internal forces and moments solved problems
TRANSCRIPT
1 Β© 2019 Montogue Quiz
Quiz SM104
Internal Forces and Moments Lucas Montogue
Problems
PROBLEM βΆ
The simply supported beam shown in the figure below carries two concentrated loads. Which of the following pairs of graphs best approximates the shear and bending moment diagrams for this system?
A)
B)
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C)
D)
PROBLEM β· (Beer et al., 2013, w/ permission)
Consider beam AB shown below. What are the maximum values for shear force and bending moment for this system?
A) |ππmax| = 365 lb and |ππmax| = 3510 lbβin. B) |ππmax| = 365 lb and |ππmax| = 5110 lbβin. C) |ππmax| = 515 lb and |ππmax| = 3510 lbβin. D) |ππmax| = 515 lb and |ππmax| = 5110 lbβin.
PROBLEM βΈ (Hibbeler, 2010, w/ permission)
Which of the following graphs best illustrates the shear force and bending moment diagrams for the beam illustrated below?
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A)
B)
C)
D)
PROBLEM βΉ (Merriam & Kraige, 2002, w/ permission)
Consider a beam with the loading pattern shown below. Find the distance x, measured from the left end, for which the bending moment is maximum.
A) π₯π₯ = 4.47 ft B) π₯π₯ = 6.12 ft C) π₯π₯ = 7.54 ft D) π₯π₯ = 10.0 ft
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PROBLEM βΊ (Beer et al., 2013, w/ permission)
Determine the maximum absolute value of the bending moment in the beam illustrated below.
A) ππmax = (π€π€πππΏπΏ2) (2ππ)β B) ππmax = (π€π€πππΏπΏ2) ππβ C) ππmax = (π€π€πππΏπΏ2) (2ππ2)β D) ππmax = (π€π€πππΏπΏ2) ππ2β PROBLEM β» (Beer et al., 2015, w/ permission)
Two small channel sections DF and EH have been welded to the uniform beam AB of weight w = 3 kN to form the rigid structural member shown. This member is being lifted by two cables attached at D and E. Knowing that ππ = 30o and neglecting the weight of the channel sections, determine the maximum values of the shear force and bending moment in the beam.
A) |ππmax| = 600 N and |ππmax| = 624 Nβm B) |ππmax| = 600 N and |ππmax| = 924 Nβm C) |ππmax| = 900 N and |ππmax| = 624 Nβm D) |ππmax| = 900 N and |ππmax| = 924 Nβm
PROBLEM βΌ (Hibbeler, 2010, w/ permission)
Determine the ratio of a/b for which the shear force will be zero at the midpoint C of the double-overhang beam.
A) ππ/ππ = 1/8 B) ππ/ππ = 1/6 C) ππ/ππ = 1/4 D) ππ/ππ = 1/2
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PROBLEM β½ (Hibbeler, 2010, w/ permission)
Determine the internal normal force, shear force, and moment at point E of the two-member frame shown. True or false?
1. ( ) The absolute value of the internal normal force at E is less than 1 kN.
2.( ) The absolute value of the shear force at E is less than 1 kN. 3. ( ) The absolute value of the internal moment at E is less than 1.2 kNβm. PROBLEM βΎ (Hibbeler, 2010, w/ permission)
If L = 9 m, the beam will fail when the maximum shear force is Vmax = 5 kN or the maximum bending moment is Mmax = 22 kNβm. Determine the largest couple moment M0 the beam will support.
A) M0 = 22 kNβm B) M0 = 44 kNβm C) M0 = 66 kNβm D) M0 = 88 kNβm
PROBLEM βΏ (Hibbeler, 2010, w/ permission)
Determine the largest intensity w0 of the distributed load that the beam can support if the beam can withstand a maximum shear force Vmax = 1200 lb and a maximum bending moment of Mmax = 600 lbβft.
A) w = 13.9 lb/ft B) w = 21.8 lb/ft C) w = 32.5 lb/ft D) w = 44.1 lb/ft
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PROBLEM β« (Beer et al., 2015, w/ permission)
For the beam and loading shown, determine the maximum normal stress due to bending in a transverse section at C.
