The minimum vertical load that must be applied at the opposite

8.0×361[kN/m
]

Q1. A hollow section with the cross section shown in Figure Q1 is subjected to an axial load of
20×361 [kN] not to exceed a maximum of 1 650 kN acting through the external corner “A”.
Calculate:

i)
ii)

The maximum compressive and tensile stresses in the strut;
The minimum vertical load that must be applied at the opposite external corner
marked “B” in order for there to be no resultant tensile stress at B caused by the two
loads acting together.

200 mm
A all wall thicknesses 16 mm
B
Figure Q1
Q2. Using Macaulay’s Method, calculate the required second moment of area (I) for the steel
beam shown in Figure Q2 with Young’s Modulus of 210 kN/mm2 and a central (mid point)
deflection of the beam that may not exceed 30 mm. Quote the answer in cm4
10.5×361 [kN]
Figure Q2
Hint. Keep all units in metres and kN in the calculations.

2.0 m

6.5 m

2.0 m

500 mm
2
Q3 Determine, using a table to lay out your calculations, the position of the centroid
for the cross sectional area shown in Figure Q1. Quote x bar and y bar in mm
from the bottom left hand corner of the area.

All dimensions in mm
2×361
1

10×361

19×361

2×361
x361

7×361

Figure Q1
3
Q4 The beam shown in Fig. Q2 below has a pinned support at A and roller supports at B
and C. Note that an internal pin is located at point P. Determine the reactions at A, B and
C when the beam is loaded as shown.
8×361 (kN/m)
Pin “P”
15×361 (kN)
A
B
C

45o

4 m 2 m 3 m 2 m
Figure Q2
4
Q5 (a) Calculate the reactions HA, VA and VB for the pin jointed plane frame shown in
Figure Q3(a).
(b)

Calculate the magnitude and sense of the axial force in members AC,
CD, CF, FD and FB using the method of joint resolution

AF,

3×361 (kN) 2×361 (kN)
C D E
3 m
F
A B
3 m 3 m
Figure Q3(a)
(c) Use the method of sections to obtain the forces in the members DE, DB and FB.
1×361 (kN)

The post The minimum vertical load that must be applied at the opposite appeared first on My Assignment Tutor.