Newton's Law of Cooling

Newton's Law of Cooling states that the temperature of a body changes at a rate proportional to the difference in temperature between its own temperature and the temperature of its surroundings.
 

We can therefore write

$\dfrac{dT}{dt} = -k(T - T_s)$

where,
T = temperature of the body at any time, t
Ts = temperature of the surroundings (also called ambient temperature)
To = initial temperature of the body
k = constant of proportionality

Problem 1010 and Problem 1011 | Investigation of timber reinforced by two steel channels

Problem 1010
A pair of C250 × 30 steel channels are securely bolted to wood beam 200 mm by 254 mm, as shown in Fig. P-1010. From Table B-2 in Appendix B, the depth of the channel is also 254 mm.) If bending occurs about the axis 1-1, determine the safe resisting moment if the allowable stresses σs = 120 MPa and σw = 8 MPa. Assume n = 20.
 

1010-timber-reinforced-withc-channel.gif

 

Problem 1011
In Problem 1010, determine the safe resisting moment if bending occurs about axis 2-2.
 

Problem 1009 | Width of aluminum plate reinforcement for the wood section to resist 14 kN-m moment

Problem 1009
A timber beam 150 mm wide by 200 mm deep is to be reinforced at the top and bottom by aluminum plates 6 mm thick. Determine the width of the aluminum plates if the beam is to resist a moment of 14 kN·m. Assume n = 5 and take the allowable stresses as 10 MPa and 80 MPa in the wood and aluminum, respectively.
 

Problem 1008 | Finding the width of steel plate reinforcement

Problem 1008
A timber beam 150 mm wide by 250 mm deep is to be reinforced at the top and bottom by steel plates 10 mm thick. How wide should the steel plates be if the beam is to resist a moment of 40 kN·m? Assume that n = 15 and the allowable stresses in the wood and steel are 10 MPa and 120 MPa, respectively.
 

Problem 1007 | Flexural stresses developed in the wood and steel fibers

Problem 1007
A uniformly distributed load of 300 lb/ft (including the weight of the beam) is simply supported on a 20-ft span. The cross section of the beam is described in Problem 1005. If n = 20, determine the maximum stresses produced in the wood and the steel.
 

Problem 1006 | Width of fastened steel plate for balanced reinforcement

Problem 1006
Determine the width b of the 1/2-in. steel plate fastened to the bottom of the beam in Problem 1005 that will simultaneously stress the wood and the steel to their permissible limits of 1200 psi and 18 ksi, respectively.
 

Problem 1005 | Maximum concentrated load at the midspan that the reinforced timber beam can carry

Problem 1005
A timber beam 6 in. by 10 in. is reinforced only at the bottom by a steel plate as shown in Fig. P-1005. Determine the concentrated load that can be applied at the center of a simply supported span 18 ft long if n = 20, fs ≤ 18 ksi and fw ≤ 1200 psi. Show that the neutral axis is 7.1 in. below the top and that INA = 1160 in.4.
 

1005-given-timber-and-steel.gif

 

Problem 1003 | Maximum stresses in wood and steel of composite beam

Problem 1003
A simply supported beam 4 m long has the cross section shown in Fig. P-1002. It carries a uniformly distributed load of 20 kN/m over the middle half of the span. If n = 15, compute the maximum stresses in the wood and the steel.
 

Problem 1004 | Increase in moment capacity due to aluminum plate reinforcement

Problem 1004
Repeat Problem 1002 assuming that the reinforcement consists of aluminum plates for which the allowable stress is 80 MPa. Use n = 5.
 

Problem 1002 | Increase in moment capacity due to steel plate reinforcement

Problem 1002
A timber beam is reinforced with steel plates rigidly attached at the top and bottom as shown in Fig. P-1002. By what amount is moment increased by the reinforcement if n = 15 and the allowable stresses in the wood and steel are 8 MPa and 120 MPa, respectively?
 

1002-given-section.gif

 

Pages

Subscribe to MATHalino RSS