A) ππ = 10.9 MPa B) ππ = 21.2 MPa C) ππ = 30.4 MPa D) ππ = 41.0 MPa
PROBLEM β¬ (Beer et al., 2015, w/ permission)
The beam AB supports two concentrated loads P and Q. The normal stress due to bending in the bottom edge of the beam is +55 MPa at D and +37.5 MPa at F. Determine the maximum normal stress due to bending that occurs in the beam.
A) ππ = 37.5 MPa B) ππ = 50.0 MPa C) ππ = 62.5 MPa D) ππ = 75.0 MPa
Solutions
P.1 β Solution
The determination of the expressions for V and M follows. First, consider segment AB, which encompasses 0 < π₯π₯ < 2 m. The figure below shows the forces and moments acting on the beam when it is divided into two segments by a point E, located between A and B, and a distance x away from the left support.
Applying an equilibrium of forces in the y-direction, we have
Ξ£πΉπΉπ¦π¦ = 0 β 18 β ππ = 0
β΄ ππ = 18 kN
and, from the equilibrium of moments relative to point E,
Ξ£πππΈπΈ = 0 β ππ β 18π₯π₯ = 0
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β΄ ππ = 18π₯π₯ kN β m (0 < π₯π₯ < 2 m)
Next, we consider the forces and moments at segment BC, 2 < x < 5 m, as shown below.
Applying the two equations of equilibrium as before, we have
Ξ£πΉπΉπ¦π¦ = 0 β 18 β 14 β ππ = 0
β΄ ππ = 4 kN
Ξ£πππΉπΉ = 0 β ππ β 18π₯π₯ + 14(π₯π₯ β 2) = ππ β 18π₯π₯ + 14π₯π₯ β 28 = 0
β΄ ππ = 4π₯π₯ + 28 kN β m (2 < x < 5 m)
Finally, we consider segment CD, 5 < x < 7 m.
Applying the two equations of equilibrium gives
Ξ£πΉπΉπ¦π¦ = 0 β 18 β 14 β 28 β ππ = 0
β΄ ππ = β24 kN
Ξ£πππΊπΊ = 0 β ππ + 28(π₯π₯ β 5) + 14(π₯π₯ β 2)β 18π₯π₯
β΄ ππ + 42π₯π₯ β 140 β 28 β 18π₯π₯ = 0
β΄ ππ + 24π₯π₯ β 168 = 0
β΄ ππ = 168 β 24π₯π₯ kN β m
We are now ready to assemble these results in moment and shear force diagrams. The shear force varies according to the laws
ππ(π₯π₯) = 18 kN (0 < π₯π₯ < 2 m)
ππ(π₯π₯) = 4 kN (2 < x < 5 m)
ππ(π₯π₯) = β24 kN (5 < π₯π₯ < 7 m)
The plot thus obtained with Mathematica is shown below.
Next, we plot the bending moment diagram, which follows the piecewise function
ππ(π₯π₯) = 18π₯π₯ (0 < π₯π₯ < 2 m)
1 2 3 4 5 6 7m
20
10
10
20kN
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ππ(π₯π₯) = 4π₯π₯ + 28 (2 < π₯π₯ < 5 m)
ππ(π₯π₯) = 168β 24π₯π₯ (5 < π₯π₯ < 7 m)
The plot thus obtained with Mathematica is given below.
Bear in mind that the moment M at any section equals the area under the shear diagram up to that section; that is,
ππ =πππππππ₯π₯ β ππ = οΏ½πππππ₯π₯
For instance, in the interval 0 < x < 2 m we have
ππ = οΏ½18πππ₯π₯ = 18π₯π₯ (0 < π₯π₯ < 2 m)
β The correct answer is B.
P.2 β Solution The reactions are easily determined with a free body diagram.
