Indian Institute of Technology Guwahati ME 101: Engineering Mechanics (2016-2017, Sem II)
Quiz-1 (30.01.2017)
Time: 8:00 AM โ 8:55 AM Full Marks: 60 Q1. A piece of sheet metal is bent into the shape shown (Fig.1) and is acted upon by three forces. If the
forces have the same magnitude P, replace them with an equivalent wrench. Determine the following:
(a) Magnitude and the direction of the resultant force [4 marks]
(b) Magnitude and the direction of moment of the wrench [8 marks]
(c) Locate the point where the axis of the wrench intersects the xy plane [8 marks]
Model Answer for Q1:
First reduce the given forces to an equivalent force-couple system (R, MOR) at the origin.
(a) We have
R = โ F = F1 + F2 + F3
= โ๐๐ฬฬ + ๐๏ฟฝฬ๏ฟฝ + ๐๐ฬฬ = ๐๏ฟฝฬ๏ฟฝ
(b)
MOR = โ Mo = โ r ร F (using Varignonโs Theorem)
= r1 ร F1 + r2 ร F2 + r3 ร F3
= ๐๐ฬฬ ร (โ๐๐ฬฬ) + (๐๐ฬฬ + ๐๏ฟฝฬ๏ฟฝ) ร (๐๏ฟฝฬ๏ฟฝ) + (5
2๐๐ฬฬ + ๐๏ฟฝฬ๏ฟฝ) ร (๐๐ฬฬ)
= โ(๐๐)๐ฬฬ โ (๐๐)๐ฬฬ + (5
2๐๐) ๏ฟฝฬ๏ฟฝ
MOR = ๐๐(โ๐ฬฬ โ ๐ฬฬ +
5
2๏ฟฝฬ๏ฟฝ)
R = P
Direction of resultant force is
along +ve z axis
Then for the wrench,
and
cos ๐๐ฅ = 0 cos ๐๐ฆ = 0 cos ๐๐ง = 1
or ๐๐ฅ = 90ยฐ ๐๐ฆ = 90ยฐ ๐๐ง = 0ยฐ
Now,
M1 = ๏ฟฝฬ๏ฟฝ๐๐ฅ๐๐ โ MOR
= ๏ฟฝฬ๏ฟฝ โ ๐๐ (โ๐ฬฬ โ ๐ฬฬ +5
2๏ฟฝฬ๏ฟฝ) =
5
2๐๐
(c) The components of the wrench are (R, M1) , where M1 = M1๏ฟฝฬ๏ฟฝ๐๐ฅ๐๐ , and the axis of the wrench is assumed to
intersect the xy-plane at point Q, whose coordinates are (๐ฅ, ๐ฆ, 0). Thus, we require
M2 = rQ ร R
Where M2 = MOR โ M1
Then
rQ ร R = MOR โ M1
(๐ฅ๐ฬฬ + ๐ฆ๐ฬฬ) ร ๐๏ฟฝฬ๏ฟฝ = ๐๐ (โ๐ฬฬ โ ๐ฬฬ +5
2๏ฟฝฬ๏ฟฝ) โ
5
2๐๐๏ฟฝฬ๏ฟฝ
โ๐ฅ๐ ๐ฬฬ + ๐ฆ๐ ๐ฬฬ = ๐๐ (โ๐ฬฬ โ ๐ฬฬ)
Equating coefficients:
๐ฬฬ: ๐ฆ๐ = โ๐๐ or ๐ฆ = โ๐
๐ฬฬ : โ ๐ฅ๐ = โ๐๐ or ๐ฅ = ๐
The axis of the wrench is parallel to the z-axis and intersects the xy-plane at
Alternate Model Answer for Q1:
First reduce the given forces to an equivalent force-
couple system (R, MOR) at the origin.
