The figure below shows a simply supported beam with an overbang. The beam is made of two types of wood, as shown in the cross section. Calculate the maximum allowable magnitude for the uniformly distributed load (w) if you know that: Allowable stresses in pine =11 MPa, Epie =8 GPA Allowable stresses in plywood = 14 MPa. Epywood 12 GPA

The figure below shows a simply supported beam with an overbang. The beam is made of two types of wood, as shown in the cross section. Calculate the maximum allowable magnitude for the uniformly distributed load (w) if you know that: Allowable stresses in pine =11 MPa, Epie =8 GPA Allowable stresses in plywood = 14 MPa. Epywood 12 GPA

Mechanics of Materials (MindTap Course List)

9th Edition

ISBN:9781337093347

Author:Barry J. Goodno, James M. Gere

Publisher:Barry J. Goodno, James M. Gere

Chapter1: Tension, Compression, And Shear

Section: Chapter Questions

Problem 1.5.2P: A steel riser pipe hangs from a drill rig located offshore in deep water (see figure). (a) What is...

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A steel riser pipe hangs from a drill rig located offshore in deep water (see figure). (a) What is the greatest length (meters) it can have without breaking if the pipe is suspended in the air and the ultimate strength (or breaking strength) is 550 MPa? (b) If the same riser pipe hangs from a drill rig at sea, what is the greatest length? (Obtain the weight densities of steel and sea water from Table M, Appendix I. Neglect the effect of buoyant foam casings on the pipe.)
Two bars AC and BC of the same material support a vertical load P (see figure). The length L of the horizontal bar is fixed, but the angle fl can be varied by moving support A vertically and changing the length of bar AC to correspond with the new position of support A. The allowable stresses in the bars are the same in tension and compression. When the angle ft is reduced, bar AC becomes shorter, but the cross-sectional areas of both bars increase because the axial forces are larger. The opposite effects occur if the angle 0 is increased. Thus, the weight of the structure (which is proportional to the volume) depends upon the angle ft. Determine the angle ft so that the structure has minimum weight without exceeding the allowable stresses in the bars. Note: The weights of the bars are very small compared to the force P and may be disregarded.
A brass plate with a modulus of elastici ty E = 16 X 106 psi and Poisson’s ratio a = 0.34 is loaded in biaxial stress by normal stresses r and cry. (sec figure). A strain gage is bonded to the plate at an angle 4 = 350 If the stress o is 10,700 psi and the strain means used by the gageise = 390 X 106.whatisthcmax- imum in-plane shear stress (Tr, ),. and shear strain (>‘ ).? What is the maximum shear strain (y ) in the x-: plane? What is the maximum shear strain (y ).. in the y-r plane?
Truss members supporting a roof are connected to a 26-mm-thick gusset plate by a 22-mm diameter pin, as shown in the figure and photo. The two end plates on the truss members are each 14 mm thick. (a) If the load P = 80 kN, what is the largest bearing stress acting on the pin? (b) If the ultimate shear stress for the pin is 190 MPa, what force Pult is required to cause the pin to fail in shear? Disregard friction between the plates.
Solve the preceding problem for sx= 11 MPa and ??y= -20 MPa (see figure).
A steel cable with a nominal diameter of 25 mm (see Table 2-1) is used in a construction yard to lift a bridge section weighing 38 kN. as shown in the figure. The cable has an effective modulus of elasticity E = 140 GPa. (a) If the cable is 14 m long, how much will it stretch when the load is picked up? (b) If the cable is rated for a maximum load of 70 kN, that is the factor of safety with respect to failure of the cable?
A retaining wall 6 ft high is constructed of horizontal wood planks 2.5 in. thick (actual dimension) that are supported by vertical wood piles of a 12 in, diameter (actual dimension), as shown in the figure. The lateral earth pressure is pt=125 lb/ft2 at the top of the wall and p2= 425 lb/ft2 at the bottom. Assuming that the allowable stress in the wood is 1175 psi, calculate the maximum permissible spacing s of the piles. Find the required diameter of the wood piles so that piles and planks (f = 2.5 in.) reach the allowable stress at the same time. Hint: Observe that the spacing of the piles may be governed by the load-carrying capacity of either the planks or the piles. Consider the piles to act as cantilever beams subjected to a trapezoidal distribution of load, and consider the planks to act as simple beams between the piles. To be on the safe side, assume that the pressure on the bottom plank is uniform and equal to the maximum pressure.
A round bar of 10 mm diameter is made of aluminum alloy 7075-T6 (see figure). When the bar is stretched by axial forces P, its diameter decreases by 0.0 16 mm. Find the magnitude of the load P. Obtain the material properties from Appendix 1.
A standard brick (dimensions 8 in. × 4 in. × 2.5 in ) is compressed lengthwise by a force P. as shown in the figure, If the ultimate shear stress for brick is 1200 psi and the ultimate compressive stress is 3600 psi. what force Pmax is required to break the brick?
Find the stresses in each direction, also find the change in volume of the block of dimension 120 mm x 45 mm x 43 mm, subjected to 3 mutually perpendicular loads. The load along length, breadth and depth directions are 10 kN (tensile), 20 kN (compressive), 16 kN (compressive) respectively. Take E as 140 GPa, Poisson’s ratio as 0.3 The stress along length direction (Unit in MN/m2)= The compressive stress along width direction (Unit in MN/m2)= The compressive stress along depth direction (Unit in MN/m2)= The change in volume of the block is (unit in mm3) =
The recessed composite beam in the figure is made of steel and aluminum It was produced and a force of 20 kN was applied to its tip. According to what is given, the maximum amount of steel and aluminum Find the stresses. (Eç = 200 GPa, EAl = 80 GPa) dimensions in mm
For the steel block (Length L = 5-meter, Width W = 1 meter, Height H = 1 meter) with compression force on its cross-sectional side of (Your Student Id)N shown in following figure; a) Calculate Stress b) Calculate Strain when its length becomes L = 4 meter after applying force. c) Calculate of modulus of Elasticity. (Hint: If your ID is 114-2020, then compression force would be 114 N)