First, we sum moments relative to point A,
Ξ£πππ΄π΄ = 0 β π΅π΅π¦π¦(32)β 400(22)β 480(6) = 0 β΄ π΅π΅π¦π¦ = 365 lb
Then, summing forces in the y-direction, we have
Ξ£πΉπΉπ¦π¦ = 0 β π΄π΄ β 400 β 480 + 365 = 0 β΄ π΄π΄ = 515 lb
and, because Bx is the only force acting on the x-direction, we immediately have π΅π΅π₯π₯ = 0.
1 2 3 4 5 6 7m
10
20
30
40
50kNm
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We proceed to investigate the internal forces and moments acting on the beam. From point A to C, we consider the portion of the beam to the left of section 1 in the figure shown above. The distributed load is replaced with an equivalent concentrated force. Writing equations for equilibrium of forces in the y-direction and for equilibrium of moments, we see that
Ξ£πΉπΉπ¦π¦ = 0 β 515β 40π₯π₯ β ππ = 0
β΄ ππ(π₯π₯) = 40π₯π₯ β 515
Ξ£ππππ = 0 β β515π₯π₯ + 40π₯π₯ οΏ½π₯π₯2οΏ½ + ππ
β΄ ππ(π₯π₯) = 515π₯π₯ β 20π₯π₯2 (0 < π₯π₯ < 12 in. )
These equations describe the variation of shear force and bending moment with distance x from the left support. Next, we consider the equilibrium of forces and moments to the left of section 2 in the previous figure. As before, we replace the distributed load with an equivalent force. The equations of equilibrium yield
Ξ£πΉπΉπ¦π¦ = 0 β 515β 480 β ππ = 0
β΄ ππ(π₯π₯) = 35 lb
Ξ£ππππ = 0 β β515π₯π₯ + 480(π₯π₯ β 6) + ππ = 0
β΄ ππ(π₯π₯) = 35π₯π₯ + 2880 (12 < π₯π₯ < 18 in. )
Finally, consider the segment that goes from D to B. Applying the equations of equilibrium gives
Ξ£πΉπΉπ¦π¦ = 0 β 515 β 480β 400β ππ = 0
β΄ ππ(π₯π₯) = 365 lb
Ξ£ππππ = 0 β β515π₯π₯ + 480(π₯π₯ β 6)β 1600 + 400(π₯π₯ β 18) + ππ = 0
β΄ ππ(π₯π₯) = β365π₯π₯ + 11680 (18 < π₯π₯ < 32 in. )
The shear and bending moment diagrams for the entire beam can now be prepared. Notice that the couple of moment 1600 lbβin applied at point D introduces a discontinuity into the bending moment diagram. Inspecting the graphs below, we verify that |ππmax| = 515 lb and |ππmax| = 5110 lb-in.
β The correct answer is D.
P.3 β Solution
A free body diagram for a segment of the beam having length x is illustrated in continuation.
Using proportional triangles, we see that the distributed loading on the beam segment has an intensity of w/x = 6/9, so that w = (2/3)x. The magnitude of the resultant force that can replace this distributed load is equal to (1/2)x(2/3)x =
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(1/3)x2. This force acts through the centroid of the distributed loading area, a distance (1/3)x from the right end, as shown in the previous figure. Writing equations for equilibrium of forces in the y-direction and for equilibrium of moments, we see that
Ξ£πΉπΉπ¦π¦ = 0 β 9 β13π₯π₯
2 β ππ = 0
β΄ ππ(π₯π₯) = 9 β13π₯π₯2
Ξ£ππππ = 0 β β9π₯π₯ +13 π₯π₯
2 οΏ½π₯π₯3οΏ½+ ππ = 0
β΄ ππ(π₯π₯) = 9π₯π₯ βπ₯π₯3
9 (0 < π₯π₯ < 9 m)
Plots of shear and bending moment are given below.
Note that the shear force diagrams in options A and C are quite similar. However, they differ in the point of zero shear, which can be easily obtained by setting the equation of V(x) to zero; that is,
ππ(π₯π₯) = 9 β13π₯π₯
2 = 0 β13 π₯π₯
2 = 9
β΄ π₯π₯ = 271 2β = 5.2 m
Hence, the graph in A is the correct shear force diagram.