(a) We have
R = โ F = F1 + F2 + F3
= โ๐๐ฬฬ + ๐๏ฟฝฬ๏ฟฝ + ๐๐ฬฬ = ๐๏ฟฝฬ๏ฟฝ
or ,
๐ฅ = ๐, ๐ฆ = โ๐
๏ฟฝฬ๏ฟฝ๐๐ฅ๐๐ = (R
Rโ ) = ๏ฟฝฬ๏ฟฝ
M1 =5
2๐๐
Direction of moment of the wrench is along +ve z axis
R = P
Direction of resultant force is along
+ve z axis
M1 =5
2 ๐๐, Direction of
moment of the wrench is
along +ve z axis
(b), (c)
Let Q (๐ฅ, ๐ฆ, 0) be the point of intersection of the line of action of the moment of the wrench and xy-plane. The
components of the wrench are (R, M1) and rQ is the position vector of point Q.
Now, MOR = M1 + rQ ร ๐ , for moment about O
MQR = MO
R โ rQ ร ๐ , for moment about Q
So, M1 = MQR
The direction of the moment of the wrench i.e. M1 and resultant force R should be same.
So, M1 = M1 ๏ฟฝฬ๏ฟฝ
MQR = โ MQ = โ r ร F (using Varignonโs Theorem)
= r1 ร F1 + r2 ร F2 + r3 ร F3
Here, r1 = [โ๐ฅ๐ฬฬ + (๐ โ ๐ฆ)๐ฬฬ] , r2 = [(๐ โ ๐ฅ)๐ฬฬ โ ๐ฆ๐ฬฬ + ๐๏ฟฝฬ๏ฟฝ] , r3 = [(5
2๐ โ ๐ฅ) ๐ฬฬ โ ๐ฆ๐ฬฬ + ๐๏ฟฝฬ๏ฟฝ]
Hence,
MQR = [โ๐ฅ๐ฬฬ + (๐ โ ๐ฆ)๐ฬฬ] ร (โ๐๐ฬฬ) + [(๐ โ ๐ฅ)๐ฬฬ โ ๐ฆ๐ฬฬ + ๐๏ฟฝฬ๏ฟฝ] ร (๐๏ฟฝฬ๏ฟฝ) + [(
5
2๐ โ ๐ฅ) ๐ฬฬ โ ๐ฆ๐ฬฬ + ๐๏ฟฝฬ๏ฟฝ] ร (๐๐ฬฬ)
= โ๐ (๐ฆ + ๐)๐ฬฬ โ ๐(๐ โ ๐ฅ)๐ฬฬ + [๐๐ฅ + ๐ (5
2๐ โ ๐ฅ)] ๏ฟฝฬ๏ฟฝ
M1๏ฟฝฬ๏ฟฝ = โ๐(๐ฆ + ๐)๐ฬฬ โ ๐(๐ โ ๐ฅ)๐ฬฬ + [๐๐ฅ + ๐ (5
2๐ โ ๐ฅ)] ๏ฟฝฬ๏ฟฝ
Equating coefficients:
Hence, ๐ฬฬ : โ ๐ฆ๐ โ ๐๐ = 0 โ ๐ฆ = โ๐
๐ฬฬ : โ ๐ ร (๐ โ ๐ฅ) = 0 โ ๐ฅ = ๐
๏ฟฝฬ๏ฟฝ: M1 = ๐ ๐ฅ + ๐ (5
2๐ โ ๐ฅ) = ๐๐ +
3
2 ๐๐ =
5
2 ๐๐
Answer (b):
Answer (c):
๐ฅ = ๐, ๐ฆ = โ๐
Q2. A weight of 1200 N is supported by a rope through a frictionless pulley in conjunction with a system of
beam and bent member as shown in Fig. 2. Please note that joints A, B and D are pinned joints.
