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A double riveted double cover butt joint in plates 20 mm thick is made with 25 mm diameter rivets at 100 mm pitch. The permissible stresses are:
= 120MPa T = 100MPa 0 150MPa Find the efficiency of joint, taking the strength of the rivet in double shear as twice than that of single shear.
A double riveted lap joint is made between 15mm thick plates. The rivet diameter and pitch are 25 mm and 75 mm, respectively. If the ultimate stresses are 400 MPa in tension, 320 MPa in shear and 640 MPa in crushing, find the minimum force per pitch which will rupture the joint. If the above joint is subjected to a load such that the factor of safety is 4, find out the actual stresses developed in the plates and rivets.
A plate 60 mm wide and 12.5 mm thick is to be welded to another plate by means of parallel fillet welds. The plates are subjected to a load of 80 kN. Find the length of the weld. Assume allowable shear strength to be 60 MPa.
Draw a graph and determine the magnitude of the two pulling forces M and S when their resultant is 65 N at 40" with R as shown in FIGURE 6. Use a scale of 1 N = 2 mm.
Draw a neat, labelled sketch of a wheel and axle lifting machine. The wheel diameter is 255 mm and the diameter of the axle 95 mm. An effort of 105 N is necessary to lift a load of 550 N.
A battery powered smoke detector has an average power consumption of 0.5mW and runoff a 9v battery with a capacity of 1200Mah.How often do you expect to change the battery
calculate the energy transfer rate in w/m² across 6 in wall of firebrick with a temperature difference across the wall of 55°C. the thermal conductivity of the firebrick is 0.65 BTU/hr ft °F.
A spring has a natural length of 8 inches. If a force of 20 lb. stretches the spring 0.5
inches, find the work done in stretching from 8 inches to 11 inches
Determine the uncorrected and corrected weights based on the following data:
A.Fine Aggregates - Natural
Fineness Modulus = 2.80
Specific Gravity (SSD) = 2.50
Moisture Content,% = 0.50
Absorption, % = 1.80
B.Coarse Aggregates - Angular
Maximum size,mm = 19.00
Spec.Gravity(SSD) = 2.65
Moisture Content,% = 1.35
Absorption,% = 0.80
C.Class of Concrete = Structural
D.Cement Factor = 11.10 bags/cu.m.
E.Slump,mm = 76.20
F.Spec. Gravity of Cement = 3.15
G.Flexural Strength,Mpa = 27.50
A horizontal "hot air blower" consists of a fan, an electrical heating element and a nozzle The power input to the fan is 50 J/s and the power input to the heating element is 1.44 kJ/s. The air enters the fan with negligible velocity at a pressure of 1 bar and a temperature of 20 C, the volumetric flow rate being 0.5 m3 /min. After passing over the heating element, the air leaves the nozzle with a velocity of 25 m/s and a pressure of 1.1 bar. Treating the complete arrangement as a steady flow system and ignoring any heat loss from the system, determine: (a) The mass flow rate of air entering the fan in kg per second (b) The kinetic energy of the air leaving the nozzle per second (c) The change in enthalpy of the air occurring per second between entering the fan and leaving the nozzle (d) The temperature of the air as it leaves the nozzle. (e) The density of the air as it leaves the nozzle (f) The flow area at the nozzle exit. Take R = 0.287 kJ/kgK and cp = 1.005 kJ/kgK
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