β The correct answer is A.
P.4 β Solution
The support reactions are most easily obtained by considering the resultants of the distributed loads on the whole beam. In doing so, the reactions are determined as R1 = 247 lb and R2 = 653 lb.
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Consider first the free body diagram for the beam segment that goes from the left end (i.e., from x = 0) to x = 4 ft, as shown.
Summing forces in the vertical direction and moments relative to the cut section, we have
Ξ£Fy = 0 β V = 247β 12.5x2
Ξ£M = 0 β M + 12.5x2 οΏ½x3οΏ½ β 247x = 0
β΄ ππ(π₯π₯) = 247xβ 4.17x3 (0 < x < 4 ft)
Next, we move on to interval 4 < x < 8 ft.
A vertical force summation and a moment summation about the right end yield, respectively,
Ξ£Fy = 0 β V + 100(xβ 4) + 200β 247 = 0
β΄ ππ + 100xβ 400 + 200β 247 = 0
β΄ ππ(π₯π₯) = 447β 100x
Ξ£M0 = 0 β M + 100(xβ 4) οΏ½xβ 4
2 οΏ½+ 200 οΏ½x β23 (4)οΏ½ β 247x = 0
β΄ ππ(π₯π₯) = β267 + 447xβ 50x2 (4 < x < 8 ft)
Consider now the beam segment such that 8 < x < 10 ft. Summing forces and moments, we have
ππ(π₯π₯) = β353 lb
ππ(π₯π₯) = 2930 β 353π₯π₯ (8 < π₯π₯ < 10 ft)
Finally, the last portion is analyzed by inspection. The shear is constant at +300 lb, and the moment follows a straight-line relation beginning with zero at the right end of the beam. The final shear force and bending moment diagrams are shown below.
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After graphing the diagrams, we note that the maximum positive bending moment occurs somewhere in the interval 0 < x < 8 ft, while the maximum negative bending moment is readily seen to be β600 lbβft at x = 10 ft. To obtain the maximum positive bending moment, recall that the internal moment is maximum when the shear force equals zero. In segment 0 < x < 8 ft, we have V(x) = 447 β 100x. Setting this to zero, we obtain x = 4.47 m. The bending moment at this position is
ππ(4.47) = β267 + 447(4.47)β 50(4.47)2 = 732.0 lb β ft
which is greater in absolute value than 600 lbβft; therefore, it is taken as the maximum bending moment acting on the structure.
β The correct answer is A.
P.5 β Solution
We begin with the assertion that the derivative of shear with respect to x equals the distributed load w, while the derivative of bending moment with respect to x yields the shear force. Mathematically,
πππππππ₯π₯
= βπ€π€ βπππππππ₯π₯ = βπ€π€0 sin
πππ₯π₯πΏπΏ
β΄ ππ(π₯π₯) = βοΏ½π€π€0 sinπππ₯π₯πΏπΏ πππ₯π₯ =
π€π€0πΏπΏππ cos
πππ₯π₯πΏπΏ + πΆπΆ1 =
πππππππ₯π₯
β΄ ππ(π₯π₯) =π€π€0πΏπΏ2
ππ2 sinπππ₯π₯πΏπΏ + πΆπΆ1π₯π₯ + πΆπΆ2
We have two boundary conditions, namely, M(0) = 0 and M(L) = 0. Applying these gives
ππ(0) =π€π€0πΏπΏ2
ππ2 sin 0 + πΆπΆ1(0) + πΆπΆ2 = 0 β πΆπΆ2 = 0
ππ(πΏπΏ) =π€π€0πΏπΏ2
ππ2 sinπποΏ½=0
+ πΆπΆ1πΏπΏ + πΆπΆ2β=0
= 0 β πΆπΆ1 = 0
Thus the expression for shear force is simply
ππ(π₯π₯) =π€π€0πΏπΏππ
cosπππ₯π₯πΏπΏ
while for bending moment,
ππ(π₯π₯) =π€π€0πΏπΏ2
ππ2 sinπππ₯π₯πΏπΏ (0 < π₯π₯ < πΏπΏ)
We set ππππ πππ₯π₯β = V = 0. Inspecting the equation for V(x), we see that V(L/2) = 0, in which case we have
ππ οΏ½πΏπΏ
2πποΏ½ =π€π€0πΏπΏππ cos οΏ½
πππΏπΏ οΏ½
πΏπΏ2οΏ½οΏ½ =
π€π€0πΏπΏππ cos οΏ½
ππ2οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½
=0
= 0
The maximum bending moment is then
ππmax = πποΏ½πΏπΏ2οΏ½ =
π€π€0πΏπΏ2
ππ2 sin οΏ½πππΏπΏ οΏ½
πΏπΏ2οΏ½οΏ½ =
π€π€0πΏπΏ2
ππ2 sinππ2 =
π€π€0πΏπΏ2
ππ2
β The correct answer is D.