(a) Draw the free body diagrams for members AB and BCD. [8 marks]
(b) Find the support reaction at A [12 marks]
Q3. Analyze the truss as shown in Fig. 3 using the method of joints. Determine the following:
(a) Find out the support reactions [3 marks]
(b) Identify the zero force members [4 marks]
(c) Calculate force in the each member with clearly identifying whether the member is
in tension (T) or compression (C). [13 marks]
Solution: (METHOD-1)
Part-a)
Total vertical force acting on Joint D = 40 sin 45 = 28.28 ๐๐
Total horizontal force acting on Joint D = 40 cos 45 = 28.28 ๐๐
+โ ๐บ๐ญ๐ = ๐ ๐ด๐ฅ + 28.28 ๐๐ = 0
โด ๐จ๐ = ๐๐. ๐๐ ๐๐ต (โ)
Due to symmetry w.r.t the vertical load,
๐จ๐ = ๐ฎ๐ =๐๐.๐๐
๐= ๐๐. ๐๐ ๐๐ต (โ)
Part-b) Using the principles of zero force member identification, referring to lecture-5 slides 8, 9 and 10,
we have:
Stepwise with correct sequence:
LB and HF are zero force members
LC and HE are zero force members
KC and IE are zero force members
KD and ID are zero force members
Part-c)
Now that the zero force members have been identified, the truss to be solved, effectively, is (The zero force
members have been grayed out) โ
Applying force equilibrium at joints, we get,
Joint A: +โ ๐บ๐ญ๐ = ๐ ๐น๐ด๐ต + ๐น๐ด๐ฟ cos 45 โ 28.28 = 0
โด ๐ญ๐จ๐ฉ = ๐๐. ๐๐ ๐๐ต (๐ป)
โ + ๐บ๐ญ๐ = ๐ ๐น๐ด๐ฟ sin 45 + 14.14 = 0
โด ๐ญ๐จ๐ณ = โ๐๐ ๐๐ต (๐ช)
Joint L:
Only members AL and LK are joined here and they are connected in a straight line (LB and LC are zero
force members). Thus,
+โ ๐บ๐ = ๐ ๐ญ๐ณ๐ฒ = ๐ญ๐จ๐ณ = โ๐๐ ๐๐ต (๐ช)
Joint K: Similar to Joint L,
+โ ๐บ๐ญ = ๐ ๐ญ๐ฒ๐ฑ = ๐ญ๐ณ๐ฒ = โ๐๐ ๐๐ต (๐ช)
Joint B: Similar to Joint L,
+โ ๐บ๐ญ๐ = ๐ ๐ญ๐ฉ๐ช = ๐ญ๐ฉ๐จ = ๐๐. ๐๐ ๐๐ต (๐ป)
Joint C: Similar to Joint L,
+โ ๐บ๐ญ๐ = ๐ ๐ญ๐ช๐ซ = ๐ญ๐ฉ๐ช = ๐๐. ๐๐ ๐๐ต (๐ป)
Joint J:
+โ ๐บ๐ญ๐ = ๐ ๐น๐ฝ๐ผ๐๐๐ 45 = ๐น๐ฝ๐พ cos 45
โด ๐ญ๐ฑ๐ฐ = โ๐๐ ๐๐ต (๐ช)
โ + ๐บ๐ญ๐ = ๐ โ๐น๐ฝ๐ผ๐ ๐๐ 45 โ ๐น๐ฝ๐พ sin 45 โ ๐น๐ฝ๐ท = 0
โด ๐ญ๐ฑ๐ซ = ๐๐. ๐๐ ๐๐ต (๐ป)
Joint D:
+โ ๐บ๐ญ๐ = ๐ ๐น๐ท๐ธ + 28.28 โ ๐น๐ท๐ถ = 0
โด ๐ญ๐ซ๐ฌ = ๐๐. ๐๐ ๐๐ต(๐ป)
Due to geometrical symmetry and vertical load symmetry, the forces in the following members are equal,
given as:
๐ญ๐ฏ๐ฎ = ๐ญ๐จ๐ณ = ๐ญ๐ณ๐ฒ = ๐ญ๐ฏ๐ฐ = ๐ญ๐ฒ๐ฑ = ๐ญ๐ฐ๐ฑ = โ๐๐ ๐๐ต (๐ช)
Due to asymmetry w.r.t. horizontal load, the forces in the following members are obtained as given below:
Joint E: +โ ๐บ๐ญ๐ = ๐ ๐ญ๐ฌ๐ญ = ๐ญ๐ซ๐ฌ = ๐๐. ๐๐ ๐๐ต (๐ป)
Joint F: +โ ๐บ๐ญ๐ = ๐ ๐ญ๐ญ๐ฎ = ๐ญ๐ฌ๐ญ = ๐๐. ๐๐ ๐๐ต (๐ป)
The summary of all the member forces is provided in the next page.
