3. G(s) = 1 4s² + 8s +1
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- 7. Using the formula R = √ (r² + s² /4); solve for r. Page #QUESTION 2 Question 2 A cross-section of a beam is shown in Figure Q2. If the shear force in this section is V = 125 KN, determine the value and the location of the maximum shear stress in the section. In Figure Q2, a = 30 mm and the origin of the coordinate system is at centroid of the cross section. 7 y= Z= A a AY S= 20 4a mm; mm; O Figure Q2 Answer The vertical coordinate (y-coordinate; the y-axis serves as the axis of symmetry of the cross- section.) and horizontal coordinate (z-coordinate) of the location where the maximum shear stress occurs in the section are ← a The vertical distance from the location where the maximum shear stress occurs in the section to the bottom side (AB cross section can be calculated as Distance = mm (units: mm) 3a Second moment of area The second moment of area employed in the equation to calculate maximum shear stress can be calculated as I₂ = a (units: mm²) Shear stress The second moment of area employed in the equation to calculate maximum shear…Find the general solution. a. y' 5y = 3ex - 2x + 1 - b. y" +4y' + 4y = e¯*cos(x) c. (D² + I)y = cos(wt), w² # 1
- Q1. (20’) Consider the shaded square region on the s-plane, as shown in Fig. Q1. Let p = -Ço, + j@, J1-5², S'- plane 4j find (a). the p location with largest @, . (b). the p location with smallest @, . (c). the p location with largest 5 (d). the p location with smallest 5. -3 Fig. Q13.1-8. The WorldLight Company produces two light fixtures (prod- ucts 1 and 2) that require both metal frame parts and electrical components. Management wants to determine how many units of each product to produce so as to maximize profit. For each unit of product 1, 1 unit of frame parts and 2 units of electrical compo- nents are required. For each unit of product 2, 3 units of frame parts and 2 units of electrical components are required. The com- pany has 200 units of frame parts and 300 units of electrical com- ponents. Each unit of product 1 gives a profit of $1, and each unit of product 2, up to 60 units, gives a profit of $2. Any excess over 60 units of product 2 brings no profit, so such an excess has been ruled out. (a) Formulate a linear programming model for this problem. D (b) Use the graphical method to solve this model. What is the resulting total profit?6. show all steps in detail, units and formulas please.
- A bent tube is attached to a wall with brackets as shown. . A force of F = 980 lb is applied to the end of the tube with direction indicated by the dimensions in the figure. a.) Determine the force vector F in Cartesian components. b.) Resolve the force vector F into vector components parallel and perpendicular to the position vector rDA Express each of these vectors in Cartesian components. 2013 Michael Swanbom CC BY NC SA g →> h BB Z NE 8° A x F kaz с b Values for dimensions on the figure are given in the table below. Note the figure may not be to scale. Be sure to align your cartesian unit vectors with the coordinate axes shown in the figure. Variable a Value 8 in b 12 in с 15 in d 36 in h 23 in 9 28 in a. F = î + k) lb -> b. F||DA = ( î + k) lb →> FIDA = 2+ k) lbA bent tube is attached to a wall with brackets as shown. . A force of F = 980 lb is applied to the end of the tube with direction indicated by the dimensions in the figure. a.) Determine the force vector F in Cartesian components. → → b.) Resolve the force vector F into vector components parallel and perpendicular to the position vector rDA. Express each of these vectors in Cartesian components. 2013 Michael Swanbom cc 10 BY NC SA g x B A א Z FK с кая b Values for dimensions on the figure are given in the table below. Note the figure may not be to scale. Be sure to align your cartesian unit vectors with the coordinate axes shown in the figure. Variable Value a 8 in 12 in с 15 in 36 in h 23 in g 28 in a. F = b. FDA = = ( + k) lb k) lb FIDA = 2 + k) lb1. Note: for part a), you can do this without a calculator; it might be easier to work out the problem symbolically, and then substitute the numbers at the end of the problem. В 2L The axial rod is fixed to the two walls, as shown above. Portion AB is composed of one material (EAB = E), while portion BD is composed of a different material (EBD = 2E). Portion AB has cross sectional area AAB = A, while portion BD has cross sectional area ABD = 2A. EAB = E = 15×106 psi AAB = A = 1 in² EBD = 2E = 30×10° psi ABD = 2A = 2 in² LAB = L = 10 in. LBC = 2L = 20 in. LcD = L = 10 in. An external force, P = 100 kips, is applied at point C as shown below. a) Determine the reaction forces at wall A and D, Ra and Rp. Specify the direction, (→) or (+). b) Deflection at point C. Specify the direction, (→) or (+). (Ans: RA = 100/7 kips →, Rp = 600/7 kips →, & = 1.429x10² in –)
- a = -3/7 b = -3/23) A point mass is attached to one end of a beam levered against a spike which acts as the pivot. The point mass weighs 250N and the beam has a uniform mass of 10kg and is 8m long. Parts a and b Part c Fapplied 3 of 6mov 1m 8m Krist 0 = 60° Scale 250 N Scale a. If a scale placed beneath the point mass reads 250N, what is the magnitude and direction for the force at the pivot point? Direction: North, 90° from x-axis Magnitude to k* 250N = 2500 N b. For the same scenario, how far must the pivot point be located relative to the left edge of the beam? (Hint, both the boxes weight and the force from the scale act at the farthest right edge of the beam.) www.acta 0672nos onl c. The pivot is then moved to be 6m from the left edge of the beam which now makes a 60 angle with the floor. How much downward force would be needed at the left-most edge of the beam to reduce the reading on the scale to zero? (Hint: the beam is still in static eq. when the scale reads zero.)(c) Water flows through a horizontal pipe at a rate of 5 x 104 m³/s. The pipe consists of two sections of diameters 50 mm and 20 mm with a smooth reducing section as shown in Figure Q1 (c). The pressure difference between the two pipe sections is measured by a mercury manometer. Neglecting frictional effects, determine the differential height of mercury between the two pipe sections.