P.6 β Solution
Consider the free body diagram for this structure.
From symmetry, we can infer that Ey = Dy. Further, Ex = Dx = Dy tanππ. Summing forces in the y-direction, we have
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Ξ£πΉπΉπ¦π¦ = 0 β π·π·π¦π¦ + πΈπΈπ¦π¦ β 3 = 0
β΄ π·π·π¦π¦ = πΈπΈπ¦π¦ = 1.5 kN
From the previous equality, we have Dx = 1.5tanππ β and Ex = 1.5tanππ β. We replace the forces at D and E with equivalent force-couple systems, as illustrated below.
The equivalent moment M0 is such that M0 = 1.5tanππ Γ 0.5 = (750 Nβm)tanππ. Note that the weight of the beam per unit length is
π€π€ =πππΏπΏ =
3 kN5 m = 0.6
kNm = 600
Nm
We are now ready to prepare the shear and bending moment diagrams. First, consider the segment that goes from A to F.
The sum of forces in the y-direction is such that
Ξ£πΉπΉπ¦π¦ = 0 β βππ β 600π₯π₯ = 0 β΄ ππ(π₯π₯) = β600π₯π₯
The sum of moments relative to the right end of this segment is
Ξ£ππ0 = 0 β ππ + 600π₯π₯ οΏ½π₯π₯2οΏ½ = 0 β΄ ππ(π₯π₯) = β300π₯π₯2 (0 < π₯π₯ < 1.5 m)
Note that, for x = 0, we have V(0) = 0 and M(0) = 0, while, for x = 1.5 m, it is seen that V(1.5) = β900 N and M(1.5) = β675 Nβm. Next, consider the segment that goes from F to H. The sum of vertical forces and the sum of moments relative to the right end yield, respectively,
Ξ£πΉπΉπ¦π¦ = 0 β 1500β 600π₯π₯ β ππ = 0
β΄ ππ(π₯π₯) = 1500 β 600π₯π₯
Ξ£ππ0 = 0 β β1500(π₯π₯ β 1.5) + 600π₯π₯ οΏ½π₯π₯2οΏ½+ ππβππ0 = 0
β΄ β1500π₯π₯ + 2250 + 300π₯π₯2 βππ0 + ππ = 0
β΄ ππ(π₯π₯) = ππ0 β 300π₯π₯2 + 1500π₯π₯ β 2250 (1.5 < π₯π₯ < 2.5 m) At x = 1.5 m, we have V(1.5) = 600 N and M(1.5) = M0 β 675 Nβm, while, at x = 2.5 m, we note that V(2.5) = 0 and M(2.5) = M0 β 375 Nβm. From G to B, the diagrams can be easily obtained through symmetry. The ensuing shear and bending moment diagrams are as shown.
It is easy to see that the largest shear force is |ππmax| = 900 N at x = 1.5 m and 3.5 m. Now, making ππ = 60o, moment M0 becomes M0 = 750Γ tan 60o = 1299 Nβm. Hence, we have, just to the right of point F, M(1.5) = M0 β 675 = 1299 β 675 =
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624 Nβm, while, just to the right of point G, we have M(2.5) = M0 β 375 = 1299 β 375 = 924 Nβm. Accordingly,|Mmax| = 924 Nβm.