๐ญ๐ฏ๐ฎ = ๐ญ๐จ๐ณ = ๐ญ๐ณ๐ฒ = ๐ญ๐ฏ๐ฐ = ๐ญ๐ฒ๐ฑ = ๐ญ๐ฐ๐ฑ = โ๐๐ ๐๐ต (๐ช)
๐ญ๐ฉ๐ณ = ๐ญ๐ญ๐ฏ = ๐ญ๐ช๐ณ = ๐ญ๐ฌ๐ฏ = ๐ญ๐ช๐ฒ = ๐ญ๐ฌ๐ฐ = ๐ญ๐ซ๐ฒ = ๐ญ๐ซ๐ฐ = ๐ (i.e., zero force members)
๐ญ๐ฌ๐ญ = ๐ญ๐ซ๐ฌ = ๐ญ๐ญ๐ฎ = ๐๐. ๐๐ ๐๐ต (๐ป)
๐ญ๐ฑ๐ซ = ๐๐. ๐๐ ๐๐ต (๐ป)
๐ญ๐ฉ๐ช = ๐ญ๐ฉ๐จ = ๐ญ๐ช๐ซ = ๐๐. ๐๐ ๐๐ต (๐ป)
Solution: (METHOD-2) (Not preferable)
Part-a)
Total vertical force acting on Joint D = 40 sin 45 = 28.28 ๐๐
Total horizontal force acting on Joint D = 40 cos 45 = 28.28 ๐๐
+โ ๐บ๐ญ๐ = ๐ ๐ด๐ฅ + 28.28 ๐๐ = 0
โด ๐จ๐ = ๐๐. ๐๐ ๐๐ต (โ)
Due to symmetry w.r.t the vertical load,
๐จ๐ = ๐ฎ๐ =๐๐.๐๐
๐= ๐๐. ๐๐ ๐๐ต (โ)
Part-b) Using the principles of zero force member identification, referring to lecture-5 slides 8, 9 and 10,
we have:
Stepwise with correct sequence:
LB and HF are zero force members
LC and HE are zero force members
KC and IE are zero force members
KD and ID are zero force members
Part c)
Joint A: +โ ๐บ๐ญ๐ = ๐ ๐น๐ด๐ต + ๐น๐ด๐ฟ cos 45 โ 28.28 = 0
โด ๐ญ๐จ๐ฉ = ๐๐. ๐๐ ๐๐ต (๐ป)
โ + ๐บ๐ญ๐ = ๐ ๐น๐ด๐ฟ sin 45 + 14.14 = 0
โด ๐ญ๐จ๐ณ = โ๐๐ ๐๐ต (๐ช)
Joint L: +โ ๐บ๐ญ๐ = ๐ ๐น๐ฟ๐พ cos 45 + ๐น๐ฟ๐ถ cos 45 โ ๐น๐ด๐ฟ cos 45 = 0
๐น๐ฟ๐พ + ๐น๐ฟ๐ถ = โ20
โ + ๐บ๐ญ๐ = ๐ ๐น๐ฟ๐พ sin 45 โ ๐น๐ฟ๐ถ sin 45 โ ๐น๐ด๐ฟ sin 45 = 0
๐น๐ฟ๐พ โ ๐น๐ฟ๐ถ = โ20
โด ๐ญ๐ณ๐ฒ = โ๐๐ ๐๐ต (๐ช) , ๐ญ๐ณ๐ช = ๐
Joint B: +โ ๐บ๐ญ๐ = ๐ ๐ญ๐ฉ๐ช = ๐ญ๐ฉ๐จ = ๐๐. ๐๐ ๐๐ต (๐ป)
Joint C: +โ ๐บ๐ญ๐ = ๐ ๐น๐ถ๐ท โ ๐น๐ถ๐ต โ ๐น๐ถ๐ฟ๐๐๐ 45 = 0
