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User: manasahegde234

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Posts by manasahegde234

<prev • 134 results • page 1 of 14 • next >
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Answer: A: Assume static conditions.
... ![enter image description here][1] The areas of three welds are as follows: $A_1 = (60 t) mm^2 $ $A_2 = (60 t) mm^2 $ $A_3 = (90 t) mm^2$ $A = A_1 + A_2 + A_3 = (210 t mm^2)$ The primary shear stress in the weld is given $\tau=\frac{P}{A}=\frac{50000}{210*t}=\frac{238.1}{t}N/mm^2 (i) $ S ...
written 5 days ago by manasahegde23410
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Answer: A: Determine the load carrying capacity of the hook.
... Step 1: Calculation of permissible tensile stress: $\sigma_{max}=\frac{S_yt}{fs}=\frac{380}{3.5}=108.57N/mm^2$ Step 2: Calculation of eccentricity(e): For the cross section XX $R_N=\frac{\frac{b_i+b_o}{2}h}{\frac{b_ir_o-b_oR_i}{h}*log_e\frac{R_o}{R_i}-(b_i-b_o)}$ $R_N=\frac{\frac{90+30}{2}*120} ...
written 8 days ago by manasahegde23410
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Answer: A: Determine the load carrying capacity of the hook.
... Step 1: Calculation of permissible tensile stress: $\sigma_{max}=\frac{S_yt}{fs}=\frac{380}{3.5}=108.57N/mm^2$ Step 2: Calculation of eccentricity(e): For the cross section XX $R_N=\frac{\frac{b_i+b_o}{2}h}{\frac{b_ir_o-b_oR_i}{h}*log_e\frac{R_o}{R_i}-(b_i-b_o)}$ $R_N=\frac{\frac{90+30}{2}*120} ...
written 8 days ago by manasahegde23410
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Answer: A: The C-frame of a 100 kN capacity press is shown in Fig.2.2 The material of the f
... Step 1: Calculation of permissible tensile stress: $\tau_max=\frac{S_ut}{fs}=\frac{200}{3}=66.67N/mm^2$ Step 2: Calculation of eccentricity (e): b_i=3t h=3t R_i=2t R_o=5t t_i=t t=0.75t $R_N=\frac{t_i(b_i-t)+th}{(b_i-t)log_e(\frac{R_i+t_i}{R_i})+tlog_e\frac{R_o}{R_i}}$ $R_N=\frac{t(3t-0.75 ...
written 22 days ago by manasahegde23410
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Answer: A: Calculate the wire diameter and the mean coil diameter of the spring.
... Assuming the outer diameter to be 38mm, $D_o=D+d=Cd+d=d(C+1)$ i.e $d=\frac{D_o}{C+1}=\frac{38}{C+1}$ (a) We have, $\tau_max=K\frac{8P_maxD}{\pi d^3}=K\frac{8P_maxC}{\pi d^2}$ Therefore, $650=K\frac{8*650*c*(c+1)^2}{\pi *38^2}$ $C(C+1)^2K=567.05$ (b) The problem is solved by trial and error ...
written 22 days ago by manasahegde23410
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Answer: A: Specify the size of the bolts.
... ![enter image description here][1] [1]: https://i.imgur.com/o9vebN9.jpg Step 1: Permissible shear stress $\tau=\frac{S_sy}{fs}=\frac{0.5S_yt}{fs}=\frac{0.5*380}{3}=63.33 N/mm^2$ Step 2: Primary and secondary shear forces: The centre of gravity of three bolts will be at the centre of bolt-2. ...
written 22 days ago by manasahegde23410
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Answer: A: A riveted joint, consisting of four identical rivets, is subjected to an eccentr
... ![enter image description here][1] [1]: https://i.imgur.com/GdJlsQl.jpg Step 1. Primary shear force: $P_1'=P_2'=P_3'=P_4'=\frac{P}{4}=\frac{5*10^3}{4}=1250N$ Step 2 Primary shear force: $r_1=r_2=r_3=r_4=100mm$ $C=\frac{Pe}{(r_1^2+r_2^2+r_3^2+r_4^2)}=\frac{(5*10^3)(200)}{4(100)^2}=25$ There ...
written 23 days ago by manasahegde23410
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Answer: A: With neat sketches for failure cross section areas check all components under di
... 1. Draw a neat sketch of knuckle joint: ![enter image description here][1] [1]: https://i.imgur.com/nrajR6C.jpg 2. FInd the diameter of the rod(d): $d=\sqrt \frac{4P}{\sigma_t \pi}=\sqrt \frac{4(150*10^3)}{\pi *75}=50.462 or 52mm$ 3. Using empirical find all dimensions: Diameter of knuckle ...
written 23 days ago by manasahegde23410
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Answer: A: Design a spigot and socket type cotter joint to transmit reversible load of 50KN
... Step 1: Selection of material: The rods are subjected to tensile force and strength is the criterion for the selection of rod material. On the basis of strength, the material of the two rods and the cotter is selected as plain carbon steel of Grade 30C8 (Syt=400MPa) Step 2: Selection of factor of ...
written 23 days ago by manasahegde23410
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Answer: A: Show that the efficiency of a self-locking screw is less than 50 percent.
... The efficiency of a square threaded screw is calculated by, $\eta=\frac{tan\alpha}{tan(\phi+\alpha} $ (a) For a self-locking screw, $\phi \geq \alpha$ Substituting the limiting value for $ \phi=\alpha $ in eqn (a) $\eta \leq\ \frac{tan\alpha}{tan(\phi+\phi}$ $\eta \leq\ \frac{tan\alpha}{tan2 ...
written 24 days ago by manasahegde23410

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