β The correct answer is D.
P.7 β Solution
Consider the following free body diagram for the beam segment that extends itself from the left end to the right of point B.
Taking moments about point B in the beam, we have
Ξ£πππ΅π΅ = 0 β12
(2ππ + ππ) Γ π€π€ οΏ½13
(ππ β ππ)οΏ½ β π΄π΄π¦π¦ππ = 0
β΄ π΄π΄π¦π¦ =π€π€2ππ (2ππ + ππ)(ππ β ππ)
Consider now the following free body diagram.
The problem requires that VC = 0. Summing forces in the vertical direction, then, we have
Ξ£πΉπΉπ¦π¦ = 0 βπ€π€6ππ
(2ππ + ππ)(ππ β ππ) β12οΏ½ππ +
ππ2οΏ½π€π€2 = 0
β΄π€π€6ππ
(2ππ + ππ)(ππ β ππ) =π€π€8
(2ππ + ππ)
β΄ππππ
=14
β The correct answer is C.
P.8 β Solution
Consider first the free body diagram for member AB.
The vertical component of the reaction at B can be obtained by taking moments about point A; that is,
Ξ£πππ΅π΅ = 0 β π΅π΅π¦π¦ Γ 4 β 1000 Γ 2 = 0 β΄ π΅π΅π¦π¦ = 500 N
Similarly, the horizontal component of B can be obtained by taking moments about point C, as shown.
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Mathematically,
Ξ£πππΆπΆ = 0 β β500 Γ 4 + 225 Γ 0.5 + π΅π΅π₯π₯ Γ 1.5 = 0 β΄ π΅π΅π₯π₯ = 1258.3 kN
Consider now the free body diagram for segment BE, as illustrated in continuation.
Summing forces in the horizontal direction, we have
Ξ£πΉπΉπ₯π₯ = 0 β βπππΈπΈ β 1258.3β 225 = 0
β΄ πππΈπΈ = β1483.3 N
Similarly, summing forces in the y-direction, we see that
Ξ£πΉπΉπ¦π¦ = 0 β πππΈπΈ β 500 = 0
β΄ πππΈπΈ = 500 N
Finally, the bending moment at E can be determined by applying the second condition of equilibrium to this point, giving
Ξ£πππΈπΈ = 0 β βπππΈπΈ + 225 Γ 0.5 + 1258.3 Γ 1.5 β 500 Γ 2 = 0
β΄ πππΈπΈ = 1000 N β m
β Statements 2 and 3 are true, whereas statement 1 is false.
P.9 β Solution
The free body diagram of the structure is illustrated below.
First, consider the segment in the interval 0 < x < L/2.
Applying the two conditions of equilibrium, we have
Ξ£πΉπΉπ¦π¦ = 0 β βππ0
πΏπΏ β ππ = 0 β΄ ππ(π₯π₯) = βππ0
πΏπΏ
Ξ£ππ0 = 0 β ππ +ππ0
πΏπΏ π₯π₯ = 0 β΄ ππ(π₯π₯) = βππ0
πΏπΏ π₯π₯ (0 < π₯π₯
< πΏπΏ 2β )
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Next, consider the segment located in the interval L/2 < x < L. Applying the two conditions of equilibrium, we see that
Ξ£πΉπΉπ¦π¦ = 0 β βππ0πΏπΏβ ππ = 0 β΄ ππ(π₯π₯) = βππ0
πΏπΏ
Ξ£ππ0 = 0 β ππ +ππ0
πΏπΏ π₯π₯ βππ0 = 0 β΄ ππ(π₯π₯)
= ππ0 βππ0
πΏπΏ π₯π₯
= ππ0 οΏ½1 βπ₯π₯πΏπΏοΏ½ οΏ½
πΏπΏ2 < π₯π₯ < πΏπΏοΏ½
When M0 = 500 Nβm and L = 9 m, and noting that Vmax = 5 kN and Mmax = 22 kNβm, we make use of the expressions
ππmax =ππ0
πΏπΏ β 5 =ππ0
9
β΄ ππ0 = 45 kN β m
and
ππmax =ππ0
2 β 22 =ππ0
2
β΄ ππ0 = 44 kN β m
The lower result controls; thus, the largest couple moment that the beam is capable of withstanding is M0 = 44 kNβm.