โด ๐ญ๐ช๐ซ = ๐๐. ๐๐ ๐๐ต (๐ป)
โ + ๐บ๐ญ๐ = ๐ ๐น๐ถ๐พ + ๐น๐ถ๐ฟ๐ ๐๐ 45 = 0
โด ๐ญ๐ช๐ฒ = ๐ ๐๐ต
Joint K: +โ ๐บ๐ญ๐ = ๐ ๐น๐พ๐ฝ๐๐๐ 45 + ๐น๐พ๐ท cos 45 โ ๐น๐พ๐ฟ cos 45 = 0
๐น๐พ๐ฝ + ๐น๐พ๐ท = โ20
โ + ๐บ๐ญ๐ = ๐ ๐น๐พ๐ฝ๐ ๐๐ 45 โ ๐น๐พ๐ท sin 45 โ ๐น๐พ๐ฟ sin 45 = 0
๐น๐พ๐ฝ โ ๐น๐พ๐ท = โ20
โด ๐ญ๐ฒ๐ฑ = โ๐๐ ๐๐ต (๐ช) , ๐ญ๐ฒ๐ซ = ๐
Joint J: +โ ๐บ๐ญ๐ = ๐ ๐น๐ฝ๐ผ๐๐๐ 45 = ๐น๐ฝ๐พ cos 45
โด ๐ญ๐ฑ๐ฐ = โ๐๐ ๐๐ต (๐ช)
โ + ๐บ๐ญ๐ = ๐ โ๐น๐ฝ๐ผ๐ ๐๐ 45 โ ๐น๐ฝ๐พ sin 45 โ ๐น๐ฝ๐ท = 0
โด ๐ญ๐ฑ๐ซ = ๐๐. ๐๐ ๐๐ต (๐ป)
Joint D: +โ ๐บ๐ญ๐ = ๐ ๐น๐ท๐ธ + 28.28 โ ๐น๐ท๐ถ + ๐น๐ท๐ผ cos 63.43 โ ๐น๐ท๐พ cos 63.43 = 0
โด ๐น๐ท๐ธ + 0.447 ๐น๐ท๐ผ = 14.14
โ + ๐บ๐ญ๐ = ๐ ๐น๐ท๐ฝ โ 28.28 + ๐น๐ท๐ผ sin 63.43 + ๐น๐ท๐พ sin 63.43 = 0
โด ๐ญ๐ซ๐ฐ = ๐
On substituting, ๐ญ๐ซ๐ฌ = ๐๐. ๐๐ ๐๐ต (๐ป)
Due to geometrical symmetry and vertical load symmetry, the forces in the following members are equal,
given as:
๐ญ๐ฏ๐ฎ = ๐ญ๐จ๐ณ = ๐ญ๐ณ๐ฒ = ๐ญ๐ฏ๐ฐ = ๐ญ๐ฒ๐ฑ = ๐ญ๐ฐ๐ฑ = โ๐๐ ๐๐ต (๐ช)
๐ญ๐ฉ๐ณ = ๐ญ๐ญ๐ฏ = ๐ญ๐ช๐ณ = ๐ญ๐ฌ๐ฏ = ๐ญ๐ช๐ฒ = ๐ญ๐ฌ๐ฐ = ๐ญ๐ซ๐ฒ = ๐ญ๐ซ๐ฐ = ๐ (i.e., zero force members)
Due to asymmetry w.r.t. horizontal load, the forces in the following members are obtained as given below:
Joint E: +โ ๐บ๐ญ๐ = ๐ ๐น๐ธ๐น โ ๐น๐ธ๐ท + ๐น๐ธ๐ป cos 45 = 0
โด ๐ญ๐ฌ๐ญ = ๐๐. ๐๐ ๐๐ต(๐ป)
Joint F: +โ ๐บ๐ญ๐ = ๐ ๐น๐น๐บ โ ๐น๐น๐ธ = 0
โด ๐ญ๐ญ๐ฎ = ๐๐. ๐๐ ๐๐ต (๐ป)
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