1. No category

Tutorial 2
ME101 (Engineering Mechanics) 2014-2015 Sem II
Ans 1. The reactions at A and B will act normally to the respective surfaces of contact. The
forces acting are the weight, W acting downwards, NA and NB acting at 150 and 600 to the
horizontal respectively.
For static equilibrium
∑ 𝐹𝑥= 0, 𝑖𝑒,
𝑁𝐴 cos 15 − 𝑁𝐵 cos 60 = 0
Also
∑ 𝐹𝑦= 0, 𝑖𝑒,
𝑁𝐴 sin 15 + 𝑁𝐵 sin 60 − 𝑊
= 0, 𝑖𝑒,
𝑁𝐴 sin 15 + 𝑁𝐵 sin 60 − 20 × 9.81 = 0
On solving we get:
The contact forces at A, 𝐍𝐀 = 𝟏𝟎𝟏. 𝟔𝐍
The contact forces at B, 𝐍 𝐁 = 𝟏𝟗𝟔. 𝟐𝐍
Ans 2.
Dimensions in ‘m’,
G is centered in
the solid
Weight of the rectangular solid acting at G = 125 × 9.81 N
̂̂
̅̅̅̅
𝐶𝐷
.125𝑖̂−.3𝑗̂ +.4𝑘
Tension 𝑇̂ along CD = T ̅̅̅̅̅̅ = T
= T (.243𝑖̂-.582𝑗̂+.776𝑘̂)
|𝐶𝐷|
√.1252 +.32 +.4 2
To keep the rectangular solid in position, resultant moment due to the forces must be 0 at the joints
Resultant moment about x axis at A, ∑Mx = -.776 T×.3 + 125 ×9.81×0.2 = 0
⇒ T = 1053. 48 kN
Ans 3.
Given BC = CD
Δ CDE and Δ BDA are similar as
𝐶𝐸
1
Hence 𝐵𝐴 = 2
𝐷𝐶
𝐷𝐵
=
𝐷𝐸
𝐷𝐴
1
2
(= ) and < CDE = <BDA
⇒ CE = 1.5 m and <CED = <BAD = 45̊
Note that BF, CG are
perpendiculars drawn
to AD, are not truss
members.
F
ig
G
ig
Ax , Ay and Dy = the reactions at the hinge and roller supports respectively
Taking moments of all the forces about A and equating to zero, we have
5 × AF + 3 × AG – Dy × AD = 0
⇒ 5× 3cos45̊ + 3× (3+1.5cos45̊) – Dy × 6 = 0
⇒ Dy = 3.80 kN
At joint D:
𝐵𝐹
3𝑠𝑖𝑛45̊
𝜃 = 𝑡𝑎𝑛−1 (𝐹𝐷) = 𝑡𝑎𝑛−1 (6−3𝑐𝑜𝑠45̊) = 28.7̊
Considering equilibrium of forces along x direction, ∑Fx = 0
̂ cos𝜃 -𝐷𝐸
̂ =0
We have 𝐶𝐷
̂ sin𝜃 = 0
Similarly, ∑Fy = 0 ⇒ 3.80 - 𝐶𝐷
Solving the above two equations, we have
̂ = +7.92 kN and 𝐷𝐸
̂ = + 6.94 kN
𝐶𝐷
̂ and 𝐷𝐸
̂ are correct
Hence the assumed direction of 𝐶𝐷
̂ = 7.92 kN , Compressive (toward the joint) & 𝐷𝐸
̂ = 6.94 kN, Tensile (away from the joint)
Hence 𝐶𝐷
At joint C:
̂ cos28.7̊ - 7.92 cos28.7̊ - 𝐶𝐸
̂ cos45̊ = 0,
∑Fx = 0 ⇒ - ̅̅̅̅
𝐵𝐶
̂ sin28.7 ̊ + 7.92sin28.7 ̊ -𝐶𝐸
̂ sin45 -̊ 3 = 0.
∑Fy = 0 ⇒ 𝐵𝐶
Solving the above equations, we have
̂ = -5.71 kN and 𝐶𝐸
̂ = -2.74 kN
𝐵𝐶
̂ and 𝐶𝐸
̂ are incorrect.
Hence the assumed directions of 𝐵𝐶
̂ = 5.71 kN, Compressive and 𝐶𝐸
̂ = 2.74 kN, Compressive
Hence 𝐵𝐶
At joint E:
β=
180 ̊−45 ̊
=
2
67.5 ̊ (from isosceles triangle ABE)
̂ sinβ - 2.74sin45 ̊ = 0
∑Fy = 0 ⇒ 𝐵𝐸
̂ = + 2.10 kN
⇒ 𝐵𝐸
̂ is correct.
Hence assumed direction of 𝐵𝐸
̂ = 2.10 kN, Tensile.
Hence 𝐵𝐸
B
5.71 kN
C
7.92 kN
2.10 kN
A
D
2.74 kN
E 6.94 kN
Ans 4.
Section cut is drawn through members DE, EL, and LK for the release of forces
̂ × 2- 8×6 = 0,
At joint L, resultant moment, ∑M = 0 ⇒ 𝐷𝐸
̂ = +24 kN, the assumed direction of 𝐷𝐸
̂ is correct
Hence 𝐷𝐸
At joint D,
̂ cos45 ̊ + 24 = 0
Considering equilibrium of forces along horizontal direction, -𝐷𝐿
̂ = + 33.94 kN, ∴ The assumed direction of 𝐷𝐿
̂ is correct.
Hence 𝐷𝐿
̂ = 24 kN, Tensile and 𝐷𝐿
̂ = 33.94 kN, Compressive.
Hence 𝐷𝐸
Ans 5.
Conditions for zero force in members are:
1. In a joint with two non-collinear members, if there is no external force then members
have zero force.
2. In a joint with three members if two members are collinear and there is no external force,
then the non-collinear member is a zero-force member.
In the given problem we find that the Joint K has 2 collinear members KL and IK and there is no
external force, hence JK is a zero force member by property2
Now at joint J, HJ and JL are collinear, and JK is zero force, hence it can be neglected. Therefore
by condition 2, IJ is a zero force member.
Now at joint I, GI and IK are collinear, and IJ is zero force, hence it can be neglected. Therefore
by condition 2, HI is a zero force member.
At the joint C, CA and CE are collinear and there is no external force, hence by property 2, BC is
a zero force member.