β The correct answer is B.
P.10 β Solution
Since the loading is discontinuous at the midspan, the shear and moment equations must be written for regions 0 < x < 6 ft and 6 < x < 12 ft of the beam. The free-body diagrams of the beamβs segment sectioned through the arbitrary point within these two regions are shown in figures (b) and (c) below. The reactions at A and B were found as Ay = 7.5w0 and By = 10.5w0.
Consider first the segment for which 0 < x < 6 ft. Summing forces in the vertical direction and moments, we obtain
Ξ£πΉπΉπ¦π¦ = 0 β 7.5π€π€0 β π€π€0π₯π₯ β ππ = 0
β΄ ππ(π₯π₯) = π€π€0(7.5 β π₯π₯) (I)
Ξ£ππ0 = 0 β π€π€0π₯π₯ οΏ½π₯π₯2οΏ½ β 7.5π€π€0π₯π₯ + ππ = 0
β΄ ππ(π₯π₯) = 7.5π€π€0π₯π₯ β 0.5π€π€0π₯π₯2 =π€π€02
(15π₯π₯ β π₯π₯2) (0 < π₯π₯ < 6 ft) (II)
Next, consider the segment that encompasses 6 < x < 12 ft. Summing forces and moments as before, we see that
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Ξ£πΉπΉπ¦π¦ = 0 β 10.5π€π€0 β 2π€π€0(12 β π₯π₯) + ππ = 0 β΄ ππ(π₯π₯)= 2π€π€0(12 β π₯π₯)β 10.5π€π€0 = 24π€π€0 β 2π€π€0π₯π₯ β 10.5π€π€0= π€π€0(13.5β 2π₯π₯) (III)
Ξ£ππ0 = 0 β 10.5π€π€0(12β π₯π₯)β 2π€π€0(12 β π₯π₯) οΏ½12
(12β π₯π₯)οΏ½ β ππ = 0
β΄ ππ(π₯π₯) = π€π€0(βπ₯π₯2 + 13.5π₯π₯ β 18) (6 < π₯π₯ < 12 ft) (IV)
The shear diagram is plotted using Equations (I) and (III). The value of the shear force at x = 6 ft can be evaluated using either equation,
ππ(6) = π€π€0(7.5β 6) = 1.5π€π€0
The location at which the shear is equal to zero is obtained by setting V = 0 in Equation (III),
π€π€0(13.5β 2π₯π₯) = 0
β΄ 13.5π€π€0 β 2π€π€0π₯π₯ = 0
β΄ π₯π₯ =13.5
2 = 6.75 ft
The moment diagram in the figure below is plotted using Equations (II) and (IV).
The value of the moment at x = 6 ft is evaluated as
ππ(6) =π€π€02 (15 Γ 6 β 62) = 27π€π€0
The value of the moment at x = 6.75 ft is evaluated using Equation (IV),
ππ(6.75) = π€π€0(β6.752 + 13.5 Γ 6.75β 18) = 27.56π€π€0
By observing the shear and moment diagrams, we verify that Vmax = 10.5w0 and Mmax = 27.56w0. We can then propose two equalities,
ππmax = 10.5π€π€0 = 1200
β΄ π€π€0 = 114.3 lb β ftβ1
and ππmax = 27.56π€π€0 = 600
β΄ π€π€0 =600
27.56 = 21.8 lb β ftβ1
The second result governs, and the largest intensity of the distributed load is taken as w0 = 21.8 lb/ft.
β The correct answer is B.
18 Β© 2019 Montogue Quiz
P.11 β Solution Consider segment CB, as shown.
