Jib Crane Pivot Pin Uprating
A jib crane in a Guwahati workshop is rated at 1.5 t and uses a 16 mm diameter pivot pin in double shear. The pin passes through two 10 mm thick mild-steel cheek plates welded to the column bracket. The pin is commercial bright bar with a yield strength of 300 MPa; the design allows a shear stress of 80 MPa and a bearing pressure of 120 MPa. The crane lifts diesel engine blocks and is occasionally subjected to snatch loads, so the workshop applies a dynamic factor of 2. An apprentice checked only the static load, found a shear stress of about 37 MPa, and concluded that the crane could safely be uprated to 2 t. The owner wants to increase the rating without changing the hoist or the wire rope. A spare 20 mm pin and a set of 12 mm cheek plates are available. The inspection record shows that the existing pin has visible wear and the hole in one cheek plate is slightly oval.
1. Check the 16 mm pin in double shear and bearing at the existing 1.5 t rating and at the proposed 2 t rating. Use the dynamic factor of 2 and state whether either rating is acceptable.
Read the worked answer
At 1.5 t: W = 1.5 × 9.81 = 14.72 kN; design load with dynamic factor 2 = 29.43 kN. Double-shear area = 2 × π × 16²/4 = 402.1 mm²; τ = 29,430/402.1 = 73.2 MPa, below 80 MPa but with only about 9% margin. Bearing area on two 10 mm plates = 2 × 16 × 10 = 320 mm²; σb = 29,430/320 = 92.0 MPa, below 120 MPa. At 2 t: design load = 39.24 kN; τ = 97.6 MPa > 80 MPa, and σb = 122.6 MPa > 120 MPa. So the existing pin is marginal at 1.5 t and fails the simple checks at 2 t, especially with the worn pin and oval hole. The apprentice used static load and ignored the dynamic factor. Do not uprate to 2 t without changing the pin and cheek plates and assessing the whole crane, including welds, hoist and rope.
2. For a 2 t rating with the same dynamic factor, determine the minimum pin diameter and cheek-plate thickness using allowable shear 80 MPa and bearing 120 MPa. Comment on the key assumption.
Read the worked answer
For 2 t with dynamic factor 2, design load Wd = 39.24 kN. Required double-shear area = Wd/80 = 490.5 mm²; d = √(4A/(2π)) = √(2 × 490.5/π) = 17.7 mm. Select a standard 20 mm pin, giving τ = 62.5 MPa and a reasonable margin for wear and misalignment. Bearing on two plates: required total thickness = Wd/(d × 120) = 39,240/(20 × 120) = 16.35 mm, so two 10 mm plates are adequate only if both share load; 12 mm plates would give more margin. The key assumption is double shear. If a missing spacer or uneven gap lets one plate carry the load, the pin becomes single shear: τ = 39,240/(π × 20²/4) = 125 MPa, which is unsafe. Maintain the spacer and check that the pin bears evenly on both cheek plates.
3. What inspection and operational controls would you require before returning the crane to service, and why can a calculated-safe pin still fail?
Read the worked answer
Before return to service: replace the worn pin and the oval-hole cheek plate; verify plate thickness, weld throat and edge distance; fit the correct spacer and a new cotter pin; check the hoist, wire rope, hook latch and SWL marking; proof-load the crane and record the result. Introduce periodic inspection of the pin, holes and welds, and prohibit snatch lifts. The calculation is only a screening check: it does not cover corrosion, wear, fatigue, hole elongation, off-centre loads or a single-shear mistake. A pin that is calculated safe can still fail if the assumed geometry is not maintained in service. The uprating should be approved by a competent engineer and reflected in the SWL plate and operating procedure.