( 4mm diameter) Maximun force which can be applied:τ=F/A F = (π / 4) * d2 * 40 = 502.65 N

F*= It is the force applied by the man to pull the trolley.

Resolution of force applied on the trolley during climbing

So the force acting on the small wheel :F cos 30 = 502.65N F = 580.41 N Vertical component

Fw :- F sin 30 = 290.20

Maximum load which can be applied:W= Fw / 9.8 = 29.6122 = 30 kg Therefore maximum load which can be carried by the trolley is 30 kg.

Checking of safety of rubber:Tensile strength = 52 N/mm2 A = π/4 (D2-d2 ) = 235.619 mm2 F= 290.20 N τ = F/A = 1.23 N/mm2 < 52 N/mm2 so safe Force to be applied by man:F= 580.41 N

Design of bearing:Available = SKF6004 Load carrying capacity = co =4500 N Load to be carried = 295 N Assuming working condition –> speed=20 rpm Life in hours= 1000 hrs From psgdb-> c/p = 1.06 C=1.06*295 = 312.7 N Since, C < Co design of ball bearing is safe.

Selection of shaft:Dimension of SKF6004 -> d=20mm Selected shaft diameter = 20mm

Material -> wrought iron σ y = 165 N/mm2 Failure type-> Double shear σ = 0.5 * 165 = 82.5 N/mm2 Factor of safety= 5 τ max = 82.5/5 =16.5 N/mm2 τ= F/2A= 294/(2*(π/4)*202 ) = 0.5 N/mm2

Since, τ < τ max shaft does not fail by shear