Question

1 and 3

ular sections, the shear stress formula can be written as follows (Eq. 12-8) 2 A ssive deflections can lead to serviceability problems-suc improper functioning of building hardware, or poor appeara to be able to calculate deflections and compare these ty problems-such as cracking in walls and cel- nce. Therefore, it is impor wpically found in a building code. Ma includin flection g standard deflection form formula for a simply supported beam under a const ny techniques can be used to calculate deflections ulas that are found in many references. The standard de- length is as follower a simply suppoulas that are fours an be used to cakate levels that are uniform load across its entire ant When the loadings on a beam are non-standard, other techniques must be utilized. The ent-area method for finding deflections is also used many times because of its simplicity Although deflection is usually considered to affect only the intended function of a buildin it can also lead to catastrophic failure-such as a structure's collapse due to the pondins of water or snowmelt. The designer must be aware of the serviceability aspects of a lar structure and adjust specification requirements accordingly rainwa cises 1. A rectangular beaim measuring 10 in. wide and 18 in. deep has an applied d compres- above the moment of 120 kip-ft acting upon it. Determine the maximum tensile an sive bending stresses on the beam and the bending stress at a distance 3 neutral axis. in. moment of 100 kN-m acting upon it. Determine the maximum tensile and compressive bending stresses on the beam and also determine the bending stress at a distance 35 mm above the neutral axis. 2. A rectangular beam measuring 100 mm wide and 300 mm deep has an applied bending of 11 ksi on its outside fiber. Determine the magnitude of applied moment on the beam to cause this stress. 3. A rectangular beam measuring 8 in. wide and 16 in. deep has a maximum bending stress

Answer 1

sigma_max = M_y/I where, M = applied Bending moment y = distance of extreme from NA. I = moment of inertia of Beam. Max compressive stress (Bending) occurs of extreme top fibre of beam and max tensile bending stress occurs at extreme bottom fibre of beam max compressive Bending stress sigma_max = M_y/I. where M = 120 kip ft = 120 times 12 kip -in = 1440 kip 0 in y = 13/2 = 9 in. I = 1/12 times 10 times 18^3 = 4860 in^4. Therefore sigma_max = 1440 times 9/4860 = 2.67 kip/in^2 = 2.67 ksi maximum tensile Bending stress sigma_max = M_y/I = 1440 times 9/4860 = 2.67 kip/in^2 = 2.67 ksi Bending stress at 3 in NA sigma = M_y/I = 1440 times 3/4860 = 0.89 kip/in^2 = 0.89 ksi \r\n sigma_max = M_y/I Given sigma_max = 11 ksi y = 16/2 = 8 in I = 1/12 times 8 times 1663 = 2730.67 in^4 Therefore M = M times 8/2730.67