0.00 19.00 kN-m Moment diagram I₂ = 22×106 mm* Beam section y 10.00 -26.00 kN-m 100.00 mm 200.00 mm For the moment diagram and cross section shown, compute the maximum tension and compression bending stresses (in MPa) produced at any location along the beam span. The maximum stresses may occur at different locations. Tension bending stress MPa Compression bending stress MPa

0.00 19.00 kN-m Moment diagram I₂ = 22×106 mm* Beam section y 10.00 -26.00 kN-m 100.00 mm 200.00 mm For the moment diagram and cross section shown, compute the maximum tension and compression bending stresses (in MPa) produced at any location along the beam span. The maximum stresses may occur at different locations. Tension bending stress MPa Compression bending stress MPa

Mechanics of Materials (MindTap Course List)

9th Edition

ISBN:9781337093347

Author:Barry J. Goodno, James M. Gere

Publisher:Barry J. Goodno, James M. Gere

Chapter8: Applications Of Plane Stress (pressure Vessels, Beams, And Combined Loadings)

Section: Chapter Questions

Problem 8.5.11P: The hollow drill pipe for an oil well (sec figure) is 6,2 in. in outer diameter and 0.75 in. in...

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The hollow drill pipe for an oil well (sec figure) is 6,2 in. in outer diameter and 0.75 in. in thickness. Just above the bit, the compressive force in the pipe (due to the weight of the pipe) is 62 kips and the torque (due to drilling) is 185 kip-in. Determine the maximum tensile, compressive, and shear stresses in the drill pipe.
A simply supported beam is subjected to point load P at mid-span. The normal stress on an element at mid-span is known to be ??x= 1.5 ksi. Determine the element stresses lilt is rotated through angle ? = 450• Show these stresses on a sketch of an element oriented at that angle.
The stresses on an element are sx= 1000 Psi. sy= 500 psi, and txy= 350 psi. Find the stresses acting on an element oriented at an angle 0 = 250. Show these stresses on the rotated element.
A sign is supported by a pipe (see figure) having an outer diameter 110 mm and inner diameter 90 mm. The dimensions of the sign are 2.0 m X 1.0 m, and its lower edge is 3.0 m above the base. Note that the center of gravity of the sign is 1.05 m from the axis of the pipe. The wind pressure against the sign is 1.5 kPa. Determine the maximum in-plane shear stresses due to the wind pressure on the sign at points /I, B, and C, located on the outer surface at the base of the pipe.
YU Fixed end 4.25 KN.m 0.3m 5KN · 100 KN 100 KN 0.4 m 3.0 KN-m 5 KN 200 KN 200KN 0.1 m X Find the axial, torsional, bending and transverse shear stresses at the four critical points (top, bottom, left and right) of the central cross section. Draw the stress state cubes and then represent them as plane stress states, if applicable. This stainless steel beam has a circular cross section and a diameter of 7 centemeter
Calcuate Shear and Bending From Point A to Point B and Point A to Point C. Draw a FBD from point A to point B and another for point A to C indicating what stresses are been calculate. Use the stress tensor and stress cube to Indicate what forces are acting at point A. Force apply at point B is 70lb Distacen from Point A to B is 12 in or 1 feet long Distacen from Point A to C is 3 in Handle diameter 0.625 in
10/10FIND STRESS IN STEEL MATERIALS IN COMPOSITE COLUMIN SHOWN BELOW. IF E st= 200 kN/MM2, Ealu.=70 kN/MM2 100 kN RALU=30 MM Aluminum Steel 1m. R ST = 20 MM
Thank you for answering this question. Please you specify each step as soon as possible. Good Luck! Thank You. For a 60° angle lamina that subjected loads shown below, 15 MPa T 3.5 MPa 1.4 MPa Ox = -3.5 MPa, oy = 15 MPa, Txy = -1.4 MPa Calculate the stresses in L-T directions (ơ1, ơr, TLt) Determine whether the lamina will fail under these applied loads by using Tsai-Hill failure criteria. Determine whether the lamina will fail under these applied loads by using maximum stress failure criteria. Mechanical properties: EL = 14 GPa; ET=3.5 GPA; GLT = 4.2 GPa; vt = 0.40 Strength (In case of tension): OLU = 250 MPa, ơTu = 10 MPa Strength (In case of compression): OLU = 200 MPa, ơTu = 0.5 MPa Shear strength: TLTU = 8 MPa OT + OLu + <1 Tsai-Hill criterion: OLu - OLu
10" 0.75 54 In the diagram above all members have cross section dimensions of 0.75"x2.5" Calculate the shear stress in ksi on member ABC between points A and B if force P=15 kips. 4 significant figures.
Three cables having an area of 0.05 in^2 each is carrying the load W as shown. With no stress, the original lengths of the cables are 74.99 ft, 75.00 ft, and 74.94 ft. E = 29 x 10° psi Calculate for the stress in the intermediate wire. Choices: a. 8,026.65 psi b. 9,999.66 psi c. 19,939.80 psi d. 10,102.52 psi 1500 lb
Find the stresses in each direction, also find the change in volume of the block of dimension 120 mm x 60 mm x 43 mm, subjected to 3 mutually perpendicular loads. The load along length, breadth and depth directions are 10 kN (tensile), 25 kN (tensile), 13 kN (compressive) respectively, Take E as 180 GPa, Poisson's ratio as 0.3. The stress along length direction (Unit in kN/m2)= _____________ The stress along width direction (Unit in kN/m2)= _____________ The compressive stress along depth direction (Unit in kN/m2)= _____________ The change in volume of the block is (unit in mm3) = ______________
Find the stresses in each direction, also find the change in volume of the block of dimension 110 mm x 55 mm x 40 mm, subjected to 3 mutually perpendicular loads. The load along length, breadth and depth directions are 14 kN (tensile), 20 kN (compressive), 10 kN (compressive) respectively. Take E as 160 GPa, Poisson's ratio as 0.3 (ENTER ONLY THE VALUES IN THE BOXES BY REFERRING THE UNIT GIVEN IN BRACKET & UPLOAD YOUR HAND WRITTEN ANSWERS IN THE LINK PROVIDED) The stress along length direction (Unit in MN/m?)=. 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) =.