Solutions to Questions on Statics and Mechanics

SOLUTIONS TO THE QUESTIONS
QN 1) Consider the following members:
Member 1: A-B
Member 2: C-D
Member 3: B-D
A force of 200N acts at 1cm away from point A
(a) Since the upward forces are applied at the handle are equal, the clamping forces will
also be equal hence the free body diagram is as shown in figure 1: B
Pc
200N
A
C
200N 7cm 2.5cm 3.5cm
Figure 1: Free body diagram of the pliers
(b) Reactions at hinges B and D
(i) Taking moments about A by considering that the system is at equilibrium
200x1-Rbx9= 0
Rb= 200/9= 22.22N
(ii) For reactions at D, we take moments about C hence:
-200x7+2.5xRd= 0
Rd= 200x7/2.5= 560N
(iii) Force in strut AC
Firstly, we can assume compression to be positive and tension to be negative
Let us assume member AC is in compression:
y
x

We now calculate both forces Fa and Fc as shown in figure 2 above and we take the larger force
as the strut force
Let us briefly consider member A-B for the purpose of determining the reaction at A
200N
Ra 22.22N
Now given static equilibrium of the member, Ra+ 22.22= 200
Ra=200-22.22= 177.78N \
Next, consider triangle AOC (O was never in the initial diagram but can figure it out)
A B
2.5cm
O C
7-1= 6cm
Tan φ= 2.5/6= 22.62o
Back to joint A, (summation) Fy= 0
Hence Ra-FaSinφ= 0
Fa= Ra/sin φ= 177.78/sin 22.62
= 462.23N
Now considering joint C, with a little consideration we can actually see that force in C-D is zero
Hence strut force Fac= 462.23N
(iv) Clamping force
Firstly, let us take moments about B and consider AB up to the pinpoint
connection Pc
A C
Ra 10cm B 3cm

Also considering static equilibrium:
200x(9+1)-Pc x3= 0
Pc= 200 x 10/3= 666.67N
(c ) Fs= 1.5
Yield stress= 500MPa
Allowable stress= Yield stress/Fs= 500/1.5= γ=333.33MPa
The bending moment M is given as: Ma= Fal= 177.78 x 0.025= 4.4445N-m
γ=32 x 4.4445/d3xπ= 333.33x106
d= {πx 333.33x106 /142.224}0.5 = 1.94mm
QN4) Consider the Free body diagram below:
A B C
1m 1m 1m D
(a) Calculation of reaction at the support
Firstly, we convert the udl into point load
P= 10x 1=10kN to act at the middle of BC
And the UVL (uniformly distributed load) , P2= 10x 0.5= 5kN to act at 1/3x1=0.33 from D
Fbd B C D
Ra Fbc= 10kN
1
1.5
2.67