Question

A triple-threaded power screw used in a screw jack, has a nominal diameter of 50 mm and a pitch of 8 mm. The threads are square and the length of the nut is48 mm. The screw jack is used to lift a load of 7.5 kn. The coefficient of friction at the threads is 0.12 and the collar friction at the threads is 0.12 and the collar friction is negligible. Calculate, i. The principal shear stress in the screw body ii. The transverse shear stresses in the screw and the nut, and iii. The unit bearing pressure. Also state whether the screw is self-locking

Answer 1

Answer:The answer is second class lever.

In second class levers, load lies between the fulcrum and the effort.  

However, the load is not positioned at the exact center. It is placed more towards the fulcrum and away from the effort. A second class lever always multiplies a force.

There are many examples of second class levers in day to day life. Some common examples are bottle openers, nut crackers, doors and wheel barrows.

Explanation:

Answer 2

Answer:

1. Maximum shear stress in screw body (τmax) = 4.3938 N/mm²
2. Direct shear stress in screw and nut (τs) and (τn)
   τs =2.3684 N/mm², τn =1.9894 N/mm²
3. Bearing pressure (P$_{b}$) =2.1624 N/mm²
4. The screw is over-hauling, since Ф<α

Explanation:

Given data:

Types of thread = square, n=3, d=50mm, p=8mm, w=7500N, μ=0.12, L=48mm

To find:

τmax, τs, τn, P$_{b}$,= ?

Mean diameter,  $d_{n}=d-\frac{p}{2}=50-\frac{8}{2}=46mm$

Core diameter, $d_{c}=d-p=50-8=42mm$

Helix angle,  α=tan⁻¹$(\frac{l}{\pi d_{m} } )$= tan⁻¹$(\frac{n.p}{\pi d_{m} } )$
                     α=tan⁻¹$({ \frac{3*8}{\pi *46} )$

                     α=9.4293⁰

Angle of friction, Ф=tan⁻¹(μ)= tan⁻¹(0.12)=6.8428⁰

Number of threads, z=$\frac{Height of nut}{Pitch of thread}$  $=\frac{48}{8}$

                                  z=6

Step 1: Effort required to raise the load against thread friction($P_{r}$)

P$_{r}$ = W . tan(Ф+α)= 7500×tan (6.8428°+9.4293°)

P$_{r}$ = 2189.19N

Step 2: Torque required to raise the load against thread friction (T$_{tr}$)

T$_{tr}$ = P$_{r}$×$\frac{d_{m} }{2}$ = 2189.19×$\frac{46}{2}$=50351.37 N-mm

T$_{tr}$ = 50.35 N-m

Step 3: Stresses in screw body

1. Direct compressive stress in screw body(σ$_{c}$)

          σ $_{c} = \frac{W}{\frac{\pi }{4}(d_{c}^{2}) } =\frac{7500}{\frac{\pi }{4}42^{2} }$

           σ$_{c}$=5.4134 N/mm²

    2. Torsional shear stress in screw body(τ)

           τ=$=\frac{T_{tr} }{\frac{\pi }{16}(d_{c^{3} }) }= \frac{7500}{\frac{x\pi }{16}(42^{3} ) }$

            τ= 3.4611 N/mm²

     3. Maximum shear stress or maximum principal stress in screw body(τmax)

        τmax=√(σc/2)²+(τ)²

                 =√(5.4134/2)²+(3.4611)²

         τmax= 4.3938 N/mm²

Step 4: Stresses in screw and nut

1. Direct shear stress in screw threads (τs)
   τs=$\frac{W}{\pi .d_{c}.t.z }= \frac{W}{\pi .d_{c}.\frac{p}{2}.2 }= \frac{7500}{\pi .42.\frac{8}{2}.6 }$

         τs=2.3684 N/mm²

     2. Direct shear stress in nut threads (τn)

          τn=$\frac{W}{\pi .d.t.z} =\frac{7500}{\pi .50.(\frac{8}{2} ).6}$

           τn=1.9894 N/mm²

Step 5: Bearing pressure between threads(Pb)

Pb=$\frac{W}{\frac{\pi }{4}(d^{2} -d_{c}^{2}).z }= \frac{7500}{\frac{\pi }{4}(50^{2} -42^{2}).6 }$

Pb=2.1624 N/mm²

Step 6: In the screw self-locking and over-hauling

In this case,  Ф= 6.8428°

and                 α= 9.4293°

Since Ф<α, the screw is over-hauling.