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1 CEGR 212 Mechanics of Materials Final Examination NAME: ______________________________________ The test has 12 questions. You have EXACTLY TWO HOURS. Point value of each problem is as shown. Answer...

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1 CEGR 212 Mechanics of Materials Final Examination NAME: ______________________________________ The test has 12 questions. You have EXACTLY TWO HOURS. Point value of each problem is as shown. Answer those questions first that you are most comfortable with. Show all work to receive partial credit. Pay close attention to units. Answer the question asked – pay attention to things highlighted in the question. SUBMIT ALL PAGES. Page 4 is a formula sheet XXXXXXXXXXpoints) Find the reaction at support A for the beam shown below XXXXXXXXXXpoints) What is the vertical coordinate of the centroid (yc) for the section shown below? All dimensions are in inches. Use the datum (y = 0) as indicated. Do not choose your own datum XXXXXXXXXXpoints) What is most nearly the centroidal moment of inertia (Ix) of the section shown below? All dimensions are in inches. 4 ft 6 ft 3 ft 20 k 40 k-ft 2 k/ft A B y = XXXXXXXXXXx XXXXXXXXXXx 1 6 x XXXXXXXXXXpoints) What is the elongation of a rectangular bar (cross section 2 inch x 2 inch, length 4 ft) when it is subjected to an axial load of 80,000 lbs. Assume the material of the bar has the following properties: G = 7,700 ksi, ? = XXXXXXXXXXpoints) A beam is loaded as shown below. What is the bending moment function in the domain 4 < x < 10? The reaction on the right support (B) has been calculated as 36.4 kips (upward XXXXXXXXXXpoints) Draw a qualitative shape for the bending moment diagram for the beam loaded as shown. Clearly state any assumptions you make XXXXXXXXXXpoints) The stress state at a point within a structure is as shown below. All stresses are in MPa. What is the normal stress on a plane inclined at 35 degrees to the horizontal? Shear stress components are shown as dashed lines XXXXXXXXXXpoints) A T-shaped section (see below) has the following properties: cross sectional area = 3000 mm2 , distance to neutral axis (centroid) yc = 74 mm, overall depth d = 125 mm, moment of inertia I = 1.65x107 mm 4 . The section is subjected to a positive bending moment M = 40 kNm as shown. What is the maximum bending stress? Indicate tension or compression XXXXXXXXXXpoints) A simply supported beam (see figure below) supports a uniformly distributed load w = 6 kip/ft. 4 ft 6 ft 3 ft 40 k-ft 4 k/ft 12 k A B ? = 400 ? = 120 ? = 200 ??? yc d M 3 The beam has an I-shaped section as shown below (All dimensions are inches) What is the maximum shear stress? XXXXXXXXXXpoints) A hollow tube is subjected to a torque of 4 kNm. Assume that the tube can be classified as a THIN-WALLED TUBE. The cross section of the tube is shown below. Use G = 80 GPa What is the twist angle if the length of the shaft is 60 cm long? XXXXXXXXXXpoints) A steel (E = 200 GPa, ? = 10 x 10-6 /o C) rod of diameter 2 cm is welded to fixed supports as shown. The temperature is raised by 100o C. What is the stress is the rod? XXXXXXXXXXpoints) A hollow circular cylinder made of steel (E = 200 GPa, G = 77 GPa) is subject to a torque T at one end and is fixed at the other end. Outer diameter of the rod is 50 mm and inner diameter is 44 mm. The length of the rod is 60 cm. What is the maximum torque so that the allowable shear stress (300 MPa) is not exceeded and the twist angle does not exceed 3 degrees? 20 ft 6 k/ ft 8 x XXXXXXXXXXx 1 8 x 1 4 mm 3 mm 6 mm 72 mm 5 mm 28 mm 18 cm STEEL 4 FORMULAS Twist Angle (circular tube) GJ TL ? ? Shear Stress due to Torsion (circular tube) J Tr ? ? Twist Angle (thin-walled tube) ? ? t ds GA TL m 2 4 ? Shear Stress due to Torsion (thin-walled tube) A t T m 2 ? ? Axial Elongation AE PL ? ? Poisson’s ratio 1 2 ? ? ? ? G E long lat ? ? ? Moment of Inertia of Rectangle 3 tan 12 1 Irac gle ? bh Moment of Inertia of Circle 4 4 I r circle ? ? Stress Transformation Equations ? ? ? ? ? ? ? ? ? ? ? ? ? ? sin 2 cos 2 2 cos 2 sin 2 2 2 ' ' ' xy x y x y xy x y x y x ? ? ? ? ? ? ? ? ? Center and Radius of Mohr’s Circle XXXXXXXXXXxy x y x y C R ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Orientation of Maximum Normal Stress x y xy p ? ? ? ? ? ? 2 tan 2 Moment Curvature relationship EI M y?? ? Curvature of function y(x) ? ? ? ? 3 / 2 2 1 y y ? ? ?? ? ? Load, Shear and Moment Functions ( ) V (x) dx dM w x dx dV ? ? ? Longitudinal Shear Stress It VQ ? ? Bending Stress I My ? ? ? Parallel Axis Theorem 2 I I Ad x ? xc ? Thermal elongation ?T ? L??T
005_galuazo-5f4xuyiz.pdf
Answered Same Day Dec 22, 2021

Solution

Robert answered on Dec 22 2021
120 Votes
1) Since at A there is a hinged support, therefore there will be horizontal as well
as vertical reaction. In the given force system since there is no any horizontal
force therefore only vertical reaction will be present.

Let the reaction at support A be RA

Now to find out the support reaction we will use 0 BM


Now taking clockwise moment as negative and counter clockwise moment as
positive we have

)(1.7
010120940
010*6*205.1*640
10*6*205.1*640
AnskR
R
R
havewezerotoitEquating
RM
A
A
A
AB






2) Since there are three different rectangle parts. So taking upper rectangle
as area 1, mid rectangle as area 2 and the lowermost rectangle as area 3

Now as we know that vertical coordinate of centroid of the I-beam will be
given as




A
yA
yc

Here yA is the first area moment of different about the given datum and A is
the area of the different parts.

Now from the given diagram

2
1
2
2
2
1
65.1*4
61*6
81*8
inA
inA
inA





Again as we know for the rectangle the C.G acts along the mid of the width.

Therefore using the given datum we have

3
33
3
22
3
11
5.40
2
5.1
6*6
18
2
6
*6
4
2
1
*8
inyA
inyA
inyA





























Therefore the vertical coordinate of the centroid will be given as

 
 
)(725.2
20
5.54
668
5.40184
321
332211
Ansiny
AAA
yAyAyA
y
A
yA
y
c
c
c














Therefore we can say that the centroid will lay 2.725 in down from the given
datum.

3) Since the given I-beam is symmetric about both the axes. Therefore it’s
centroid will be located at the mid-point.

Now from the given figure the vertical coordinate of the centroid will be

 
inchyc 6
2
1101




Now like previous problem again we can divide this given I-beam into three
ectangles. Upper rectangle is take as part 1 the mid rectangle is taken as part
2 and the lowermost rectangle is taken as part 3

We will find out the area moment of inertia of each of these rectangles about
the centroid of the given I-beam. For this we will use the parallel axes-
theorem

So for the first rectangle
...
SOLUTION.PDF
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