The sum of moments relative to point C (the left end of the cut off segment) must equal zero; that is,
Ξ£πππΆπΆ = 0 β βππ + 2.2 Γ 3 Γ 103 Γ 1.1 = 0
β΄ ππ = 7.26 kN
Next, we compute the section modulus for the rectangular section, which is given by
ππ =ππβ2
6 =100 Γ 2002
6 = 6.67 Γ 105 mm3 = 6.67 Γ 10β4 m3
Finally, the normal stress can be obtained as the ratio of moment to section modulus (i.e., the flexure formula),
ππ =ππππ =
72606.67 Γ 10β4 = 1.088 Γ 107 Pa = 10.9 MPa
β The correct answer is A.
P.12 β Solution As in Problem 11, the cross-section of the beam is rectangular, and its
section modulus equals
ππ =ππβ2
6=
24 Γ 602
6= 14,400 mm3 = 1.44 Γ 10β5 m3
Using the flexure formula, we can write ππ = ππ ππβ , so that ππ = ππππ. At point D, we have πππ·π· = 1.44 Γ 10β5(55 Γ 106) = 792 N β m, while at point F we have πππΉπΉ =1.44 Γ 10β5(37.5 Γ 106) = 540 N β m. Now, consider segment FB of the beam.
The sum of moments relative to point F must equal zero. Hence,
Ξ£πππΉπΉ = 0 β β540 + 0.3π΅π΅ = 0 β΄ π΅π΅ =5400.3
= 1800 N
Next, using free body DEFB, we have the sum of moments
Ξ£πππ·π· = 0 β β792 β 0.3ππ +0.8(1800) = 0
β΄ ππ =(β792 + 0.8 Γ 1800)
0.3= 2160 N
Then, consider the beam as a whole.
Ξ£πππ΄π΄ = 0 β β0.2ππ β 0.7 Γ 2160 + 1.2Γ 1800 = 0
ππ =β0.7 Γ 2160 + 1.2 Γ 1800
0.2= 3240 N
Summing forces in the y-direction, we obtain the value of reaction A,
Ξ£πΉπΉπ¦π¦ = 0 β π΄π΄β 3240β 2160 + 1800 = 0 β΄ π΄π΄ = 3600 N
We are now ready to consider the shear diagram and its areas, from which the moments can be determined.
19 Β© 2019 Montogue Quiz
From A to C- (just to the left of C), ππ(π₯π₯) = 3600 N and π΄π΄π΄π΄πΆπΆ = 0.2(3600) =720 N β m = ππ; from C+ (just to the right of C) to E-, we have ππ(π₯π₯) = 3600β 3240 =360 N and π΄π΄πΆπΆπΈπΈ = 0.5(360) = 180 N β m = ππ; finally, from E+ to B, it follows that ππ(π₯π₯) = 360β 2160 = β1800 N and π΄π΄EB = 0.5(β1800) = β900 N β m. Now, the bending moments are found to be πππ΄π΄ = 0,πππΆπΆ = 0 + 720 = 720 N βm, πππΈπΈ = 720 +180 = 900 N βm, and πππ΅π΅ = 900β 900 = 0. Clearly, the maximum bending moment is |ππmax| = 900 N βm at point E. The maximum normal stress follows from the flexure formula,
ππmax =|ππmax|ππ =
9001.44 Γ 10β5 = 62.5 MPa
β The correct answer is C.
Answer Summary
Problem 1 B Problem 2 D Problem 3 A Problem 4 A Problem 5 D Problem 6 D Problem 7 C Problem 8 T/F Problem 9 B
Problem 10 B Problem 11 A Problem 12 C
References BEER, F., JOHNSTON, E., DEWOLF, J. and MAZUREK, D. (2015). Mechanics of
Materials. 7th edition. New York: McGraw-Hill. BEER, F., JOHNSTON, E., MAZUREK, D. and CORNWELL, P. (2013). Vector
Mechanics for Engineers: Statics. 10th edition. New York: McGraw-Hill. HIBBELER, R. (2013). Engineering Mechanics: Statics. 13th edition. Upper
Saddle River: Pearson. MERIAM, J. and KRAIGE, L. (2002). Engineering Mechanics: Statics. 5th
edition. Hoboken: John Wiley and Sons.
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