EBK ENGINEERING MECHANICS
15th Edition
ISBN: 9780137616909
Author: HIBBELER
Publisher: VST
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[Q2]: The cost information supplied by the cost accountant is as follows:Sales 20,00 units, $ 10 per unitCalculate the (a/ newsale guantity and (b) new selling price to earn the sameVariable cost $ 6 per unit, Fixed Cost $ 30,000, Profit $ 50,000profit ifi) Variable cost increases by $ 2 per unitil) Fixed cost increase by $ 10,000Ili) Variable cost increase by $ 1 per unit and fixed cost reduces by $ 10,000
can you please help me perform Visual Inspection and Fractography of the attatched image: Preliminary examination to identify the fracture origin, suspected fatigue striation, and corrosion evidences.
can you please help[ me conduct Causal Analysis (FTA) on the scenario attatched: FTA diagram which is a fault tree analysis diagram will be used to gain an overview of the entire path of failure from root cause to the top event (i.e., the swing’s detachment) and to identify interactions between misuse, material decay and inspection errors.
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- hi can you please help me in finding the stress intensity factor using a k-calcluator for the scenario attathced in the images.arrow_forwardHi, can you please help me .Identify and justify suitable analytical techniques of the scenario below, bearing in mind the kinds of information being handled to reach a conclusion (methodology). A child swing set was discovered to have failed at the fixing at the top of the chains connecting the seat to the top of the swing set. A 12 mm threaded steel bolt, connecting the shackle to the top beam, failed at the start of the threaded region on the linkage closest to the outside side of the swing set . The linkage and bolts were made of electro galvanised mild steel . The rigid bar chain alternatives and fixings were of the same material and appeared to be fitted in accordance with guidelines. The yield strength of the steel used is 260 MPa and the UTS is 380 MPa. The bolt that failed was threaded using a standard thread with a pitch (distance between threads) of 1.75 mm and a depth of approximately 1.1 mm. The swing set in question had been assigned to ‘toddlers’ with the application of…arrow_forwardHi, can you please define and calculate the failure mode of the linkage that failed on the swing (images added) : A child swing set was discovered to have failed at the fixing at the top of the chains connecting the seat to the top of the swing set. A 12 mm threaded steel bolt, connecting the shackle to the top beam, failed at the start of the threaded region on the linkage closest to the outside side of the swing set . The linkage and bolts were made of electro galvanised mild steel . The rigid bar chain alternatives and fixings were of the same material and appeared to be fitted in accordance with guidelines. The yield strength of the steel used is 260 MPa and the UTS is 380 MPa. The bolt that failed was threaded using a standard thread with a pitch (distance between threads) of 1.75 mm and a depth of approximately 1.1 mm. The swing set in question had been assigned to ‘toddlers’ with the application of a caged-type seat. However, the location was within the play area not…arrow_forward
- Page 11-68. The rectangular plate shown is subjected to a uniaxial stress of 2000 psi. Compute the shear stress and the tensile developed on a plane forming an angle of 30° with the longitud axis of the member. (Hint: Assume a cross-sectional area of unity) 2000 psi 2000 psi hparrow_forward11-70. A shear stress (pure shear) of 5000 psi exists on an element. (a) Determine the maximum tensile and compressive stresses caused in the element due to this shear. (b) Sketch the element showing the planes on which the maximum tensile and compressive stresses act.arrow_forward11-20. An aluminum specimen of circular cross section, 0.50 in. in diameter, ruptured under a tensile load of 12,000 lb. The plane of failure was found to be at 48° with a plane perpendicular to the longitudinal axis of the specimen. (a) Compute the shear stress on the failure plane. (b) Compute the maximum tensile stress. (c) Compute the tensile stress on the failure plane. hparrow_forward
- A long flat steel bar 13 mm thick and 120 mm wide has semicircular grooves as shown and carries a tensile load of 50 kN Determine the maximum stress if plate r= 8mm r=21mm r=38mmarrow_forwardProblem 13: F₁ = A =250 N 30% Determine the moment of each of the three forces about point B. F₂ = 300 N 60° 2 m -3 m B 4 m F3=500 Narrow_forward3 kN 3 kN 1.8 kN/m 80 mm B 300 mm D an 1.5 m-1.5 m--1.5 m- PROBLEM 5.47 Using the method of Sec. 5.2, solve Prob. 5.16 PROBLEM 5.16 For the beam and loading shown, determine the maximum normal stress due to bending on a transverse section at C.arrow_forward
- 300 mm 3 kN 3 kN 450 N-m D E 200 mm 300 mm PROBLEM 5.12 Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment.arrow_forwardCORRECT AND DETAILED SOLUTION WITH FBD ONLY. I WILL UPVOTE THANK YOU. CORRECT ANSWER IS ALREADY PROVIDED. I REALLY NEED FBD. The cantilevered spandrel beam shown whose depth tapers from d1 to d2, has a constant width of 120mm. It carries a triangularly distributed end reaction.Given: d1 = 600 mm, d2 = 120 mm, L = 1 m, w = 100 kN/m1. Calculate the maximum flexural stress at the support, in kN-m.2. Determine the distance (m), from the free end, of the section with maximum flexural stress.3. Determine the maximum flexural stress in the beam, in MPa.ANSWERS: (1) 4.630 MPa; (2) 905.8688 m; (3) 4.65 MPaarrow_forwardCORRECT AND DETAILED SOLUTION WITH FBD ONLY. I WILL UPVOTE THANK YOU. CORRECT ANSWER IS ALREADY PROVIDED. I REALLY NEED FBD A concrete wall retains water as shown. Assume that the wall is fixed at the base. Given: H = 3 m, t = 0.5m, Concrete unit weight = 23 kN/m3Unit weight of water = 9.81 kN/m3(Hint: The pressure of water is linearly increasing from the surface to the bottom with intensity 9.81d.)1. Find the maximum compressive stress (MPa) at the base of the wall if the water reaches the top.2. If the maximum compressive stress at the base of the wall is not to exceed 0.40 MPa, what is the maximum allowable depth(m) of the water?3. If the tensile stress at the base is zero, what is the maximum allowable depth (m) of the water?ANSWERS: (1) 1.13 MPa, (2) 2.0 m, (3) 1.20 marrow_forward
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