AXIAL TENSIONAND COMPRESSION Example 1 The forces with magnitudes 5P, 8P 4P and P act respectively at points a B C\ Dof the rod. Their directions are shown in the figure. Try to plot the diagram of the axial force of the rod OA B D A B D B C D Solution: Determine the internal force Ni in segment OA Take the free body as shown in the figure ∑X=0N1-P4+P-P-P=0 N1-5P+8P-4P-P=0N1=2P
Example 1 The forces with magnitudes 5P、8P、4P and P act respectively at pointsA、B、C、D of the rod. Their directions are shown in the figure. Try to plot the diagram of the axial force of the rod. Solution： Determine the internal force N1 in segment OA. Take the free body as shown in the figure. A B C D PA PB PC PD O A B C D PA PB PC PD N1 X = 0 N1 − PA + PB − PC − PD = 0 N1 −5P +8P − 4P − P = 0 N1 = 2P 21
例图示杆的A、B、C、D点分别作用着大小为5P、8P、4P P的力,方向如图,试画出杆的轴力图 OA B D A B D B C D A B D 解:求O4段内力N1:设置截面如图 2X=0-M +P-P+P+P=0 N1-5P+8P-4P-P=0N1=2P 22
22 [例1] 图示杆的A、B、C、D点分别作用着大小为5P、8P、4P、 P 的力，方向如图，试画出杆的轴力图。 解： 求OA段内力N1：设置截面如图 A B C D PA PB PC PD O X = 0 1 0 − + − + + = N P P P P A B C D N1 −5P +8P − 4P − P = 0 N1 = 2P A B C D PA PB PC PD N1 x
AXIAL TENSIONAND COMPRESSION Similarly, we get the N B D internal forces in segment AB、BC、CD. They are B D C D respectively: N2=-3P 2 D N3=5P N4 D Ni= P D The diagram↑ 5P of the axial N force is shown 2P in the right l figure BP 23
Similarly ，we get the internal forces in segment AB 、BC 、CD. They are respectively ： N2= – 3 P N3= 5 P N4= P The diagram of the axial force is shown in the right figure . B C D PB P C PD N2 C D P C PD N3 DPD N4 N x 2 P 3 P 5 P+ P + – 23
N B D 同理,求得AB、 BC、CD段内力分 B D 别为: C D N2=-3P D N2=5P N4 D N 4 = P D 轴力图如右图 N 5P 2P BP 24
24 同理，求得AB 、 BC 、CD段内力分 别为： N2= – 3 P N3= 5 P N4= P 轴力图如右图 B C D PB P C PD N2 C D P C PD N3 DPD N4 N x 2 P 3 P 5 P+ P + –
AXIAL TENSIONAND COMPRESSION Characteristic of the diagram of the axial force Value of sudden change=concentrated load Simple method to plot the diagram of axial force: From the left to the right If meeting the force P to the left e, the increase of the axial force N is positive If meeting the force to the right > the increase of the axial force N is negative 8kN 5kN 3kN 5kN 8KN 3kN 25
Simple method to plot the diagram of axial force： From the left to the right: Characteristic of the diagram of the axial force：Value of sudden change = concentrated load If meeting the force P to the left ，the increase of the axial force N is positive； If meeting the force to the right → ，the increase of the axial force N is negative. 5kN 8kN 3kN + – 3kN 5kN 8kN 25